thermodynamics problem solving in physical chemistry · thermodynamics problem solving in physical...

Thermodynamics Problem Solving in Physical Chemistry

Study Guide and Map

Kathleen E. Murphy

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Thermodynamics Problem Solving in Physical Chemistry – Full Solutions

1

Full Solutions:

PART 1: Gases and Gas Laws

1.1 A) Calculate the values the equations have in common first:

no.molCH4

= 25.0g1.00molCH

4

16.0g

é

ë

êê

ù

û

úú

= 1.163mol Vm,CH

4=

2.0L

1.563mol= 1.28

L

mol

Given RT = 0.08206 L-atm

mol-K(303K) = 24.86 L-atm

mol , then calculate the pressure from each equation of state:

(a) Pideal

=RT

Vm

=24.86 L-atm

mol

1.28 L

mol

= 19.4atm

(b) PVdW

,CH4

=RT

Vm

- b-

a

Vm

2=

24.86 L-atm

mol

(1.28 - 0.04278)L

mol

-2.283 L2-atm

mol2

(1.28)2 L2

mol2

= (20.10 -1.394)atm = 18.7atm

(c) Pvirial

=RT

Vm

1 +B

Vm

é

ë

êê

ù

û

úú

=24.86 L-atm

mol

1.28 L

mol

1 +

-43.9 cm3

mol( ) 1.0L

1000cm3

1.28 L

mol

é

ë

êêêê

ù

û

úúúú

= 19.40(0.9657)atm = 18.7atm

B) (a) Vm,CH

4=

0.20L

1.563mol= 0.128

L

mol P

ideal=

RT

Vm

=24.86 L-atm

mol

0.128 L

mol

= 194atm

(b) PVdW

,CH4

=RT

Vm

- b-

a

Vm

2=

24.86 L-atm

mol

(0.128 - 0.04278)L

mol

-2.283 L2-atm

mol2

(0.128)2 L2

mol2

= (291.7 -139.3)atm = 152atm

(c) Pvirial

=RT

Vm

1 +B

Vm

é

ë

êê

ù

û

úú

=24.86 L-atm

mol

0.128 L

mol

1 +

-43.9 cm3

mol( ) 1.0L

1000cm3

0.128 L

mol

é

ë

êêêê

ù

û

úúúú

= 194.0(0.657)atm = 127.5atm

1.2 A) Pvirial

=RT

Vm

1 +B

Vm

é

ë

êê

ù

û

úú Multiply both sidesby

Vm

RTsothat : P

virial´

Vm

RT=

Vm

RT´

RT

Vm

1 +B

Vm

é

ë

êê

ù

û

úúÞ Z = 1 +

B

Vm

é

ë

êê

ù

û

úú

B) Z(in 2.0L) = 1 +-0.043.9 L

mol( )1.28 L

mol

é

ë

êê

ù

û

úú

= 0.966 and Z(in 200mL) = 1 +-0.043.9 L

mol( )0.128 L

mol

é

ë

êê

ù

û

úú

= 0.657

1.3 A) (a)PV = nRT =mass

MWgas

é

ë

êê

ù

û

úúRT Þ d =

mass

V=

P(MWgas

)

RT

(b) dgas

= MWgas

P

RT

æ

èç

ö

ø÷ = MW

gas,

g

mol

1

Vm

,mol

L

æ

èç

ö

ø÷ = d

gas,g

L

B) (a) Z =V

m,obs

Vm,ideal

=MW

gas

dobs

é

ë

êê

ù

û

úú´

dideal

MWgas

é

ë

êê

ù

û

úú

=d

ideal

dobs

(b) dobs

=d

ideal

Zso that d

obs< d

idealwhen Z > 1.0

(c) dobs

=d

ideal

Zso that d

obs> d

idealwhen Z < 1.0

1.4 A) dgas

= MWgas

P

RT

æ

èç

ö

ø÷ =

16.0g

mol

130atm

0.08206 L-atm

mol-K(323K)

æ

èç

ö

ø÷ = 78.5g / L

B) dobs

=d

ideal

Z=

78.5g / L

0.8808= 89.1g / L


2

1.5. A) P

1V

1

P2V

2

=n

1T

1

n2T

2

ÞP

1

P2

=T

1

T2

Þ P2

= P1

T1

T2

æ

èç

ö

ø÷ = 100atm

500K

300K

æ

èç

ö

ø÷ = 167atm

B) Need Vm in L/mol so need to convert mass to moles, and volume to liters.

nN

2

= 92.4X103 g1.00molN

2

28.0g

é

ë

êê

ù

û

úú

= 3300mol VN

2

= 0.500m3 1000L

1.0m3

æ

èç

ö

ø÷ = 500L V

m,N

2=

500L

3300mol= 0.1515

L

mol

PVdW

,N2

=RT

Vm

- b-

a

Vm

2=

0.08206 L-atm

mol-K(500K)

(0.1515 - 0.0357)L

mol

-1.352 L2-atm

mol2

(0.1515)2 L2

mol2

= (354.3 - 58.9)atm = 295atm

1.6 A) Z(I) =PV

m

RT=

10.0atm(2.606 L

mol)

0.08206 L-atm

mol-K(340K)

= 0.934 Z(II) =PV

m

RT=

10.0atm(0.9082 L

mol)

0.08206 L-atm

mol-K(340K)

= 0.814

B) Both values less than 1.0 indicating attractive forces between NH3 molecules dominating, causing

lower than ideal molar volumes. Since NH3 molecules hydrogen bond with each other, this behavior is not

unexpected, since a strong attractive force. Increasing pressure causes an increase in attractive forces

since number of collisions increases, making it more likely the molecules will aggregate or group

C) TBoyle

=a

Rb=

4.169 L2-atm

mol2

0.08206 L-atm

mol-K(0.0371 L

mol)

= 1368K

1.7 A) TBoyle

=a

Rb=

a L2 - atm

mol2( )L - atm

mol -K( L

mol)

so R must be 0.08206 L-atm

mol-Kin equation

B) TB,CH

4

=2.283 L2

-atm

mol2

0.08206 L-atm

mol-K(0.0428 L

mol)

= 650K TB,N

2

=1.408 L2-atm

mol2

0.08206 L-atm

mol-K(0.03913 L

mol)

= 438.5K

TB,H

2

=0.2476 L2

-atm

mol2

0.08206 L-atm

mol-K(0.0266 L

mol)

= 113K TB,Ar

=1.355 L2

-atm

mol2

0.08206 L-atm

mol-K(0.0320 L

mol)

= 516K

So only TB for H2 comes close to tabled value, while in all others the calculation overestimated true value

by about 100 K.

1.8 MWgas

=mass(RT)

PV= d

gas

RT

P

æ

èç

ö

ø÷ = 1.881

g

L

0.08206 L-atm

mol-K(298K)

1.00 atm

æ

èç

ö

ø÷ = 46.0

g

mol

1.9 MWgas

= 3.71g

L

0.08206 L-atm

mol-K(773K)

699torr1.0 atm

760torr

æ

èç

ö

ø÷

æ

è

ççççç

ö

ø

÷÷÷÷÷

=3.71(63.43)

0.9197

g

mol= 256

g

mol

Then no.atoms per molecule =MW

gas

AW=

256 g

mol

32 g

mol

= 8.0 B) Molecule = S8

1.10 A) To identify the diatomic gas, need to determine the molecular weight of the gas from the data.

Given it is an ideal gas: MWgas

= 3.864g

L

0.08314 L-bar

mol-K(298K)

1.35bar

æ

èç

ö

ø÷ = 70.9

g

mol

So 2(AW) X = 70.9, and AW gas X = 35.45, so gas is Cl2(g).

B) To define mass % will need mass of Ar(g) in 3.864 g. Know that:

nmix

= nAr

+ nXe

=PV

mix

RT= mass Ar

1mol Ar

39.94g

æ

èç

ö

ø÷ + mass Xe

1mol Ar

131.3g

æ

èç

ö

ø÷

but then need second equation relating moles of Ar and Xe in mixture to solve for mass of Ar(g).


3

Also true that: nmix

= nAr

+ nHe

=PV

mix

RT=

1.35 bar(1.0L)

0.08314 L-bar

mol-K(298K)

æ

èç

ö

ø÷ = 0.0545mol so that:

nmix

= 0.0545mol = x g Ar1mol Ar

39.94g

æ

èç

ö

ø÷ + (3.864 - (x g Ar))

1mol Ar

131.3g

æ

èç

ö

ø÷ = 0.0254x + 0.02943 - 0.00752x

0.0545= (0.01742x + 0.02943) Þ x =0.02507

0.01742= 1.439g = mass Ar, then % Ar =

1.439g

3.864g´100 = 37.2%

1.11 A) Reduced variables must have NO units since:P reduced =P

Pc

;Vreduced =V

Vc

;Treduced =T

Tc

B) Both the terms in the reduced form of the Van der Waals equation must not have any units.

C) Vreduced

=V

Vc

=15.0L / mol

0.0752L / mol= 199 T

reduced=

T

Tc

=300K

151.5K= 1.985

Preduced

=8T

r

3Vr

-1-

3

Vr

2=

8(1.985)

3(199) -1-

3

(199)2= 0.0266 - 7.57X10-5 = 0.0266

D) Since Tr ≈ 2.0 and Pr ≈ 0.03, the value of Z should be close to 1.0 and the gas is acting as an ideal

gas.

E) Pideal

=RT

Vm

=0.08206 L-atm

mol-K(300K)

15.0 L

mol

= 1.64atm

1.12 A) By definition: Vc

= 3b Pc

=a

27b2T

c=

8a

27Rb so that: V

c= 3b = 3(0.0226 L

mol) = 0.0678 L

mol

Pc

=a

27b2=

0.751 L2-atm

mol2

27(0.0226)2 L2

mol2

= 54.5atm Tc

=8a

27Rb=

8 0.751 L2-atm

mol2

æè

öø

27(0.08206 L-atm

mol-K)(0.0226) L

mol

= 120K

B) Zc

=P

cV

c

RTc

=54.5 atm 0.0678 L

mol( )0.08206

L - atm

mol - K( )(120 K )= 0.375

1.13 A) Need PVdW

=RT

Vm

- b-

a

Vm

2 where b in L/mol so easiest to convert m3 → L and Pa→ kPa first.

Since 1m3 = 1000L and 1 kPa = 1000 Pa then: Vm

= 5.00X10-4 m3

mol

1000L

1.0m3

æ

èç

ö

ø÷ = 0.500

L

mol

and a = 0.500m6 - Pa

mol2´

1000L

1.0m3

æ

èç

ö

ø÷

2

´1 kPa

1000Pa= 500

L2 - kPa

mol2 so that:

3000kPa =0.08314 L-kPa

mol-K(298K)

0.50 L

mol- b

æ

èç

ö

ø÷ -

500 L2-kPa

mol2

0.0025 L2

mol2

æ

è

ççç

ö

ø

÷÷÷

Þ 5000kPa =22.70 L-kPa

mol

(0.50 L

mol- b)

Þ (0.50 L

mol- b) =

22.70 L-kPa

mol

5000kPa= 0.454 L

molÞ (0.50 L

mol- 0.454 L

mol) = b Þ b = 0.0460 L

mol

B) Z =PV

m

RT=

3000kPa(0.500 L

mol)

8.314 kPa-L

mol-K(273K)

= 0.661

1.14 A) Vm,Xe =

1.0L

131g Xe1mol

131.3

æ

èç

ö

ø÷

= 1.002L

mol P

ideal=

RT

Vm

=0.08206 L-atm

mol-K(298K)

1.002 L

mol

= 24.4atm

• So answer is NO, it is not an ideal gas since Pideal not close to 20 atm.


4

B) Van der Waals constants for Xe(g): a = 4.25L2 - atm

mol2b = 0.05105

L

mol

PVdW

,Xe(g) =RT

Vm

- b-

a

Vm

2=

0.08206 L-atm

mol-K(298K)

(1.002 - 0.05105)L

mol

-4.25 L2-atm

mol2

(1.002)2 L2

mol2

= (25.72 - 4.24)atm = 21.5atm

• Result is closer, but still not equal to 20 atm, so NOT a Van der Waals gas either.

C) (a) T =PV

m

R=

20atm 1.002 L

mol( )0.08206 L-atm

mol-K

= 243.7K = -29.3°C

(b) T = P +a

Vm

2

é

ë

êê

ù

û

úú´

Vm

- b

R

é

ë

êê

ù

û

úú

= 24.25atm0.9490 L

mol

0.08206 L-atm

mol-K

æ

èç

ö

ø÷ = 280.4K = 7.3°C

1.15 A) Given:P

1V

1

P2V

2

=n

1T

1

n2T

2

Þ P1

=P

2V

2

V1

= 3780torr1atm

760torr

æ

èç

ö

ø÷

4.65L

6.85L

æ

èç

ö

ø÷ = 3.37atm

B) Cannot determine P1 as a Van der Waals gas since know only P2, V1 and V2 values.

• Can’t divide equations to cancel like terms, but could possibly subtract equations to collect like terms:

P1

=RT

V1 - b-

a

V1

2 and P

2=

RT

V 2 - b-

a

V 2

2 If we assume b<<V1 or V2, then can say by subtracting P’s,

Leads to: P1

- P2

=nRT

V1

-nRT

V2

é

ë

êê

ù

û

úú

+n2a

V2

2-

n2a

V1

2

é

ë

êê

ù

û

úúÞ P

1= P

2+ nRT

V2

- V1

V1V

2

é

ë

êê

ù

û

úú

+ an2V

1

2 - V2

2

V1

2V2

2

é

ë

êê

ù

û

úú

• Cannot cancel n and T, so must know T and number of moles to solve, and

• Must also know chemical identity of gas since need value of a, the Van der Waals constant

• Since these values not given, cannot solve for P1 if a Van der Waals gas.

1.16 A) First add 2nd term to both sides: PVdW

=RT

Vm

- b-

a

Vm

2Þ P

VdW+

a

Vm

2=

RT

Vm

- b

Subtract righthand term to produce zero on the righthand side: PVdW

+a

Vm

2-

RT

Vm

- b= 0

Multiply all terms by (Vm-b):P(Vm

- b) +a(V

m- b)

Vm

2-

RT(Vm

- b)

Vm

- b= 0 Þ PV

m- Pb +

a

Vm

-ab

Vm

2- RT = 0

And then multiply equation by Vm2 and divide by P:

PVm

Vm

2

P

é

ë

êê

ù

û

úú

- PbV

m

2

P

é

ë

êê

ù

û

úú

+a

Vm

Vm

2

P

é

ë

êê

ù

û

úú

-ab

Vm

2

Vm

2

P

é

ë

êê

ù

û

úú

- RTV

m

2

P

é

ë

êê

ù

û

úú

= 0 ÞVm

3 - b +RT

P

æ

èç

ö

ø÷Vm

2 +a

PV

m-

ab

P= 0

• Combining like terms leads to a polynomial in Vm as the final result (in blue).

B) (a) Vm,O

2

=RT

P=

(0.08314 L-bar

mol-K)(298K)

200bar= 0.124

L

mol

(b) Given the form: Ax3 + Bx2 + Cx + D = 0 then A = 1.0, B = - b +RT

P

æ

èç

ö

ø÷ , C =

a

P, D = -

ab

P

For roots of equation need a, b values for O2(g) a = 1.378 L2-atm-mol-2, b = 0.03183 L-mol-1

Then: A = 1.0L

mol, B = - 0.03183 L

mol+

0.08314 L-bar

mol-K(298K)

200bar

æ

èç

ö

ø÷ = -(0.03183 + 0.124) = -0.1557

L

mol

C =a

P=

1.378 L2-atm

mol2

200bar ´1.0atm

1.013bar

é

ëê

ù

ûú

= 6.96X10-3 L2

mol2D = -

ab

P=

1.378 L2-atm

mol20.03183 L

mol( )

200bar ´1.0atm

1.013bar

é

ëê

ù

ûú

= 2.215X10-4 L3

mol3The

only possible root that solves the equation is: Vm,VdW

= 0.110L

mol


5

C) Since value for Van der Waals gas is very different than the ideal molar volume, expect that O2 is

acting as real gas in the tank, not an ideal gas.

1.17 A) First assume b << V then substitute for n as mass/MW in modified Van der Waals equation

PVdW

=mass

MWgas

é

ë

êê

ù

û

úú

RT

V-

mass

MWgas

é

ë

êê

ù

û

úú

2

a

V2Þ

mass

V

é

ëê

ù

ûú

RT

MWgas

-mass

V

é

ëê

ù

ûú

2

a

MWgas

2Þ P

VdW=

RT

MWgas

é

ë

êê

ù

û

úúd

gas-

a

MWgas

2

é

ë

êê

ù

û

úúd

gas

2

The equation has the form of a quadratic equation: Ax2 + Bx + C = 0 where x =-B ± B2 - 4AC

2A

•

BUT the terms A, B and C would have units of pressure, so we must divide all terms in equation by

P BEFORE rearranging into quadratic so x = density in units of g/L and no units exist on either side of

the equation, which produces:

1.0 =RT

P(MWgas

)

é

ë

êê

ù

û

úúd

gas-

a

P(MWgas

2 )

é

ë

êê

ù

û

úúd

gas

2 Þa

P(MWgas

2 )

é

ë

êê

ù

û

úúd

gas

2 -RT

PMWgas

é

ë

êê

ù

û

úúd

gas+1.0 = 0

Then: Ax2 + Bx + C = 0 and d =-B ± B2 - 4AC

2Awhere : A

VdW=

a

P(MWgas

)2B

VdW= -

RT

P(MWgas

)C

VdW= 1.0

B) For the Virial equation use same approach as for Van der Waals equation to develop a unit-less

quadratic, with d in g/L. The coefficients differ in that the second virial coefficient, B, replaces “a” in C, in

the virial equation polynomial, the sign of the second term changes, and the 1.0 is subtracted.

Pvirial

=nRT

V1 +

nB

V

é

ëê

ù

ûú =

nRT

V+

n2BRT

V2=

mass

V

é

ëê

ù

ûú

RT

MWgas

+(mass)2

V 2

é

ëêê

ù

ûúú

BRT

MWgas( )

2Þ P

virial= d

gas

RT

MWgas

é

ë

êê

ù

û

úú

+ dgas

2 BRT

MWgas( )

2

é

ë

êêê

ù

û

úúú

Þ1.0 = dgas

RT

P(MWgas

)

é

ë

êê

ù

û

úú

+ dgas

2 BRT

P(MWgas

)2

é

ë

êê

ù

û

úúÞ d

gas

2 BRT

P(MWgas

)2

é

ë

êê

ù

û

úú

- dgas

RT

P(MWgas

)

é

ë

êê

ù

û

úú

-1.0 = 0

The coefficients of the quadratic equation are: Avirial

=BRT

P(MWgas

)2

é

ë

êê

ù

û

úú

Bvirial

=RT

P(MWgas

)

é

ë

êê

ù

û

úú

Cvirial

= -1.0

C) Ideal gas density: dCl

2(g)

(ideal) = MWgas

P

RT

æ

èç

ö

ø÷ =

71.0g

mol

1.0atm

0.08206 L-atm

mol-K(323K)

æ

èç

ö

ø÷ = 2.88g / L

For the Van der Waals gas density:

AVdW

=a

P(MWgas

)2=

6.579L2-atm

mol2

1.0atm(71.0 g

mol)2

= 1.305X10-3 L2

g2B

VdW= -

RT

P(MWgas

)= -

1

dideal

= -0.347L

gC

VdW= 1.0

dVdW

=-B ± B2 - 4AC

2A=

-(-0.347) ± (0.347)2 - 4(1.0)(1.305X10-3)

2(1.305X10-3)=

0.347 ± 0.3394

2.61X10-3= 263

g

Lor 2.91

g

L

For the Virial gas density:

AVirial

=BRT

P(MWgas

)2=

-0.314L

mol(0.08206

L-atm

mol-K(300K))

1.0atm(71.0 g

mol)2

= -1.53X10-3 L2

g2

BVdW

=RT

P(MWgas

)= -

1

dideal

= 0.347L

gC

VdW= -1.0

dVirial

=-B ± B2 - 4AC

2A=

-(0.347) ± (0.347)2 - 4(-1.0)(-1.53X10-3)

2(-1.53X10-3)=

-0.347 ± 0.338

-3.06X10-3= 2.94

g

Lor 224

g

L

• The density values are very close which proves the derivations are valid equations.

1.18 A) As ideal gases:

Only 2nd root

makes sense

Only 1st root

makes sense.


6

no.molCl2

= 250g1.00mol

71.0g

é

ëê

ù

ûú = 3.52molCl

2 P

Cl2

=nRT

V=

3.52mol(0.08206 L-atm

mol-K)(300K)

10.0L= 8.75atm

no.molC2H

6= 150g

1.00mol

30.0g

é

ëê

ù

ûú = 5.00molC

2H

6 P

C2H

6

=nRT

V=

5.00mol(0.08206 L-atm

mol-K)(300K)

10.0L= 12.43atm

Ptotal

= PCl

2

+ PC

2H

6

= 8.75 +12.43 = 21.2atm

B) As Van der Waals gases (real gases) need a, b and Vm:

Vm,Cl

2=

10.0L

3.52mol= 2.84

L

mol a

Cl2

= 6.579L2 - atm

mol2b

Cl2

= 0.05632L

mol

PVdW

,Cl2(g) =

RT

Vm

- b-

a

Vm

2=

0.08206 L-atm

mol-K(303K)

(2.84 - 0.05622)L

mol

-6.579 L2-atm

mol2

(2.84)2 L2

mol2

= (8.93 - 0.816)atm = 8.11atm

Vm,C

2H

6=

10.0L

5.00mol= 2.00

L

mol a

C2H

6

= 5.562L2 - atm

mol2b

C2H

6

= 0.0638L

mol

PVdW

,C2H

6(g) =

RT

Vm

- b-

a

Vm

2=

0.08206 L-atm

mol-K(303K)

(2.00 - 0.0638)L

mol

-5.562 L2-atm

mol2

(2.00)2 L2

mol2

= (12.84 -1.39)atm = 11.45atm

Ptotal,VdW

= PCl

2

+ PC

2H

6

= 8.11+11.45 = 19.6atm

1.19 Since the volume is constant, the partial pressure ratio will equal the mole ratio in the mixture.

n

CO

ntotal

=P

CO

Ptotal

andn

O2

ntotal

=P

O2

Ptotal

so thatn

O2

nCO

=P

O2

PCO

• Also need to convert the ppm units to g/L for partial pressure of CO, assuming ideal gas:

50 ppm =50gCO

1.0X106 gsoln=

50gCO

1000L= 0.050

gCO

L 800 ppm = 0.800

gCO

L 3200 ppm = 3.200

gCO

L

PCO

@50ppm = dgas

RT

MWgas

é

ë

êê

ù

û

úú

= 0.050g

L

0.08206 L-atm

mol-K(298K)

28.0g

mol

é

ë

êê

ù

û

úú

= 0.0437atm n

O2

nCO

=0.200atm

0.0437atm= 4.58@50ppm

PCO

@800ppm = 0.800 g

L

0.08206 L-atm

mol-K(298K)

28.0 g

mol

é

ë

êê

ù

û

úú

= 0.699atm n

O2

nCO

=P

O2

PCO

=0.200atm

0.699atm= 0.286@800ppm

PCO

@3200ppm = 3.200 g

L

0.08206 L-atm

mol-K(298K)

28.0 g

mol

é

ë

êê

ù

û

úú

= 2.81atm n

O2

nCO

=P

O2

PCO

=0.200atm

2.81atm= 0.0712@3200ppm

• So to be safe we need over 4.6 molecules of O2 for every 1 molecule of CO in the air we breathe.

1.20 nH

2O

=0.062 atm(500L)

0.08206 L-atm

mol-K(310K)

æ

èç

ö

ø÷ = 1.22mol

18.0g

1 mol

æ

èç

ö

ø÷ = 21.9gH

2O

1.21 Reaction: 2H2(g) + O2(g) → 2 H2O(l)

Know: Ptotal

- Pexcess

= Preacted

= (1.0-0.40)atm = 0.60 atm = PH

2reacted

+ PO

2reacted

and P

O2

PH

2

=1

2Þ P

O2

=1

2P

H2

then:

0.60 atm = PH

2reacted

+ PO

2reacted

= PH

2reacted

+ 0.5PH

2reacted

= 1.5PH

2reacted

Þ PH

2reacted

=0.60

1.5= 0.40atm

PH

2initial

= PH

2excess

+ PH

2reacted

= 0.40- 0.40 = 0.80atm so that PO

2initial

= 1.0 - 0.80 = 0.20atm

• The molar ratio will

equal the molecule ratio


7

Mol % O2

= cO

2

´100 =P

O2

Ptotal

´100 = 20.0% Mol % H2

= 80.0%

1.22 A) • The number of moles will not change when gases mix so the mole fraction of CO2 constant.

(a) no.molCO2

= 500g1.00mol

44.0g

é

ëê

ù

ûú = 11.36molCO

2 no.mol Ar = 35.5g

1.00mol

39.95g

é

ëê

ù

ûú = 0.889mol Ar

cCO

2

=n

CO2

ntotal

=11.36mol

(11.36 + 0.889)mol= 0.927, c

Ar= 1.0 - 0.927 = 0.073

(b) PCO

2

=nRT

V=

11.36mol(0.08206 L-atm

mol-K)(293.5K)

5.50L= 49.75atm

PAr

=nRT

V=

0.889mol(0.08206 L-atm

mol-K)(293.5K)

5.50L= 3.89atm P

total= (49.75 + 3.89)atm = 53.64atm

B) For CO2:Vm,CO

2=

5.50L

11.36mol= 0.484

L

mol a = 3.640 L2-atm/mol2 and b = 0.04267 L/mol

PVdW

,CO2(g) =

RT

Vm

- b-

a

Vm

2=

0.08206 L-atm

mol-K(293.5K)

(0.484 - 0.05622)L

mol

-3.640 L2-atm

mol2

(0.484)2 L2

mol2

= (56.30 -15.54)atm = 40.8atm

For Ar: Vm,Ar =

5.50L

0.889mol= 6.19

L

mol a = 1.355 L2-atm/mol2 and b = 0.0320 L/mol

PVdW

,Ar(g) =0.08206 L-atm

mol-K(293.5K)

(6.19 - 0.0320)L

mol

-1.355 L2-atm

mol2

(6.19)2 L2

mol2

= (3.91 - 0.0354)atm = 3.87atm

Ptotal,VdW

= (40.8 + 3.87)atm = 44.67atm

• So Ptotal changes significantly since P CO2 is much smaller as a Van der Waals gas.

C) The partial pressure of Argon is unchanged, when treated as a van der Waals gas, since there are far

fewer moles of gas in the large volume (0.161 mol/L), compared to CO2(g) which has 2.16 mol/L.

Consequently, the corrections for “a” and “b” are much smaller so Ar acts as an ideal gas.

1.23 • Can determine ntotal from ideal gas law. Then apply %’s to get moles of each gas.

A) nTOT

=P

TOTV

mix

RTmix

=1.0atm(40.0L)

(0.08206 L-atm

mol-K)(298K)

= 1.64 mol B) dmix

=(62.1 + 4.74)g

40.0L= 1.67

g

L

nC

4H

10

= 1.64mol5.0molC

4H

10

100mol total

æ

èç

ö

ø÷

58.0gC4H

10

1.0molC4H

10

æ

èç

ö

ø÷ = 4.74gC

4H

10

nAr

= 1.64mol95.0mol Ar

100mol total

æ

èç

ö

ø÷

39.95g Ar

1.0mol Ar

æ

èç

ö

ø÷ = 62.1g Ar

1.24 A) Applying Dalton’s Law, given Tmix and Vmix same for all gases, then

PNe

Vmix

Ptotal

Vmix

=n

NeT

mix

ntotal

Tmix

Þ Ptotal

= PNe

ntotal

nNe

æ

èç

ö

ø÷ • Only need to know nTOT and nNe to determine PTOT.

nNe

= 0.225g1.00mol

20.18g

é

ëê

ù

ûú = 0.01115molNe n

CH4

= 0.320g1.00mol

16.0g

é

ëê

ù

ûú = 0.0200molCH

4

nAr

= 0.175g1.00mol

39.95g

é

ëê

ù

ûú = 0.00438mol Ar n

total= n

Ne+ n

CH4

+ nAr

= 0.0355mol


8

Ptotal

= PNe

ntotal

nNe

æ

èç

ö

ø÷ = 66.5torr

0.0355mol

0.01115mol

æ

èç

ö

ø÷ = 218torr

B) Vmix

=n

TOTRT

mix

PTOT

=0.0355mol (0.08206 L-atm

mol-K)(300K)

212torr1.0atm

760torr

æ

èç

ö

ø÷

= 3.13 L

1.25 A) nSO

2

= 20X106 tons2000 lb

1.0ton

æ

èç

ö

ø÷

454g

1.00 lb

æ

èç

ö

ø÷

1.0mol SO2

64.0gSO2

æ

èç

ö

ø÷ = 2.84X1011 mol SO

2

PSO

2

=nRT

V=

2.84X1011 mol(0.08206 L-atm

mol-K)(300K)

8.0X1018m3 1000L

1m3

æ

èç

ö

ø÷

= 8.74X10-10 atm

B) Mol % SO2

=P

SO2

Ptotal

æ

è

çç

ö

ø

÷÷´100 =

8.74X10-10 atm

1.00atm

æ

èç

ö

ø÷ ´100 = 8.74X10-8%

1.26 A) The % by mass will allow the calculation of the mol ratio of each gas to the total moles.

Assuming a mixture of 100 g (mol ratio will be the same, even though mass in tank not the same).

nHCl

= 5.00g1.00mol

36.5g

é

ëê

ù

ûú = 0.136mol HCl n

Ne= 94.0g

1.00mol

20.18g

é

ëê

ù

ûú = 4.66molNe

nH

2

= 1.00g1.00mol

2.02g

é

ëê

ù

ûú = 0.500mol H

2 n

total= n

Ne+ n

H2

+ nHCl

= 5.30mol

cHCl

=n

HCl

ntotal

=0.136

5.30= 0.0257 c

Ne=

nNe

ntotal

=4.66

5.30= 0.879 c

H2

=n

H2

ntotal

=0.500

5.30= 0.0945

B) • Since mol ratio equals pressure ratio, need only Ptotal to define partial pressures.

PHCl

= Ptotal

nHCl

ntotal

æ

èç

ö

ø÷ = 138kPa(0.0257) = 3.55kPa P

Ne= P

total

nNe

ntotal

æ

èç

ö

ø÷ = 138kPa(0.879) = 121kPa

PH

2

= Ptotal

nH

2

ntotal

æ

è

çç

ö

ø

÷÷

= 138kPa(0.0945) = 13.0kPa

C) • Calculate actual ntotal from ideal gas law. nTOT

=P

TOTV

mix

RTmix

=138kPa(49.0L)

(8.314 L-KPa

mol-K)(298K)

= 2.728 mol

Then apply mol ratio to ntotal and use MW to get mass: massof gas = ntotal

´ (cgas

) ´ (MWgas

)

massHCl = 2.728mol ´ (0.0257) ´36.5g

1.0mol

æ

èç

ö

ø÷ = 2.56gHCl

massNe = 2.728mol ´ (0.880) ´20.18g

1.0mol

æ

èç

ö

ø÷ = 48.50gNe

massH2

= 2.728mol ´ (0.0945) ´2.02g

1.0mol

æ

èç

ö

ø÷ = 0.52gH

2

Total actual mass = (2.56 + 48.50 + 0.52)g = 51.57g

1.27 A) Since n, T constant for each gas can start with: P1V1 = P2V2

P2,CO

2= P

1

V1

V2

æ

èç

ö

ø÷ = 2.13atm

1.50L

4.50L

æ

èç

ö

ø÷ = 0.710atm P

2,H

2= 0.956atm

1.00L

4.50L

æ

èç

ö

ø÷ = 0.212atm


9

P2,Ar = 1.15atm

2.00L

4.50L

æ

èç

ö

ø÷ = 0.511atm P

total= P

CO2

+ PH

2

+ PAr

= 0.710 + 0.212 + 0.511 = 1.43atm

B) cCO

2

=P

CO2

Ptotal

=0.71

1.43= 0.495 c

H2

=P

H2

Ptotal

=0.212

1.43= 0.148 c

Ar=

PAr

Ptotal

=0.511

1.43= 0.357

1.28 A) • Can calculate actual the density of each gas by applying mass % to density, then ideal gas

law to the partial pressure.

dN

2

=0.7808gN

2

1.00g mix

1.146gmix

1.0L

é

ëê

ù

ûú =

0.895gN2

1.00L P

N2

= dgas

RT

MWgas

é

ë

êê

ù

û

úú

=0.895g

1.00L

0.08206 L-atm

mol-K(300K)

28.0g

mol

é

ë

êê

ù

û

úú

= 0.787atm

dO

2

=0.2095gO

2

1.00g mix

1.146g mix

1.0L

é

ëê

ù

ûú =

0.240gO2

1.00L P

O2

= dgas

RT

MWgas

é

ë

êê

ù

û

úú

=0.240g

1.00L

0.08206 L-atm

mol-K(300K)

32.0g

mol

é

ë

êê

ù

û

úú

= 0.185atm

• Since gases are in the same mixture, the partial pressure ratio equals the molar ratio.

Ptotal

= 740torr1.00atm

760torr

é

ëê

ù

ûú = 0.974atm c

N2

=P

N2

Ptotal

=0.787

0.974= 0.808 c

O2

=P

O2

Ptotal

=0.185

0.974= 0.190

1.29 nNH

4NO

3

= 40.0 lb454g

1.00 lb

æ

èç

ö

ø÷

1.0molNH4NO

3

80.0g

æ

èç

ö

ø÷ = 227molNH

4NO

3

2.0mol N2

2.0molNH4NO

3

æ

èç

ö

ø÷ = 227mol N

2

VN

2

=nRT

P=

227mol (0.08206 L-atm

mol-K)(723K)

736torr1.0atm

760torr

æ

èç

ö

ø÷

= 1.39X104 L

V

H2O

VN

2

=2

1Þ V

H2O

= 2VN

2

= 2(1.39X104L) = 2.78X104L V

O2

VN

2

=1

2Þ V

O2

= 0.5VN

2

= 0.5(1.39X104L) = 6.95X103L

Vtotal

= VN

2

+VH

2O

+VO

2

= (1.39X104) + (2.78X104) + (0.695X104) = 4.87X104 L

1.30 A) rateforVCO

2

=4.50L

1.00min

æ

èç

ö

ø÷

60min

1.00hr

æ

èç

ö

ø÷

24hr

1.00da

æ

èç

ö

ø÷ =

220LCO2

1.00da

B) rateforVO

2

=220LCO

2

1.00da

æ

èç

ö

ø÷

1.0LO2

2.0LCO2

æ

èç

ö

ø÷ =

110LO2

1.00da C) n

CO2

=PV

RT=

0.9684atm(220.0L / da)

(0.08206 L-atm

mol-K)(298K)

= 8.72 mol / da

Then for 1 year: Annual VCO

2

= 1.0yr365.25da

1.0yr

æ

èç

ö

ø÷

8.72molCO2

1.00da

æ

èç

ö

ø÷ = 3186molCO

2

mass Na2O

2needed = 3186molCO

2

2.0molNa2O

2

2.0molCO2

æ

èç

ö

ø÷

78.0g

1.0mol Na2O

2

æ

èç

ö

ø÷

1.0kg

1000g

æ

èç

ö

ø÷ = 249kgNa

2O

2

• Since gases are at the same P, T, the

Law of Combining Volumes applies.

• Use stoichiometry

to determine mass

Na2O2 needed.


10

PART 2: FIRST LAW OF THERMODYNAMICS - Work (PV), heat, ∆U and ∆H

2.1 (1) Work done in STEPS where volume changes:

STEP 2:w = -PdV = -20atm ´ (12 - 4)L ´101.3 J

L-atm= -1.62X104 J = -16.2kJ

STEP 4: w = -PdV = -12atm ´ (24 -12)L ´101.3 J

L-atm= -1.46X104 J = -14.6kJ

Total work lost by system= -30.8 kJ

(2) • Need to consider T changes to determine to calculate q:

STEP 1: T1

= T0

P2

P1

æ

èç

ö

ø÷ = 250K

20

28

æ

èç

ö

ø÷ = 203.1K q

(STEP1)= C

m,vDT = (1mol)12.48 J

mol-K(203.1- 250)K = -585 J

STEP 2: T2

= T1

V3

V2

æ

èç

ö

ø÷ = 203.1K

12

4.0

æ

èç

ö

ø÷ = 609.3K q

(STEP2)= C

m,PDT = (1mol)20.79 J

mol-K(609.3 - 203.1)K = 8445 J

STEP 3: T3

= T2

P2

P1

æ

èç

ö

ø÷ = 609.3K

12

20

æ

èç

ö

ø÷ = 356.6K

q(STEP3)

= Cm,v

DT = (1mol)12.48 J

mol-K(365.6 - 609.3)K = -3041J

STEP 4: T4

= T3

V5

V4

æ

èç

ö

ø÷ = 365.6K

24

12

æ

èç

ö

ø÷ = 731.2K

q(STEP4)

= Cm,P

DT = (1mol)20.79 J

mol-K(731.2 - 365.6)K = +7601J

STEP 5: Tfinal

= T4

P5

P4

æ

èç

ö

ø÷ = 731.2K

4

12

æ

èç

ö

ø÷ = 243.7K

q(STEP5)

= Cm,v

DT = (1mol)12.48 J

mol-K(243.7 - 731.2)K = -6082 J

Total work = -30.8 kJ Total q = +6339.4 J = 6.34 kJ ∆U = -24.5 kJ

2.2 STEP 1: The gas is expanded from isothermally and reversibly until the volume doubles. So can say

T and n constant: Assumeideal :P

1V

1

n1T

1

=P

2V

2

n2T

2

® P2

=P

1V

1

V2

= 1000kPa ´1

2= 500kPa Since isothermal, dw=

-dq

wrev

= -nRT lnV

2

V1

= -1.0mol ´ 8.314J

mol - K(293K)ln 2é

ëùû

= -1689J = -1.69kJ then q = +1.69 kJ

∆U = q + w = 0 → If isothermal heat flows in or out to compensate for work done, keeping T constant.

STEP 2: Then the temperature of the gas is raised to 80°C at a constant volume.

Since n, V constant: P

3

T3

=P

2

T2

® P3

=P

2T

3

T2

= 500kPa ´353K

293K= 602kPa Since ∆V= 0, then w = 0

Assuming Cv constant T1 → T2: q = nCv∆T = 1mol ´ (12.472 J / mol - K) ´ (60K) = 748 J ∆U = q = 748 J

2.3 A) Isothermal, irreversible expansion (since against constant P) from 20L → 30 L, know work done

= -5065.8 J

wirrev

= -PV1

V2

òdV = -P(V2

- V1) ´101.3

J

L - atm= w and

P =- -5065.8J( )

(V2

- V1)

´1L - atm

101.3J=

5065.8J

10.0L´

1L - atm

101.3J= 5.00atm.

B) Since isothermal, qirrev = -wirrev to keep T the same, and ∆H = ∆U +∆(PV) = ∆U + nR(∆T)

so ∆U = 0, q = +5065.8 J = 5.06 kJ, and ∆H = 0 since ∆T =0


11

C) Since T1 = T2 could choose either set of conditions and use T =PV

nR but NOTE that the initial P is not

5.00 atm, only P2 is, so the final set is most convenient to use. T2

=P

2V

2

nR=

5.00atm(30.0L)

(2.0mol)0.08206 L-atm

mol-K

= 914K

2.4 A) ∆U = q

V= nC

V ,m∆T C

P,m O2

- R = CV ,m O

2

= 29.96J

mol-K- 8.314

J

mol-K= 21.65

J

mol-K

qV

= 2.0mol 21.65 J

mol-K( ) (373 - 273)K( ) = 4329 J = ∆U

B) ∆H = qP

= nCP,m∆T q

P= 2.0mol 29.96

J

mol - K

æ

èç

ö

ø÷ (373 - 273)K( ) = 5992 J = ∆H

C) The difference of 5992-4329 J = 1663 J and can be explained by the definition of ∆H from H = U + PV

∆H = ∆U + ∆(PV) = ∆U + ∆(nRT) = ∆U + nR(∆T) and so∆H - ∆U = nR(∆T)

nR(∆T) = 2.0mol(8.314 J

mol-K)(373 - 273)K = 1663J

2.5 Since compression is done with a constant P, the work is irreversible work:

wirrev

= -

V1

V2

òPdV = -PV1

V2

òdV = -P(V2

- V1) ´101.3 J

L-atm= -5.00atm(20 - 60)L ´101.3 J

L-atm= +2.062X104 J = +20.6kJ

Because the T must be kept the same, heat must flow out of system to compensate for work done on

system, q=-20.6 kJ so that ∆U = 0 = q + w and ∆H = 0 since ∆T =0.

2.6 A) The maximum amount of work lost would be through an isothermal reversible expansion, since

that is when the area under the curve would be a maximum.

wrev

= -

V1

V2

òPdV = -

V1

V2

ònRT

V

æ

èç

ö

ø÷ dV = - nRT

V1

V2

òdV

V= -nRT ln

V2

V1

æ

èç

ö

ø÷

Since ideal,n and T constant :P

1

P2

=V

2

V1

wrev

= -nRT lnP

1

P2

æ

èç

ö

ø÷

wrev

= -nRT lnP1

P2

æ

èç

ö

ø÷ = -5.25mol 8.314

J

mol - K

æ

èç

ö

ø÷ (450K)ln

15.0bar

3.50bar

æ

èç

ö

ø÷

= -1.964X104 J(1.455) = -2.858X104 J = -28.6kJ

B) The minimum amount of work lost would be through an isothermal

irreversible expansion against a constant P, since that when the area

under the curve from P2×∆V would be a minimum (see figure below).

wirrev

= -

V1

V2

òPdV = -PV1

V2

òdV = -P(V2

- V1) ´100 J

L-bar

V2

=nRT

P2

=5.25mol(0.08314 L-bar

mol-K)(450K)

3.50bar= 56.1L V

1=

nRT

P1

=

5.25mol(0.08314L - bar

mol - K)(450K)

15.0bar= 13.1L

wirrev

= -P2(V

2- V

1) ´100

J

L - bar= -3.5bar ´ (56.1-13.1)L ´100

J

L - bar= -1.505X104 J = -15.1kJ

2.7 A) Vplasma = 0.550L = V1,gas Know V2 = 0.945V1 so ∆V = -0.055V1= -0.03025L

w = -P∆V = -95.2bar ´ (-0.03025)L ´100 J

L-bar= +288 J

B) Given V1 N2(g) = 0.550L then


12

∆V = -w

P=

-288J

10atm ´101.3 J

L-atm

= -0.284L V2

= V1

+ ∆V = 0.550 + (-0.284)L = 0.394L

2.8 When ideal gas considered: wrev

= -

V1

V2

òPdV = -

V1

V2

ònRT

V

æ

èç

ö

ø÷ dV = - nRT

V1

V2

òdV

V= -nRT ln

V2

V1

æ

èç

ö

ø÷

w

rev, ideal = -nRT ln

V2

V1

= 2.0mol 8.314J

mol - K

æ

èç

ö

ø÷ (273K)ln

30L

10L

æ

èç

ö

ø÷ = -4987 J

Work with virial gas, given produces an equation with 2 terms: Pvirial

=nRT

V1 +

nB

V

é

ëê

ù

ûú

wrev

= -

V1

V2

òPdV = - nRTV

1

V2

ò1

V+

nB

V2

é

ëê

ù

ûúdV = -nRT

V1

V2

òdV

V+ nB

V1

V2

òdV

V2

é

ë

êê

ù

û

úú

= -nRT lnV

2

V1

æ

èç

ö

ø÷ - n2BRT

1

V1

-1

V2

é

ë

êê

ù

û

úú

• So the difference when a real (virial) gas is considered is the second term,

∆wrev

= -nRT nB1

V1

-1

V2

é

ë

êê

ù

û

úú

é

ë

êê

ù

û

úú

= - 2.0mol(8.314 J

mol-K)(273K)é

ëùû´ 2.0mol 13.7

cm3

mol

é

ëêê

ù

ûúú

1.0L

1000cm3

é

ëê

ù

ûú 0.0667L-1éë

ùû

é

ë

êê

ù

û

úú

= -8.30J

wrev

,real(virial) = -4987 - 8.30J = -4995J = -5.00kJ

2.9 A) P1

=n

1RT

V1

=2.25mol(0.08314 L-bar

K-mol)(306.75K)

26L= 2.22atm

B)

wirrev

= - ò PdV = -PV

1

V2

ò dV = -P(V2

- V1) = -0.825bar(70 - 26)L ´

100J

1L - bar= -3630J = -3.63kJ

C) Since gas is ideal and the expansion isothermal:

wrev

= - ò PdV = -V

1

V2

ònRT

VdV = -nRT

V1

V2

òdV

V= -nRT(lnV

2- lnV

1) = -nRT ln(

V2

V1

)

wrev

= - 2.25mol ´ (8.314 J

mol-K) ´308.75Ké

ëùûln

70

26

é

ëê

ù

ûú = -5776J (ln(2.69) = -5718J = -5.72kJ

D) Since changing conditions for an ideal gas:

P1V

1

P2V

2

=n

1T

1

n2T

2

= 1.0 since n1=n

2, T

1 = T

2so P

1V

1= P

2V

2and

P1

P2

=V

2

V1

Then need to calculate P2:

P2

=n

1RT

V2

=2.25mol(0.08314 L-bar

K-mol)(306.75K)

70L= 0.825atm

wrev

== -nRT ln(V

2

V1

) = -nRT ln(P1

P2

) = -5776J(ln2.22atm

0.825atm

æ

èç

ö

ø÷) = -5776J(ln 2.69( )) = -5718J = -5.72kJ

2.10 A) CP

=dH

dT

é

ëê

ù

ûúP

=q

p

n∆T=

229J

(3.0mol)2.55K= 29.93

J

mol - K

B) CV

= CP

- R = 29.93 - 8.314( ) J

mol - K= 21.62

J

mol - K

2.11 A) (a) Reaction: C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(g) So ∆n=(5 - 4.5) = + 0.5 mol gas.

So the magnitude of reversible work is not significantly

different from the ideal gas value unless B is very large.


13

When the ∆n is positive, expansion must occur and will be against a constant P since the weight on the

piston exerts a constant force in Flask A. If you had chosen H2O(l) as product state, the opposite,

contraction occurs, since ∆n is negative. In either case, the piston moves.

(b) If ∆n is a positive value, expansion occurs, piston moves up. (If ∆n is a negative value contraction

occurs, piston moves down.)

(c) Know that Ptotal remains constant, so true that:

Pbefore

= ntotal reactants

RT1

V1

æ

èç

ö

ø÷ = P

after= n

total products

RT2

V2

æ

èç

ö

ø÷

So since ntotal, prod ≠ ntotal, reactants T could stay constant (as well as P) ONLY as long as a compensating

volume change occurs.

B) Answers:

(a) In Flask B, the volume is fixed, so P must change when the combustion occurs, since ∆n not equal to

zero, if T constant. If choosing H2O(g) as product, ∆n (+) so P increases. If choosing H2O(l) as product,

∆n (-) so P decreases.

(b) In Flask A, heat flow is under constant P, so qp=∆H, but in Flask have constant V conditions so qV =

∆U. Since ∆H = ∆U + ∆(PV) = ∆U + ∆n(RT), the two terms cannot be equal if ∆n is not equal to zero. So

the heat flow cannot be the same.

C) ∆U is a state function and independent of pathway, so ∆U the same in both Flask A and B changes.

The work done in Flask A volume change will compensate for qP being larger than qV, so that ∆UA=qP +w

= ∆UA=qV.

2.12 Important to note adiabatic expansion, then: wadiab

= nCv∆T

A) wadiab

= nCv∆T = 1.0mol 1.5(8.314

J

mol - K)

æ

èç

ö

ø÷ (322 - 475K) = -1908 J

B) Isothermal reversible defined as wrev

= -nRT lnV

2

V1

but will need to rewrite in terms of P1 and P2.

Since ideal gas, and n, T constant then: P1V

1= P

2V

2 and true that: w

rev= -nRT ln

P1

P2

Given the value of work can then find P2:

lnP

2

P1

=w

rev

nRT=

-1908J

1.0mol(8.314J

mol - K)(300K)

= -0.7127

P2

= P1e-0.7127 = 2.04(2.25bar) = 4.59bar

2.13 A) When irreversible and adiabatic: dU = -dw leads to:nCV(T

2- T

1) = -P(V

2-V

1) [See map]

(T2

- T1) = -P(V

2- V

1) Þ ∆T =

-P(V2

- V1)

nCV ,N

2

= -1.00atm(2.0 - 6.0)L

2.0mol(20.811J

mol - L)

´101.3J

1L - atm= 9.74K

∆T = 9.74K = 9.74°C = T2

- 30.0°C Þ T2

= 39.7°C

B) When reversible and adiabatic: dU = -dw leads to :CVln

T2

T1

= -RlnV

2

V1

[See map]

lnT

2

T1

= -R

Cv

lnV

2

V1

= -8.314 J

mol-K

20.81 J

mol-K

ln2

6

æ

èç

ö

ø÷ = -0.3995(-1.0996) = +0.439

T2

T1

= e+0.439 = 1.55 T2

= 1.55T1

= 1.55(303K) = 470K Þ T2

= 197°C

C) Although adiabatic means that the work for both would be defined the same way, the calculated work

terms are very different since the T2’s are different,.


14

For (A): wadiab

= nCv∆T = 2.0mol(20.811

J

mol - K)(9.74K) = 405 J for irreversible, adiabatic process

For (B): wadiab

= nCv∆T = 2.0mol(20.811

J

mol - K)(167K) = 6951J for reversible, adiabatic process

2.14 A) Need to determine V1

V1

=nRT

1

P1

=1.0mol (0.08314 L-bar

mol-K)(475K)

2.25bar= 17.55L

and since

Cvln

T2

T1

= -RlnV

2

V1 then:

-C

v

Rln

T2

T1

= lnV

2

V1

Þ lnV

2

V1

= -12.47 J

mol-K

8.314 J

mol-K

ln322K

475K

é

ëê

ù

ûú = -1.59(-0.389) = 0.583

V2

V1

= e+0.583 = 1.79 V2

= 1.79(17.55L) = 31.4L

B) wadiab

= nCv∆T = 1.0mol(12.47 J

mol-K)(322 - 475)K = -1908 J = -1.91kJ

C) wrev

= -nRT lnV

2

V1

= -1.0mol ´ 8.314 J

mol-K(293K)ln 1.79é

ëùû

= -2302J = -2.30kJ

2.15 A) Solving for Cv:

Given:Cvln

T2

T1

= -RlnV

2

V1

(from map) Þ Cvln

(521.59K)

(571.30K)= -8.314(0.693)

J

mol - K

Cvln(0.913) = -5.762

J

mol - KÞ C

v=

-5.762J

mol-K

-0.0916= 62.9

J

mol - K

B) Solving for gamma, ϒ: lnT

2

T1

= -(C

p- C

v)

Cv

lnV

2

V1

Þ lnT

2

T1

= (1 - g )lnV

2

V1

Þ -0.0916 = (1 - g )(0.693)

So 1- g = -0.132 and g = 1+ 0.132 = 1.132

2.16 Given:

P1V

1

g = P2V

2

g so thatP

1

P2

=V

2

g

V1

gÞ

P1

P2

é

ë

êê

ù

û

úú

1/g

=V

2

V1

ÞV

2

V1

=P

1

P2

é

ë

êê

ù

û

úú

Cv/C

p

Substitute for

V2

V1

in Cvln

T2

T1

= -RlnV

2

V1

Then: Cvln

T2

T1

= -RlnP

1

P2

é

ë

êê

ù

û

úú

Cv/C

pé

ë

êêê

ù

û

úúú

Þ Cvln

T2

T1

=C

v

Cp

-RlnP1

P2

é

ë

êê

ù

û

úú

Multiply both sides by C

p

Cv

and result is: Cpln

T2

T1

= -RlnP

1

P2

é

ë

êê

ù

û

úú

B)

Cpln

(521.59K)

(571.30K)= -8.314 J

mol-Kln

1522.2torr

613.5torr

é

ëê

ù

ûú Þ C

p(-0.0916) = -8.314 J

mol-Kln(2.481) = -7.555 J

mol-KÞ

Cp

=-7.555

-0.0916

J

mol - K= 82.48

J

mol - Kcompared to :

Cp

Cv

= 1.132 Þ Cp

= 1.132(62.9) = 71.20J

mol - K

C) Since final molar volume and temperature would be the same, then:

Zobs

=P

obsV

m

RTand Z

ideal=

Pideal

Vm

RTÞ

Zobs

Zideal

=P

obs

Pideal

ÞZ

obs

1.0=

613.5torr

634torr= 0.9677

Given: ln(x y) = y ∙ ln(x)


15

2.17 A) From Problem 2.13 can apply: Cpln

T2

T1

= -RlnP

1

P2

Assume CP, air = CP, N2(g) = 29.1 J/mol-K.

P2

= 65psi1.00atm

14.7psi

é

ëê

ù

ûú = 4.42atm P

1= 755torr

1.00atm

760torr

é

ëê

ù

ûú = 0.993atm

lnT

2

T1

= -R

Cp

lnP

1

P2

é

ë

êê

ù

û

úú

=-8.314 J

mol-K

29.1 J

mol-K

ln0.993atm

4.42atm

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= -0.2857 ´ (-1.493) = 0.4265

T

2(K)

T1(K)

= e+0.4265 = 1.53 T2(K) = 1.53(295K) = 452K(= 179°C)

B) Since work is done on the air, the internal energy ∆U increases. Since the additional energy does not

transfer through tire walls it increases the temperature of the air and tire by adding to the kinetic energy

of the molecules.

2.18 A) dq = 0 since adiabatic

B) wadiab

= nCv∆T = 3.0mol(27.5

J

mol - K)(250 - 200)K = 4125J = 4.125kJ

C) ∆U = w = 4.125 KJ

D) Can use: Cvln

T2

T1

= -RlnV

2

V1

for V2 if V1 is first calculated:

V1

=nRT

1

P1

=3.0mol (0.08206 L-atm

mol-K)(200K)

2.0atm= 24.62L

• Then lnV

2

V1

= -C

v

Rln

T2

T1

= -27.5 J

mol-K

8.314 J

mol-K

ln250K

200K

æ

èç

ö

ø÷ = -3.308(0.223) = -0.738 gives V

2= 0.478(24.62L) = 11.8L

E) The final P can be gotten from ideal gas law once T2, V2 and n are known.

P2

=nRT

1

V2

=3.0mol (0.08206 L-atm

mol-K)(250K)

11.8L= 5.23atm

F) Even though dq = 0, ∆H is not zero within the system, since: ∆H = ∆U + ∆(nRT) = ∆U + nR(∆T) and if

ideal gas, ∆H defined as :

∆H = (4.125kJ) + (3.00mol)(8.314 J

mol-K)(50K)

1kJ

1000J

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= (4.125kJ) +1.247 = 5.37kJ

2.19 A) -qlost

= qgain

Þ -qreaction

= - ∆U°comb

´ (masssample)éë

ùû

= Ccal∆T + C

H2O(l)

(massof H2O)

∆U°comb

(kJ / g) =- C

cal∆T + C

H2O(l)

(massof H2O)∆Té

ëêùûú

mass sample

∆U = -2.753 kJ

K´ 3.11K + 4.184 kJ

kg-K´ 0.9892kg ´ 3.11K( )

0.700g= -

8.56kJ +12.87kJ

0.700g= -30.6kJ / g

B) Need reaction to determine ∆n: C3H6O(l) + 4 O2(g) → 3 CO2(g) + 3 H2O(l) ∆n = -2, then:

∆Hcomb

= ∆Ucomb

(kJ / mol) + ∆n(RT) and

∆Hcomb

= -58.0g

1.0mol

30.6kJ

1.00g

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

+ [-1.0) ´ (8.314 J

mol-K(298K)] = -1776

kJ

mol- 2.48

kJ

mol= -1778

kJ

mol


16

C) Tabled value is -1790 kJ/mol for propanone (acetone), so calculated value quite close.

2.20 A) Reaction: C3H6(g) + 9/2 O2(g) → 3 CO2(g) + 3 H2O(g) Given ∆Hcomb = -1946 kJ (from ∆Hf°

values)

∆U°comb

= ∆H°comb

- ∆n(RT) = -1946kJ - (0.5mol) 8.314J

mol - K(298K)

é

ëê

ù

ûú = -1946kJ -1.239kJ = -1947kJ

Then: -qlost

= qgain

Þ -qreaction

= - ∆U°comb

´ (no.molC3H

6reacted)é

ëùû

= Ccal+contents

∆T

Ccal+contents

=-∆U°

comb´ (no.molC

3H

6reacted)

∆T=

1947kJ ´7.00gC3H

6

13.45°C(K)´

1.0molC3H

6

42.0g

é

ë

êê

ù

û

úú

=324.5kJ

13.45°C(K)= 24.13

kJ

°C(K)

B) Produces a different value for ∆H°comb = -2058 kJ and also for ∆U°comb since ∆n changes.

∆U°comb

= ∆H°comb

- ∆n(RT) = -2058kJ - (-2.5mol) 8.314J

mol - K(298K)

é

ëê

ù

ûú = -2058kJ + 6.2kJ = -2052kJ

Ccal+contents

=-∆U°

comb´ (no.molC

3H

6reacted)

∆T=

2052kJ ´7.00gC3H

6

13.45°C(K)´

1.0molC3H

6

42.0g

é

ë

êê

ù

û

úú

=342.1kJ

13.45°C(K)= 25.43

kJ

°C(K)

• So calorimeter constant is increased slightly if water liquid is formed instead of a gaseous water.

2.21 A) Reaction: CaCl2(s) → Ca+2(aq) + 2 Cl-(aq) ∆Hsoln = -81.3 kJ -qlost

= -qreaction

= qgain

= qwater

-qreaction

= - ∆Hsoln,CaCl

2

´ (no.molCaCl2)é

ëêùûú

= qwater

= Cp,H

2O(l)

´ (mol H2O(l)) ´ ∆Té

ëêùûú

∆T =- ∆H

soln,CaCl2

´ (no.molCaCl2)é

ëêùûú

Cp,H

2O(l)

´ (mol H2O(l))

=

- -81.3kJ

mol

1000 J

1.0kJ

æ

èç

ö

ø÷ ´ (40.25g)

1.00mol

18.0g

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

75.3J

mol - K(°C)´ 99.7g

1.00mol

18.0g

æ

èç

ö

ø÷

æ

èç

ö

ø÷

=2.95X104 J

75.3(5.54)J / °C= 70.7°C

Then:T2

= T1

+ ∆T = 22.0 + 70.7 = 92.7°C

B) Would want T2 to be 0°C, but not start the freezing process, by dissolving more solid.

Reaction: CaCl2(s) → Ca+2(aq) + 2 Cl-(aq) ∆Hsoln = 27.7 kJ

-qwater

= - Cp,H

2O(l)

´ (massH2O(l)) ´ ∆Té

ëêùûú

= qreaction

= ∆Hsoln,NH

4NO

3

´ (no.mol NH4NO

3)é

ëêùûú

- 75.3J

mol - K(°C)(0.3067mol)(-22)°C

é

ëê

ù

ûú = 25.7X104 J

mol´ (no.mol NH

4NO

3)

é

ëê

ù

ûú

no.mol NH4NO

3= 0.357mol ´

80.0g

1.00mol

é

ëê

ù

ûú = 28.6gNH

4NO

3

Since solubility is 119 g NH4NO3 per 100 mL of water at 0°C, all of the solid would dissolve.

C) There are no gases involved in the reaction, so ∆n =0 and ∆H = ∆U. Can also assume the flexible

pouch will just change volume slightly to keep P the same for air trapped in pouch.

2.22 • Same P, T means the volume ratio equals the mole ratio of gases, so to volume % will need to

find moles of each gas in the mixture.

• The tabled combustion enthalpies can be used since they apply to combustions that produce CO2(g) and

liquid water


17

qtotal

= -45.6kJ = qCH

4

+ qC

2H

6

= nCH

4

∆Hcomb,CH

4

+ nC

2H

6

∆Hcomb,C2H6

qtotal

= -45.6kJ = nCH

4

(-890.3)kJ

mol+ n

C2H

6

(-1559.7)kJ

mol (1)

• Have an equation with 2 unknowns so need a second defined equation that relates moles of CH4(g) and

C2H6(g).

• Know P,T, V of mixture so can define ntotal:

nTotal

=P

totalV

mix

RTmix

=1.0atm(1.0L)

0.08206 L-atm

mol-K(273K)

= 0.04464mol and

0.04464mol = nCH

4

+ nC

2H

6

Þ nC

2H

6

= 0.04464mol - nCH

4

(2)

Then substituting (2) into (1) produces:

qtotal

= -45.6kJ = nCH

4

(-890.3)kJ

mol+ (0.4464 - n

CH4

)(-1559.7)kJ

molÞ -45.6= (-890.3 +1559.7)n

CH4

- 69.62

-45.6= (669.4)nCH

4

- 69.62 Þ69.62 - 45.6

669.4= n

CH4

= 0.0365mol

nC

2H

6

= 0.04464 - 0.0365 = 0.00814mol

Vol%CH4

=V

CH4

Vtotal

´100 =n

CH4

ntotal

´100 =0.0365

0.04464´100 = 81.8 % ThenVol%C

2H

6= 18.2%

2.23 A) C3H

8(g) +5O

2(g)® 3CO

2(g) + 4H

2O(l)

∆Hcomb

= 3mol∆Hf ,CO

2(g)

o + 4mol∆Hf ,H

2O(l)

oéëê

ùûú

- 1mol∆Hf ,C

3O

8(g)

o + 5mol∆Hf ,O

2(g)

oéëê

ùûú

= 3mol(-393.5kJ

mol) + 4(-285.8

kJ

mol)

é

ëê

ù

ûú - 1mol(-104.5

kJ

mol) + 0

é

ëê

ù

ûú = (-2314.7)kJ +104.5kJ = -2210.2kJ / molC

3O

8(g)

B) nC

3H

8

reacted =(P

2- P

1)V

RT=

(2.35 -1.10)atm(200L)

0.08206 L-atm

mol-K(300K)

= 10.19mol

-qcomb

= - ∆Hcomb

´ (no.molC3H

8reacted)é

ëùû

= -2210.2 kJ

mol(10.19mol) = -2.252X104kJ

qabsorbed

= Cp,H

2O(l)

J

g-K(°C)

éë

ùû´ (massH

2O(l)) ´ ∆Té

ëêùûú

= 4.184 J

g -K(°C)( ) 1.325X105 g( ) 36.2°C( )

= 2.007X107 J1.0kJ

1000 J

æ

èç

ö

ø÷ = 2.01X104kJ

%heat absorbed =2.00X104kJ

2.26X104kJ´100 = 88.9%

2.24 A) qcalories

= ∆Hfus

´ (no.mol H2Omelted)é

ëùû

= ∆Hfus

´ (mass icemelted) ´1mol H

2O

18.0g

æ

èç

ö

ø÷

é

ë

êê

ù

û

úú

mass icemelted =q

calories´ 4.184kJ

1.0Cal

æ

èçç

ö

ø÷÷ ´18.0g

∆Hfus

kJ

mol(1mol)

=500Cal ´ 4.184 kJ

Cal´18.0 g

mol

6.00 kJ

mol´ (1mol)

= 6276g = 6.276kg

B) qcalories

(kJ) = qmelting

+ qheating liquid

= ∆Hfus,H

2O(s)

´ (mass ice) ´1mol

18.0g

é

ëê

ù

ûú + C

p,H2O(l)

J

g - K(°C)

é

ëê

ù

ûú ´ (mass ice) ´ ∆T

é

ëêê

ù

ûúú


18

qcalories

(kJ) = (mass ice)∆H

fus,H2O(s)

18.0

kJ

g

æ

èç

ö

ø÷ +

Cp,H

2O(l)

J

g-K(°C)

é

ë

êê

ù

û

úú´ ∆T

liquid

1000J

kJ

é

ë

êêê

ù

û

úúú

mass ice =q

calories(kJ)

6.00

18.0+

4.184 ´ 37°C

1000

æ

èç

ö

ø÷

æ

èç

ö

ø÷

kJ

g

é

ë

êê

ù

û

úú

=2092kJ

0.333 + 0.155éë

ùûkJ

g

= 4284g = 4.284kg

• The energy absorbed by heating the melted ice to 37°C decreases the amount of ice needed to

consume 500 Calories by about one-third. So it should not be neglected.

2.25 Have to three things consider:

The heat released will be dependent on the number of moles of Ca(OH)2 produced by the limiting

reactant, so a limiting reactant calculation must be done first.

The heat produced will be absorbed by the water and will change its T. The question then is ill it heat the

water enough to boil it?

(3) If it does boil, then how much of the liquid water will be converted to water vapor?

Limiting reactant calculation:

56.0gCaO1molCaO

56.0g

æ

èç

ö

ø÷

1molH2O

1molCaO

æ

èç

ö

ø÷ = 1.0molH

2Orequired tocompletely react

100mLH2O

1.00g

1mLH2O

æ

èç

ö

ø÷

1mol H2O

18.0g

æ

èç

ö

ø÷ = 5.55molH

2Oavailabletoreact

So 1.0 mol CaO will be limiting the amount of product reacted 1.00 mol Ca(OH)2 if completely reacted.

(That leaves 4.55 moles of H2O un-reacted to absorb the heat released as qgain.)

• We will need the ∆H for the reaction to calculate qlost.

∆Hr

= 1mol∆Hf ,Ca(OH)

2(s)

oéëê

ùûú

- 1mol∆Hf ,CaO(s)

o +1mol∆Hf ,H

2O(l)

oéëê

ùûú

= 1mol(-987kJ

mol)

é

ëê

ù

ûú - 1mol(-635

kJ

mol) +1mol (-285.8

kJ

mol)

é

ëê

ù

ûú = -987kJ + 920.8kJ = -66.2kJ / mol

Then since - qlost = qgain can be defined as:

-qr

= -qlost

= - -66.2kJ

molCa(OH)2

(1.0molCa(OH)2)

é

ë

êê

ù

û

úú

= 66.2kJ = qgain

(2) Determine how much heat is required to heat H2O(l) from 26°C to 100°C and compare it to qgain.

qneeded

= qH

2O(l)excess

= 4.45molH2O 75.3 J

mol-K(°C)( ) 100 - 26( )°C = 2.480X104 J = 28.4kJ

• So there is more than enough heat to bring all remaining water to 100°C.

The question now is how much water will be converted at 100°C by the remaining heat (66.0 - 28.4) =

37.6 kJ

qleft

= ∆Hvap

´ (no.molH2Oconverted)é

ëùû

Þ no.molH2Oconverted =

(66.2 - 28.4)kJ

40.66 kJ

mol

= 0.924mol

So at the end of the reaction, there will be 0.924 mol of water as steam (H2O(g)), 3.526 mol of liquid

water and 1.0 mol of Ca(OH)2 all at 100°C.

2.26 A) Using Hess’s Law for combustion: C16

H34

(l) + 24.5O2(g)®16CO

2(g) +17H

2O(l)

∆Hcomb

= -10,700kJ = 16mol∆Hf ,CO

2(g)

o +17mol∆Hf ,H

2O(l)

oéëê

ùûú

- 1mol∆Hf ,C

16O

34(l)

o + 24.5mol∆Hf ,O

2(g)

oéëê

ùûú

∆Hf ,C

16O

34(l)

o =+10,700kJ + 16mol∆H

f ,CO2(g)

o +17mol∆Hf ,H

2O(l)

oéëê

ùûú

- 24.5mol∆Hf ,O

2(g)

o

1molC16

H34


19

∆Hf ,C

16O

34(l)

o = +10,700kJ - 6296kJ - 4859kJ - 0 = -455kJ / mol

B) C16

H34

(l) + 24.5O2(g)®16CO

2(g) +17H

2O(l) ∆n = 17.0 -24.5 = -7.5

∆Ucomb

= ∆Hcomb

+ ∆nRT = (-10,700kJ) + -7.5mol ´ 8.314J

mol - K(298K)

é

ëê

ù

ûú

∆Ucomb

= (-10,700kJ) + (-21.1kJ) = -10,721kJ per mol ´1molC

16H

34

226g

é

ë

êê

ù

û

úú

= -47.4kJ / g

C) Given 1 kCal = 4.184 kJ, then:

Caloric content = ∆Ucomb

(kCal

g) =

-47.4kJ

1.0gC16

H34

1kCal

4.184kJ

é

ëê

ù

ûú = 11.3

kCal

g

The hydrocarbon has a higher caloric value than the average fat molecule 11.3 versus 9.0 kCal/g.

2.27 A) Write the balanced reaction: C6H

12O

6(s) +O

2(g)® 2CH

3COCO

2H(l) + 2H

2O(l)

• Need to calculate ∆Hreaction from data using calorimetry: -qlost

= -qreaction

= qgain

= qwater

-qreaction

= - ∆Hr´ (no.molC

6H

12O

6)é

ëùû

= qwater

= Cp,H

2O(l)

J

g - K(°C)

é

ëê

ù

ûú ´ (massH

2O(l)) ´ ∆T

é

ëêê

ù

ûúú

∆Hr

= -

4.184kJ

kg - K(°C)´ (0.9550kg) ´ (7.03°C)

é

ëê

ù

ûú

10.52g1.0molC

6H

12O

6

180.16g

æ

èç

ö

ø÷

= -28.10kJ

0.05839mol= -481.2

kJ

molC6H

12O

6

• Then use Hess’s Law for balanced reaction to get ∆H°f, pyruvic acid solid:

∆Hr

= -481.2kJ = 2mol∆Hf ,CH

3COCO

2H(s)

o + 2mol∆Hf ,H

2O(l)

oéëê

ùûú

- 1mol∆Hf ,C

6H

12O

6(s)

o +1mol∆Hf ,O

2(g)

oéëê

ùûú

∆Hf ,CH

3COCO

2H(s)

o =-481.2kJ + 1mol∆H

f ,C6H

12O

6(s)

o +1mol∆Hf ,O

2(g)

oéëê

ùûú

- 2mol∆Hf ,H

2O(l)

o

2 mol

∆Hf ,CH

3COCO

2H(s)

o =-481.2kJ + (-1273.3)kJ + 0 - 571.6kJ

2 mol=

-1183kJ

2mol= -591.5kJ / mol

2.28 Want ∆Hr° for: NaC2H

3O

2(s) + 3H

2O(l)® NaC

2H

3O

2·3H

2O(s)

Given the hydrate must be on the product side after the addition and the anhydrous form must be on the

reactant side, then:

NaC2H

3O

2(s)®

H2O

Na+(aq) + C2H

3O

2

-(aq) ∆Hsoln

= -17.32kJ

+ Na+(aq) + C2H

3O

2

-(aq) + 3H2O(l)®NaC

2H

3O

2·3H

2O(s) -1 ´ ∆H

soln= -19.66kJ

NaC2H

3O

2(s) + 3H

2O(l)®NaC

2H

3O

2·3H

2O(s)

B) So the enthalpy of hydration is ∆Hhydration

= -17.32+ (-19.66) = -36.98kJ

2.29 Combustion reactions: C6H

12O

6(s) +6O

2(g)® 6CO

2(g) + 6H

2O(l) ∆H = -2805 kJ

C2H

5OH(l) +3O

2(g)® 2CO

2(g) + 3H

2O(l) ∆H = -1367.3 kJ

To produce overall reaction C6H

12O

6(s)® 2CH

3CH

2OH(l) + 2CO

2(g) will need to keep C6H12O6(s) on reactant

side, but C2H5OH(l) must appear on the product side, so the second reaction must be reversed before it is


20

added to the first. Also to get correct coefficient for C2H5OH(l), the second reaction must be multiplied by

2.

Applying these changes:

C6H

12O

6(s) + 6O

2(g) ® 6CO

2(g) + 6H

2O(l) ∆H = -2805kJ

+ 4CO2(g) + 6H

2O(l) ® 2C

2H

5OH(l) + 6O

2(g) ∆H = -2(-1367.3) = +2734.6kJ

C6H

12O

6(s)® 2CO

2(g) + 2C

2H

5OH(l) ∆H

r= -2805kJ + 2734.6kJ = -70.4kJ

2.30 Want ∆H for the reaction: H2O(g) → H(g) + OH(g) by adding the appropriate reactions.

Reaction 1: ½ H2(g) + ½ O2(g) → OH(g) ∆H = 38.95 kJ

Reaction 2: H2(g) + ½ O2(g) → H2O(g) ∆H = -241.8 kJ

Reaction 3: H2(g) → 2 H(g) ∆H = 496.0 kJ

Reaction 4: O2(g) → 2 O(g) ∆H = 498.3 kJ

• Focus on OH, H and H2O as the unique substances, which will tell you how to use reactions 1-3.

• Will need to get H2O(g) on the reactant side, so Reaction 2 must be reversed, and its ∆H multiplied by -

1.

• OH is already on correct side and has the needed coefficient so that reaction can be added as is.

• H is already on correct side but does not have the needed coefficient, so the third reaction will be

added after multiplying it (and its ∆H) by ½.

• We will not need to use the fourth reaction.

H2O(g)®H

2(g) +1 /2O

2(g) -1 ´ ∆H

2= 241.8kJ

+ 1 /2H2(g) +1 /2O

2(g)® OH(g) ∆H

1= 38.95kJ

+ 1 /2 H2(g)® 2H(g)é

ëùû

1 /2 ´ ∆H3

= 218kJ

H2O(g) + 1 /2H

2(g) + 1 /2O

2(g) + 1 /2H

2(g) ® H

2(g) + 1 /2O

2(g) + OH(g) + H(g)

Leads to: H2O(g)®OH(g) + H(g) ∆H

r= 241.8 + 38.95 + 218( )kJ = 499kJ

B) Reaction 1 and 2 represent one mole of a compound being formed from elements in their most

common state, so those reactions are formation reactions. However, reaction 3 and 4 describe breaking a

bond in a gaseous molecule so they are dissociation reactions and the enthalpy of reaction is equal to the

respective bond energies of H-H and O=O.

2.31 Combustion equations: B2H

6(g) +3O

2(g)® B

2O

3(s) + 3H

2O(l) ∆H =-2034 kJ

2B(s) +3 /2 O2(g)® B

2O

3(s) ∆H = -1264 kJ

A) Reaction sought: 2B(s) +3 H2(g)® B

2H

6(g) ∆H = ∆H

f ,B2H

6(g)

o = ?

• B2H6(g) must be on product side so first combustion reaction must be reversed and the ∆H multiplied

by -1.

• In the 2nd reaction, B is already on correct and has the needed coefficient, so that reaction can be

added as is.

2B(s) +3 /2 O2(g)® B

2O

3(s) ∆H = -1274kJ

+ B2O

3(s) + 3H

2O(l)® B

2H

6(g) +3O

2(g) -1 ´ ∆H

comb= -(-2034)kJ = 2034kJ

2B(s) +3 /2 O2(g) + B

2O

3(s) + 3H

2O(l)® B

2O

3(s) + B

2H

6(g) +3O

2(g)

Produces :2B(s) + 3H2O(l)® B

2H

6(g) +3 /2O

2(g) ∆H = -1274 + 2034 = 760kJ

• H2O(l) is a reactant and must be removed (and H2(g) added), so the best way to address this will be to

add the formation reaction for H2O(l) (and its ∆H).

H2(g) +1 /2 O

2(g)® H

2O(l) ∆H

f

o = -285.8kJ


21

• H2O(l) needs to be on the product side and it is, but have 3 for the coefficient, so the reaction will be

multiplied by 3 before adding. (The same will have to done to ∆H).

2B(s) + 3H2O(l)® B

2H

6(g) +3 /2O

2(g) ∆H = 760kJ

+ 3(H2(g) +1 /2 O

2(g)® H

2O(l)) 3 ´ ∆H

f ,H2O(l)

o = 3(-285.8) = - 857.4kJ

2B(s) + 3H2O(l) + 3H

2(g) + 3 /2 O

2(g) ® B

2H

6(g) + 3 /2O

2(g) + 3H

2O(l)

2B(s) + 3H2(g)® B

2H

6(g) ∆H = ∆H

f ,B2H

6(g)

o = 760 + (-857.4)kJ = -97.4kJ

B) Hess’s Law for combustion:

∆Hcomb

B2H

6(g) = 1mol∆H

f ,B2O

3(s)

o + 3mol∆Hf ,H

2O(l)

oéëê

ùûú

- 1mol∆Hf ,B

2H

6(g)

o + 3mol∆Hf ,O

2(g)

oéëê

ùûú becomes:

∆Hf ,B

2H

6(g)

o =1mol∆H

f ,B2O

3(s)

o + 3mol∆Hf ,H

2O(l)

oéëê

ùûú

- 3mol∆Hf ,O

2(g)

oéëê

ùûú

- ∆Hcomb

,B2H

6(g)

1mol

= (-1274) + 3(-285.8) - 3(0) - (-2034) = -1264 - 857.4 + 2034 = -97.4kJ / mol

• So adding reactions to get a ∆H and Hess’s law are often equivalent processes.

2.32 (1) Picking the “unique” substances that only appear once in the listed reactions and are in the

overall reaction sought, CS2 and S2Cl2, shows how 2 of the reactions - Reactions I and II- should be

added:

(2) From the sum CO2 and SO2 will need to be eliminated and CCl4 needs to be added.

• Could introduce CCl4 either by subtracting (reversing) Reaction VII or adding Reaction III. Using

Reaction VII would introduce 2 substances we could not get rid of, so must use reaction III.

CS2(l) + 3O

2(g) + 2 S(s) +Cl

2(g)® CO

2(g) + 2SO

2(g) + S

2Cl

2(l) ∆H

I+ ∆H

II

+ C(s) +2Cl2(g)® CCl

4(l) ∆H

III= -135.4kJ

CS2(l) + 3O

2(g) + 2 S(s) +3Cl

2(g) + C(s)® CO

2(g) + 2SO

2(g) + S

2Cl

2(l) + CCl

4(l)

• Add the reverse of Reactions IV, multiplied by 2, and VI to get rid of SO2, CO2

CS2(l) + 3O

2(g) + 2 S(s) +3Cl

2(g) + C(s)® CO

2(g) + 2SO

2(g) + S

2Cl

2(l) + CCl

4(l)

+ 2(SO2(g)® S(s) + O

2(g)) - 2 ´ ∆H

IV= 593.6kJ

+ CO2(g)® C(s) +O

2(g) -1 ´ ∆H

VI= 393.5kJ

CS2(l) + 3O

2(g) + 2 S(s) +3Cl

2(g) + C(s) + 2SO

2(g) + CO

2(g) ®

CO2(g) + 2SO

2(g) + S

2Cl

2(l) + CCl

4(l) + 2S(s) + 3O

2(g) + C(s)

leads to :CS2(l) +3Cl

2(g)® S

2Cl

2(l) + CCl

4(l)

∆H = ∆HI

+ ∆HII

+ ∆HIII

- 2∆HIV

- ∆HVI

= -1077 + (-58.2) + (-135.4) - 2(-296.8) - (-393.5) = -1270.6 + 987.1 = -283.5kJ

2.33 A) Reaction: 2 H2O2(g) → 2 H2O(g) + O2(g)

∆H298

o = 2 mol∆Hf ,H

2O(g)

o +1mol∆Hf ,O

2(g)

oéëê

ùûú

- 2mol∆Hf ,H

2O

2(g)

oéëê

ùûú

= 2(-241.8) + 0éë

ùû

- 2(-136.3)éë

ùû

= -211kJ

B) Pathway involving bond energies:

CS2(l) + 3O

2(g)®CO

2(g) + 2SO

2(g) ∆H

I= -1077kJ

+ 2 S(s) +Cl2(g)® S

2Cl

2(l) ∆H

II= 38.95kJ

CS2(l) + 3O

2(g) + 2 S(s) +Cl

2(g)® CO

2(g) + 2SO

2(g) + S

2Cl

2(l)


22

∆H298

o = q1

+ q2

+ q3

+ q4

Þ ∆H298

o = 4(B.E.O-H

) + 2(B.E.O-O

) - 4(B.E.O-H

) - (B.E.O=O

)

-211kJ = 2(B.E.O-O

) - 498kJ Þ (B.E.O-O

) =(498 - 211)kJ

2mol= 143.5kJ / mol

C) Normally, using a catalyst would not change the ∆H for the reaction, since largely only affecting

energy barrier, but if the states for H2O2 and H2O change, the ∆H for the reaction is affected.

2 H2O2(g) → 2 H2O(l) + O2(g)

∆H298

o = 2 mol∆Hf ,H

2O(l)

o +1mol∆Hf ,O

2(g)

oéëê

ùûú

- 2mol∆Hf ,H

2O

2(aq)

oéëê

ùûú

= 2(-285.8) + 0éë

ùû

- 2(-191.7)éë

ùû

= -188kJ

D) Since bond energies are only defined for to gaseous molecules forming gaseous atoms, the inclusion

of H2O2 as an aqueous species and liquid water in the enzyme reaction would make the estimation of the

bond energy much more uncertain. More heat changes would have to be included in the pathway and the

conversion energy for H2O2 (aq) to H2O2(g) would be very difficult to estimate.

2.34 Pathway and calculation:

∆Hr

= q1

+ q2

+ q3

= 3(B.E.F -F

) + (-6B.E.Xe-F

) + (-∆Hsub,XeF

6(s)

)

-368.2kJ = 3(158kJ) - 6(B.E.Xe-F

) - 62.3kJ

B.E.Xe-F

=411.7 + 368.2

6kJ = 130kJ

• NOTE: Since bond energies are for gaseous species only, must condense the gas to the solid to

complete pathway.

2.35 A) q1 = B.E. for N º N (or ∆Hf° N(g)) q2 = B.E. for H-H (or ∆Hf° H(g)) q3 = B.E. for N-H

B)∆H

r= q

1+ q

2+ q

3= 0.5(B.E.

NºN) +1.5(B.E.

H-H) + (-3B.E.

N-H)

∆Hr

= 0.5(945kJ) +1.5(432kJ) - 3(391kJ) = -52.5kJ

So the value is higher than the published value of -46.11 kJ. It

overestimates the actual ∆H by about 14%

C) The bond energies can’t account for interactions between molecules or atoms that affect ∆H. In this

case, H-bonding between NH3(g) molecules could not be accounted for in bond dissociation.

Break 4 O-H bonds

q1= 4 B.E. O - H

Free gaseous atoms: 4 H (g) + 4 O(g)

Form 1 O =O bond

Break 2 O -O bonds

q2= 2 B.E. O - O

Form 4 O-H bonds

q3= - 4 B.E. O - H

q4= - B.E O=O

2 H2O2(g) à 2 H2O (g) + O2(g) ∆H = -211 kJ

Convert (g) à (s)q1= 3 B.E. F – F

= 3 ∆HdissF2(g)

Form 6 Xe - F bonds q2= - 6 B.E. Xe - F

q3= -∆Hsub, XeF6(g)

Xe(g) + 3 F2(g) à XeF6(s) ∆H = -368.2 kJ

Xe(g) + 6 F (g) à XeF6(g)

Break 3 F - F bonds

1/2 N2(g) + 3/2 H2(g) à NH3(g)

N (g) + 3 H(g)

∆Hr°

q2 = 3/2 B.E. H - H

Break H-H bondBreak N º N bond

q3 = 3 B.E. N - H


23

2.36 A) A) q1 = ∆Hsublimation C(s,graphite) q2 = B.E. for H-H (or ∆Hf° H(g))

q3 = B.E. for C-C and B.E. for C-H q4 = ∆Hvaporization C8H18

Pathway and calculation:

∆Hr

= q1

+ q2

+ q3

+ q4

= 8(∆Hsub

,C(s)) + 9(B.E.H-H

) - (7B.E.C-C

+18B.E.C-H

) - (∆Hvap

,C8H

18(l))

∆Hr

= 8(716kJ) + 9(432kJ) - (7(347kJ) +18(413kJ)) - 41.4kJ = 9621.6 - 9904kJ = -283kJ

B) ∆Hcomb

C8H

18(l) = 8mol∆H

f ,CO2(g)

o + 9mol∆Hf ,H

2O(l)

oéëê

ùûú

- 1mol∆Hf ,C

8H

18(l)

o +12.5mol∆Hf ,O

2(g)

oéëê

ùûú

∆Hf ,C

8H

18(l)

o =8mol∆H

f ,CO2(g)

o + 9mol∆Hf ,H

2O(l)

oéëê

ùûú

- 12.5mol∆Hf ,O

2(g)

oéëê

ùûú

- ∆Hcomb

,C8H

18(l)

1mol

∆Hf ,C

8H

18(l)

o = 8(-393.5) + 9(-285.8) -12.5(0) - (-5470.2) = -5720.2 + 5470.2 = -250kJ / mol

C) Replacing q1 with:q1

= 8(2B.E.C-C

+ B.E.C=C

) = 8(2(347) + (614))kJ = 8(961)kJ = 7688kJ versus 5733 kJ for

the first term results in a huge change in ∆Hf ,C

8H

18(l)

o . It becomes + 1676 kJ, which doesn’t make sense.

This tells you that the assumption that bond energies can describe the change C(s, graphite → C(g) is

incorrect.

D) The calculated total change in bond energy for the reaction will not match the true ∆H for the reaction,

but we may get a reasonable approximation of the magnitude of the bond energy, using the pathways

with as many known ∆H’s as possible, so it can only be an estimate.

2.37 A) Cycle will involve the ∆Hf° KF(s), ∆Hf° K(g), First ionization energy (I.E.) of K, Bond energy of

F-F, Electron affinity (E.A.) for F(g) and the lattice energy for K+ and F- in KF(s).

B) Given:

∆Hf° K(g) = 89.24 kJ/mol

B.E. F-F = 159 kJ/mol

I.E (1st) K = 418.8 kJ/mol

E.A. F(g) = -328 kJ/mol

Then:

q5

= ∆Hf

o KF(s) - q1

+ q2

+ q3

+ q4

éë

ùû

q5

= -567.2 - [89.2 + 79.5 + 418.8 - 328]

= -567.2 - 259.5 = -826.8kJ

So the calculated value is close to the tabled values

for the lattice energy.

2.38 A) Units: a = J/mol-K b = J/mol-K2 c = = J/mol-K3

B) Polynomial from Graph 1 a = 211.99, b= -0.7425 and c = 1.5 X 10-3

C

p= a + bT + cT 2 = (211.99) + (-0.7425 ´ 298) + (1.5X10-5 ´ (298)2)é

ëùûJ / mol - K

= 211.9 + (-221.3) +133éë

ùûJ / mol - K = 123.7 J / mol - K

• The calculated value is very close to the tabled value of 124.3 J/mol-K

Form gaseous

neutral atoms

q1= ∆Hf° K(g)

K(s) + ½ F2(g) à KF(s) ∆Hf° = -567.2 kJ

Form ionic solid

from gaseous

ions

Ionize the gaseous

atoms

q2=½ B.E. F - F q5= lattice energy

for gaseous ions

forming solidK(g) + F(g)

K+(g) + F-(g)

q3= 1st I.E. K(g)

q4= E.A. F(g)


24

C) Polynomial from Graph 2: A(1) = 717, A(2) = -5.9176, A(3) = 0.0191 and A(3) = - 1.00 X 10-5

Cp

= A(1) + A(2)T + A(3)T 2 + A(4)T 3 = (717) + (-5.9176 ´298) + (0.0191 ´ (298)2) + (-2.00X10-5 ´ (298)2)éë

ùû

= 717 + (-1763.4) +1696.3 + (-529.3)éë

ùûJ / mol - K = 120.5 J / mol - K

• This calculated value is not as close to the tabled value of 124.3 J/mol-K, and probably not worth the

effort in having to include the cube term.

D) No, all the terms in either polynomial significantly affect the value of Cp and should not be neglected.

2.39 A) ∆H

f(T

2) = ∆H

f(T

1) +

T1

T2

ò CpdT = ∆H

f(298) + C

p T1

T2

ò dT = ∆Hf(298) + C

p,298(T

2- T

1)

= -248.1 kJ

mol+ 0.1243 kJ

mol-K(400 - 298)Ké

ëùû

= -248.1 +12.7 = -235.4 kJ

mol@400K

B)

∆Hf(T

2) = ∆H

f(T

1) +

T1

T2

ò CpdT = ∆H

f(T

1) +

T1

T2

ò (a + bT + cT 2)dT

= ∆Hf(T

1) +

T1

T2

ò adT +T1

T2

ò (bT)dT +T1

T2

ò (cT 2)dT = ∆Hf(298) + a(T

2- T

1) +

b

2(T

2

2 - T1

2) +c

3(T

2

3 - T1

3)

Where a(T2

- T1) = 211.99

J

mol - K´ (102)K ´

1kJ

1000J= 21.62kJ ,

b

2T

2

2 - T1

2éë

ùû

=-0.7425

2

J

mol - K2(400)2 - (298)2éë

ùûK2 ´

1.0kJ

1000 J= 26.43

kJ

mol , and

c

3T

2

3 - T1

3éë

ùû

=1.5X10-3

3

J

mol - K3(400)3 - (298)3éë

ùûK3 ´

1.0kJ

1000 J= 18.77

kJ

mol so that:

∆H(400K) = -248.1 kJ

mol+ (21.62 - 26.43 +18.77) kJ

mol= -234.1 kJ

mol@400K

• The difference is very slight, so unless the temperature change is very large, holding Cp constant is a

good approximation of the change in the ∆H value for chemical substances.

2.40 A)

∆H298

o = 2 mol∆Hf ,CH

4(g)

o +1mol∆Hf ,O

2(g)

oéëê

ùûú

- 2mol∆Hf ,CH

3OH(g)

oéëê

ùûú

= 2(-74.86) + 0éë

ùû

- 2(-200.7)éë

ùû

= 250.7kJ

B) (a) Pathway for ∆Hr at 500 °C:

(b) ∆Hr(500°C) = q

2+ ∆H

r

°(25°C) + q3

+ q4 where:

q2

= 2mol ´ CpCH

3OH(g)

´ ∆T = 2mol ´ 44.101J

mol - K´ (298 - 773)K ´

1.0kJ

1000 J= -41.90kJ

q3

= 2mol ´ Cp,CH

4(g)

´ ∆T = 2mol ´ 35.695J

mol - K´ (773 - 298)K ´

1.0kJ

1000 J= 33.91kJ and

q4

= 1mol ´ Cp,O

2(g)

´ ∆T = 1mol ´ 29.378J

mol - K´ (773 - 298)K ´

1.0kJ

1000 J= 13.96kJ

So that: ∆Hr(500°C) = -41.90kJ + 250.7kJ + 33.91kJ +13.96kJ = 257.7kJ

C) Applying the integration and using the coefficients gotten from plotting the data yields:

∆Hf(T

2) = ∆H

f(T

1) + C p dT

T1

T2

ò = ∆Hf(298) + (a + bT)dT

T1

T2

ò = ∆Hf(298) + a(T

2- T

1) +

b

2(T

2

2 - T1

2)

a b T2 (K) T1(K) ∆Hf (T2) New q term

CH3OH(g), q1 22.811 0.0726 298 773 -29.30 kJ -58.60 kJ

Cool CH3OH(g), q2 T2 à T1

∆Hr° 298 q2

Heat CH4(g), q3 T1 à T2

∆H at 500°C (773K) 2 CH3OH(g) à 2 CH4(g) + O2(g)

∆H at 25°C (298K) 2 CH3OH(g) à 2 CH4(g) + O2(g)

Heat O2(g), q4 T1 à T2


25

CH4, q3 18.347 0.0572 773 298 23.26 kJ 46.53 kJ

O2 (g), q4 26.677 0.0089 773 298 14.94 kJ 14.94 kJ

∆Hr(500°C) = q

2+ ∆H

r

°(25°C) + q3

+ q4

= (-58.60 + 250.7 + 46.53 +14.94)kJ = 253.6kJ

• So taking into account the T dependence of the individual substances shows an even smaller change in

∆H for the reaction than the value in (B). Nonetheless, the value gotten by assuming Cp’s constant is not

very different, representing a 2.8% decrease instead of a 1.2% decrease.

2.41 A) Adding the reactions appropriately, as shown below, produces the final result:

1 /2 2W(s) + 3O2(g)® 2WO

3(s)é

ëùû

1 /2 ´ ∆H1

= -840.3kJ

+ C(graphite) + O2(g)® CO

2(g) ∆H

2= -393.5kJ

+ 1 /2 2WO3(s) + 2CO

2(g)® 2WC(s) + 5O

2(g)é

ëùû

-1 /2 ´ ∆H3

= 1195.5kJ

W(s) +3

2O

2(g) + C(graphite) + O

2(g) + WO

3(s) + CO

2(g) ® WO

3(s) + CO

2(g) +WC(s) +

5

2O

2(g)

leads to :W(s)+ C(graphite)®WC(s) ∆Hr

= ∆Hf ,WC(s)

o = -840.3 + -393.5 +1195.5( )kJ = -37.7kJ

B) B) (a) Pathway for ∆Hr at 1400 °C:

(b) where: ∆Hf ,WC(s)

(1400°C) = q1

+ +q2

+ ∆Hf ,WC(s)

° (25°C) + q4

q1

= 1mol ´ Cp,W(s)

´ ∆T = 1mol ´ 26.31J

mol - K´ (298 -1673)K ´

1.0kJ

1000 J= -36.2kJ

q2

= 1mol ´ Cp,C(s)

´ ∆T = 1mol ´ 8.54J

mol - K´ (298 -1673)K ´

1.0kJ

1000 J= -11.74kJ and

q4

= 1mol ´ Cp,WC(s)

´ ∆T = 1mol ´ 48.60J

mol - K´ (1673 - 298)K ´

1.0kJ

1000 J= 66.8kJ

So that: ∆Hr(1400°C) = -36.2kJ -11.74kJ - 37.7kJ + 66.8kJ = -18.8kJ

As an alternative, you can describe the change in ∆H by the difference in Cp’s =(∆Cp

´ ∆T ), as long as

Cp’s are constant T1 → T2, where ∆Cp

=

products

å Cp

-

reactants

å Cp.

For :aA + bB ® cC and∆T = T2

- T1: ∆H

r

o(T2) = ∆H

r

o(T1) + cC

p,C- aC

p,A+ bC

p,B( )éë

ùû∆T

∆Hr

o(1400°C) = ∆Hr

o(25°C) + 1molCp,WC(s)

- 1molCp,W(s)

+1molCp,C(s)( )é

ëùû∆T

= -37.7kJ + 1molWC(48.61) J

mol-K(°C)- 1molW(26.31) J

mol-K(°C)+1molC(8.62) J

mol-K(°C)éë

ùû

éë

ùû´1375°C

∆Hr

o(1400°C) = -37.7kJ + 48.61 - 34.93( ) J ´1375 ´1kJ

1000 J

é

ëê

ù

ûú = -37.7kJ +18.81kJ = -18.9kJ

C) Calculating the change in the q2 term:

q2

=

298

1673

ò Cp,C(s)

dT =T1

T2

ò (a + bT + cT 2)dT =T1

T2

ò adT +T1

T2

ò (bT)dT +T1

T2

ò (cT 2)dT = a(T2

- T1) +

b

2(T

2

2 - T1

2) +c

3(T

2

3 - T1

3)

Cool W(s), q1 T2 à T1

∆Hr° 298 q3

∆H at 1400°C (1673K) W(s) + C(s,graphite) à WC(s)

∆H at 25°C (298K) W(s) + C(s,graphite) à WC(s)

Heat WC(s), q4 T1 à T2

Cool C(s), q2 T2 à T1


26

which is significantly different. So in this case, assuming Cp constant is NOT a good assumption for C(s).

The correction the q2 term affects ∆Hr markedly, so that:

q2

= 1mol -0.4493 J

mol-K(1375K) + 0.01777 J

mol-K2(1673)2 - (298)2éë

ùûK2 - 4.333X10-5 J

mol-K3(1673)3 - (298)3éë

ùûK3é

ëêùûú

= 27.35kJ

and ∆Hr(1400°C) = -36.2- 27.35- 37.7+ 66.8é

ëùûkJ = -34.45kJ instead of -18.9 kJ.

So although most times assuming Cp is constant is a good approximation, it will not always be true and

the only way to evaluate is to know the proper function of Cp is for the T range applied to the reaction.

2.42 A) One major assumption will be to assume that you can use the solid state values for the ∆H’s to

calculate the ∆Hr° although the conversion likely takes place in solution phase. It is reasonable to expect

that any shift in the ∆H value for alanine and aspartic acid, due to changes in interactions between solute

and solvent, are similar and would cancel out.

Another assumption would be that the Cp’s stay constant from 25 to 50°C. It is unlikely that you easily

find data for the temperature dependence of alanine and aspartic acid, so that this assumption is

necessary.

B) You would need the enthalpies of formation of alanine(s), aspartic acid(s) and CO2(g) and their

respective Cp, 298 values. You would also need to use Hess’s law and the temperature dependence of ∆H

as:

B) ∆Hr

o(T2) = ∆H

r

o(T1) + 1mol C p,CO

2(g) +1molC

p,alanine( ) -1molCp,aspartic acid

éëê

ùûú∆T

∆H298

o = 1 mol∆Hf ,alanine(s)

o +1mol∆Hf ,CO

2(g)

oéëê

ùûú

- 1mol∆Hf ,aspartic acid(s)

oéë

ùû

= -563.6 + -393.5éë

ùûkJ - -973.3é

ëùûkJ = +16.3kJ

∆Hr

o(50°C) = ∆Hr

o(25°C) + 1mol C p,CO2(g) +1molC

p,alanine( ) -1molCp,aspartic acid

éëê

ùûú∆T

= (16.3kJ) + (37.11 +115)J -155.2 Jéë

ùû´ (25)K ´

1.0kJ

1000 J= (16.3 - 0.077)kJ = 16.2kJ

2.43 A)

∆H298

o = 2 mol∆Hf ,lactic acid(s)

oéë

ùû

- 1mol∆Hf ,glucose(s)

oéë

ùû

= 2mol (-694.0)kJ

mol

é

ëê

ù

ûú - 1mol (-1274.5)

kJ

mol

é

ëê

ù

ûú = -113.1kJ

B) Assuming Cp’s are constant 25°C → 37°C:

∆Hr

o(37°C) = ∆Hr

o(25°C) + 2mol C p,lactic acid(s)( ) -1molCp,glucose(s)

éëê

ùûú∆T

= (-113.5kJ) + 2mol (127.6)J

mol - K-1mol (219.2)

J

mol - K

é

ëê

ù

ûú ´ (12)K ´

1.0kJ

1000 J= -113.1kJ

C) The ∆Hr is not very sensitive to this change in T, since the T difference is very small and ∆Cp is also

very small on the kilojoule scale.

2.44 A) Applying Hess’s Law:

∆H298

o = 2 mol∆Hf ,NH

3(g)

oéëê

ùûú

- 1mol∆Hf ,N

2(g)

o + 3mol∆Hf ,H

2(g)

oéëê

ùûú

= 2mol (-46.11)kJ

mol

é

ëê

ù

ûú - 0 + 0é

ëùû

= -92.22kJ

B) ∆Hr

o(425°C) = ∆Hr

o(25°C) + 2mol C p,NH3(g)( ) - 1molC

p,N2(g)

+ 3molCp,H

2(g)( )é

ëêùûú∆T

∆Hr

o(425°C) = (-92.22kJ) + 2mol(33.06J

mol - K)

æ

èç

ö

ø÷ - 1mol (29.13

J

mol - K) + 3mol(28.82

J

mol - K)

æ

èç

ö

ø÷

é

ëêê

ù

ûúú(400K)

= (-92.22kJ) + (-49.98J

K) ´ 400K ´

1.0kJ

1000J

é

ëê

ù

ûú = (-92.22 -19.79)kJ = -111.9kJ = ∆H

r

o(425°C)


27

• So for this reaction the ∆H for the reaction changes significantly with the T change, but then ∆Cp as well

as ∆T is much larger than in previous examples.

2.45 A) The pathway would need to something like:

B) Method in (A):

∆Hvap

o (298K) = Cp,CH

3OH(l)

(337.7 - 298)K + ∆Hvap

o (337.7K) + Cp,CH

3OH(g)

(298 - 337.7)K

= 1mol (81.1J

mol - K)(39.7K) ´

1.0kJ

1000 J

é

ëê

ù

ûú + 35.27kJ + 1mol (44.1

J

mol - K)(-39.7K) ´

1.0kJ

1000 J

é

ëê

ù

ûú

= 3.248kJ + 35.27kJ + (-1.750)kJ = 36.77kJ

Hess’s law: ∆Hvap,298

o = ∆Hr

o = 1mol∆Hf ,CH

3OH(g)

oéëê

ùûú

- 1mol∆Hf ,CH

3OH(l)

oéëê

ùûú

= -201.0éë

ùû

- -239.2éë

ùû

= 38.2kJ

Since the tabled value is 37.99 kJ/mol, the method using the heats of formation of CH3OH in the two

states produces the better result. But the cyclic method produces a value that is only off by about 1.0 kJ

(an error of about 3.2%), so it is reasonably close.

2.46 • For alkanes, only need to find “n” the number of moles of carbon that are in one mole of the

compound to determine the molecular formula.

Based on one mole reacting the combustion reaction would have been:

CnH

2n+2+

3n +1

2O

2(g)® nCO

2(g) + (n +1)H

2O(g)

• Kirchhoff’s law ∆Hr

o(T2) = ∆H

r

o(T1) + ∆C

p T1

T2

ò dT can be applied where n can be determined from ∆Cp since

∆Hr is known at two different T’s

∆Hr

o(T2) = ∆H

r

o(T1) + ∆C

p T1

T2

ò dT Þ ∆Hr

o(T2) = ∆H

r

o(T1) + ∆C

p∆T Þ ∆C

p=∆H

r

o(T2) - ∆H

r

o(T1)

∆T

For :aA + bB ® cC + dD then∆Cp

= cCp,C

+ dCp,D( ) - aC

p,A+ bC

p,B( )éë

ùû

Then for this reaction:

∆Cp

= nCp,CO

2(g)

+ (n +1)Cp,H

2O(g)( ) - C

p,CnH

2n+2

+ (1.5n + 0.5)Cp,B( ) = 37.11n + 33.58n + 33.58( ) - 120.2 + 44.04n +14.68( )

∆Cp

= 26.66n -101.3

∆Cp

=∆H

r

o(T2) - ∆H

r

o(T1)

∆TÞ 26.66n -101.3( ) J

mol - K=

(-3532.8 + 3536.1)

(400 - 298)KkJ ´

1000J

1kJ= 32.35

26.66n -101.3( ) = 32.35 Þ n =133.65

26.66= 5.01 so C

5H

12= alkane

∆Hvap

o (298K) = q1

+ ∆Hvap

o (337.7K) + q3

Given: (CRC Handbook)

Cp, CH3OH(l) = 81.1 J/mol-K

Cp, CH3OH(l) = 44.1 J/mol-K

Cool CH3OH(g), q3 T2 à T1

∆Hvap° q2

∆Hvap at 298K CH3OH(l) à CH3OH(g)

∆Hvap at 337.7 CH3OH(l) à CH3OH(g)

Heat CH3OH(l), q4 T1 à T2


28

PART 3: SECOND AND THIRD LAWS - ENTROPY

3.1 Answer: ∆Svap

=∆H

vap

o

Tbp

∆Svap

,CCl4

=30.0 kJ

mol

349.7K´

1000J

1kJ

æ

èç

ö

ø÷ = 98.8

J

mol - K ∆S

vap,CH

4=

8.18 kJ

mol

111.5K´

1000J

1kJ

æ

èç

ö

ø÷ = 73.4

J

mol - K

∆Svap

,H2S =

18.7 kJ

mol

212.6K´

1000J

1kJ

æ

èç

ö

ø÷ = 87.9

J

mol - K ∆S

vap,H

2O =

40.66 kJ

mol

373K´

1000J

1kJ

æ

èç

ö

ø÷ = 109

J

mol - K

B) The values for ∆Svap for CCl4 and H2S are very similar, even though their melting points are very

different. When boiling, all the attractive intermolecular forces have to be overcome to create the random

gaseous array. The forces in solid CCl4 would be London Dispersion forces (LDF), while H2S would have

dipolar intermolecular forces which are generally stronger attractive forces than LDF. CH4 would also only

have LDF like CCl4, but shows that much less order is destroyed when CH4 is converted to a gas. The

higher molar mass of CCl4 would increase the strength of the LDF forces, due to increased contact area

due to greater size.

C) Water has H-bonding forces that produce a very ordered system in liquid water. Therefore, the

entropy changes much more when all the H-bonding interactions are severed to produce the gas state.

3.2 A) ∆Sfus

,CH3C

6H

5=∆H

fus

Tfp

=6.64X103 J

mol

178K= 37.3

J

mol - K ∆S

fus,C

6H

6=

9.90X103 J

mol

278.8K= 35.5

J

mol - K

So the value of the molar ∆Sfus is slightly higher for toluene than benzene.

B) ∆Svap

,CH3C

6H

5=∆H

vap

Tbp

=

361.9J

g

92.0gCH3C

6H

5

1mol

é

ë

êê

ù

û

úú

384K= 86.7

J

mol - K

∆Svap

,C6H

6=

394J

g

78.0gC6H

6

1mol

é

ë

êê

ù

û

úú

353.2K= 87.2

J

mol - K

3.3 • We know that in order for the body to keep T constant it must lose heat to the surroundings and

that qlost by body = q gain by surroundings.

A) ∆Ssurr

=q

surr

Tsurr

=-(q

lost by body)

Tsurr

qsurr

= -(qlost by body

) = 7.20kJ

kg - hr80.0kg( ) 24.0hr( ) = 13,824kJ

So that: ∆Ssurr

=q

surr

Tsurr

=13,874kJ

293K= 47.2

J

K

B) There will be no observable ∆T in the surroundings so can’t apply ∆S =C

p∆T

T .

3.4 A)∆Sv

=

T1

T2

òC

v

TdT =C

v

T1

T2

òdT

TÞ ∆S = C

vln

T2

T1

assuming Cv constant T

1® T

2

∆SV

= Cvln

T2

T1

= 1.0mol(12.48 J

mol-K)ln

500K

300K

æ

èç

ö

ø÷ = (12.48 J

K)(0.5108) = 6.375 J / K

B) ∆SP

=

T1

T2

òC

p

TdT =C

p

T1

T2

òdT

TÞ ∆S = C

pln

T2

T1

assuming Cp constant T

1® T

2

∆Sp

= Cpln

T2

T1

= 1.0mol(20.79 J

mol-K)ln

500K

300K

æ

èç

ö

ø÷ = (20.79 J

K)(0.5108) = 10.62 J / K

The value of molar ∆Svap is

slightly higher for benzene

than toluene, the opposite

of ∆Sfus. But in both cases

the values are very similar

values.


29

C) Since qV ≠ qP then ∆S constant V≠ ∆S constant P for the same temperature change for gases.

3.5 A) ∆S = nCPln

T2

T1

= 1.0mol(37.14 J

mol-K)ln

1000K

298K

é

ëê

ù

ûú = 37.14(1.211) = 45.0 J / K

B) ∆S =

T1

T2

òC

p

TdT =

T1

T2

òa + bT + cT 2

TdT = a

T1

T2

òdT

T+ b

T1

T2

òdT + cT1

T2

òTdT = alnT

2

T1

+ b(T2

- T1) +

c

2T

2

2 - T1

2( ) so that:

∆S = 18.86 ln1000

298

é

ëê

ù

ûú + 0.07937(702) +

-6.783X10-5

2

é

ëêê

ù

ûúú

1.00X106 - 8.88X104( ) +2.443X10-6

3

é

ëêê

ù

ûúú

1.00X109 - 2.65X107( )

= 22.83 + 55.72 - 30.90 + 9.55éë

ùû

= 57.2 J / K

C) We would expect that the value from the integration of Cp is more accurate.

3.6 A) ∆S(A)

= Cp,solid

lnT

mp

T1

= 54.44J

mol - Kln

394.8K

303K

é

ëê

ù

ûú = 14.6

J

mol - K

∆S(B)

=∆H

fus

o

Tmp

=15.57 kJ

mol

391.8K´

1000 J

1kJ= 39.6

J

K - mol

∆S(C)

= Cp,solid

lnT

mp

T1

= 80.33J

mol - Kln

458.4K

391.8K

é

ëê

ù

ûú = 12.6

J

mol - K

∆S(D)

=∆H

Vap

o

Tbp

=41.8 kJ

mol

458.4K´

1000 J

1kJ= 91.2

J

K - mol

∆S(E)

= Cp,gas

lnT

4

Tbp

= 36.9J

mol - Kln

473.1K

394.8K

é

ëê

ù

ûú = 1.2

J

mol - K

B)∆Stotal

= 14.0 + 39.6 +12.6 + 91.2 +1.2( ) J

mol - K= 159

J

mol - K

C) ∆S(C) = ∆Svap = 55.2% of total

3.7 A) ∆SC(s)

=

T1

T2

òC

p

TdT =

T1

T2

òa + bT + cT 2

TdT = a

T1

T2

òdT

T+ b

T1

T2

òdT + cT1

T2

òTdT = alnT

2

T1

+ b(T2

- T1) +

c

2T

2

2 - T1

2( )

∆SC(s)

= -0.4493J

mol-Kln

873K

298K

é

ëê

ù

ûú + 0.03553

J

mol-K2(873 - 298)K +1.3X10-5 J

mol-K3(7.62X105 - 8.88X104)K2

= -0.4493(1.075) + 0.03553(575) + 6.5X10-5(6.73X105)éë

ùû

J

mol - K= 24.3

J

mol - K= ∆S

C(s)25°C ® 600°C

B)

∆SFe(s)

=

T1

T2

òC

p

TdT =

T1

T2

òa + bT + cT 2 + dT 3

TdT = a

T1

T2

òdT

T+ b

T1

T2

òdT + cT1

T2

òTdT + dT1

T2

òT2dT

= alnT

2

T1

+ b(T2

- T1) +

c

2T

2

2 - T1

2( ) +d

3T

2

3 - T1

3( )

∆SFe(s)

= 18.429(1.075) + 0.02464(575) + -4.6X10-6(6.73X105) + 3.11X10-9(6.39X108)

= 19.81 +14.17 - 3.00 +1.98( ) J

K - mol= 33.0

J

K - mol= ∆S

Fe(s)25°C ® 600°C

• The value of ∆S for Fe(s) in the same temperature range is greater than that of C(s), indicating the

atoms have more freedom of choice for position in Fe(s) than C(s).

T1

T

T2

Heat added à


30

C) If Cp for Fe taken as constant, then ∆SFe(s)

= Cp,solid

ln873

298= 25.10

J

mol - K(1.075) = 27.0

J

mol - K which is

significantly lower than value calculated from the integrated function, so it is necessary to do the

integration.

3.8 Apply ∆Sdiss

= 2SX(g)

o - SX

2(g)

oéëê

ùûú

Even though the individual S° values are very different, the entropy change that occurs when the

diatomic gaseous molecules dissociate into 2 gaseous atoms are very similar. This indicates that the

chemical properties are not as important as the fact that you get 2 gaseous atoms for every one molecule

in each case.

3.9 • Will need to know the number of moles in sample, so use the ideal gas law to calculate:

nAr

=1.00atm(0.500L)

0.08206 L-atm

mol-K(298K)

æ

èç

ö

ø÷ = 0.02045mol

Two reversible steps:

(1) First let the volume change at a constant T, calculate ∆S for volume change

∆S(1)

= nRlnV

2

V1

= 0.02045mol(8.314J

mol - K)ln

1.0L

0.50L

é

ëê

ù

ûú = 0.1700(0.693) = 0.118

J

K

(2) Then let T change - Heat gas at V2 from T1 → T2

∆S(2)

= nCvln

T2

T1

= 0.02045mol(12.65J

mol - K)ln

373K

298K

é

ëê

ù

ûú = 0.2587(0.2245) = 0.0581

J

K

∆Stotal

= ∆S(1)

+ ∆S(2)

= 0.118 + 0.0581 = 0.176J

K

3.10 A) (a)

(b) wrev

= -nRT lnV

2

V1

= -0.50mol ´ 8.314 J

mol-K(298K)ln 2.5é

ëùû

= -1135J = -1.14kJ

(or could use: wrev

= -T ∆S = 298K 3.81 J

K( ) = -1135kJ = -1.14kJ ).

(c) Since reversible and isothermal, ∆U = 0, work compensated by heat flow from surroundings, so that:

qsurroundings

= -wrev

and ∆Ssurr

=q

surr

T= -

T ∆Sgas

T= -∆S

gas= -3.81

J

K

(d) ∆Stotal

= ∆Ssurroundings

+ ∆Sgas

= 3.81- 3.81J

K= 0.0

B) (a) ∆Sgas = 3.81 J/K stays the same since only V2, V1 important, not how the expansion occurred.

(Since ∆S is a state function, it is independent of pathway.)

(b) Work changes to zero since wirrev

= -PdV = (0)dV = 0.0

(c) Since no work done, then ∆Ssurroundings

=q

surr

T=

-wirrev

T= -

0

T= 0.0

J

K

(d) ∆Stotal

= ∆Ssurr

+ ∆Sgas

= 3.81 - 0J

K= 3.81

J

K

3.11 • Have both a ∆S due to change in P and change in T, so break process down into two reversible

steps:

A) Break down change into two reversible steps:

∆Sgas

= nRlnV

2

V1

= 0.050mol (8.314 J

mol-K)ln

25.0L

10.0L

é

ëê

ù

ûú = 4.157

J

K(0.916) = 3.81

J

K

Element S°X(g) S°X2(g) ∆S°diss

Br 175 245.5 104.5

Cl 165.2 223.1 107

I 180.8 260.7 101

H 114.7 130.7 98.7


31

(1) First let the pressure change for 1 breath at a constant T, calculate ∆S for pressure change per 1.0

breath:

Assuming air is an ideal gas: P

1V

1

nT 1

=P

2V

2

nT 2

®P

1

P2

=V

2

V1

so that ∆S = nRlnV

2

V1

= nRlnP1

P2

∆S(∆P)

= nRlnP

1

P2

= 0.0208mol(8.314 J

mol-K)ln

760torr

756torr

é

ëê

ù

ûú = 0.1729(0.00528) = 9.124X10-4 J

K

(2) Then let the temperature change for 1.0 breath from T1 → T2 at a constant P:

∆S(∆T )

= nCvln

T2

T1

= 0.0208mol(29.05 J

mol-K)ln

310K

298K

é

ëê

ù

ûú = 0.0208(0.0564) = 0.0341

J

Kper breath

∆Stotal

= ∆S(1)

+ ∆S(2)

= (0.03411+ 9.12X10-4)J

K= 0.0350

J

K per breath

B) Taking into account, there is a breath every 3 seconds:

∆Sperday

=0.0350 J / K

1breath

é

ëê

ù

ûú ´

1breath

3.0sec

é

ëê

ù

ûú ´

3600sec

1hr

é

ëê

ù

ûú ´

24hr

1day

é

ëê

ù

ûú = 1008

J

K= 1.01

kJ

K

C) The T change in (2). The pressure change contributes very little to the ∆Stotal- only about 2.6%.

3.12 • Need to determine the number of moles of each component

A) no.molCH4

= 10.0g1.00mol

16.0g

é

ëê

ù

ûú = 0.625mol no.molC

2H

6= 100.0g

1.00mol

30.0g

é

ëê

ù

ûú = 3.33mol

cCH

4

=0.625

3.96= 0.158 lnc

CH4

= ln(0.158) = -1.153 cC

2H

6

= 1.0 - 0.158 = 0.842 lncC

2H

6

= ln(0.842) = -0.573

∆Smix

= -R n1lnc

1+ n

2lnc

2éë

ùû

= -8.314J

mol - K0.625mol(-1.153) + 3.33mol(0.573)éë

ùû

= 14.35J

K

B) Changing the volume will not affect the ∆Smix since the number of moles of each component doesn’t

change.

3.13 • Need to determine the number of moles of each component, to get number of moles total.

• Can use the volume ratio to define the molar ratio of each component and use.

Assuming gases act as ideal gases:PV

1

n1T

=PV

total

ntotal

T®

n1

ntotal

=V

1

Vtotal

∆Smix

=∆S

mix

ntotal

= -R c1lnc

1+ c

2lnc

2éë

ùû

A) cCH

4

=10.0L

110L= 0.09091 lnc

CH4

= ln(0.09091) = -2.398 cC

2H

6

= 0.9099 lncC

2H

6

= ln(0.9099) = -0.0953

nCH

4

=1.00bar(10.0L)

0.08314 L-atm

mol-K(293K)

æ

èç

ö

ø÷ = 0.4105mol

nC

2H

6

=1.00bar(100.0L)

0.08314 L-atm

mol-K(293K)

æ

èç

ö

ø÷ = 4.105molC

2H

6 n

total= 0.4105 + 4.015 = 4.515mol

∆Smix

= ∆Smix

´ ntotal

= -R c1lnc

1+ c

2lnc

2éë

ùû

= -8.314 J

mol-K0.09091(-2.398) + 0.9099(-0.0953)éë

ùû

= 2.529 J

mol-K´ 4.515mol = 11.42 J / K

B) The total ∆S mix is lower in this example because the mol fraction of C2H6(g) increases while that for

CH4 gets smaller. Adding a small amount of the second gas to a large amount of the first produces lower

values of ∆Smix.

3.14 A) Ptotal

= PO

2

+ PAr

+ PN

2

= (nO

2

+ nAr

+ nN

2

)RT

Vmix

nO

2

= 100.0g1.00mol

32.0g

é

ëê

ù

ûú = 3.125mol

nAr

= 100.0g1.00mol

39.95g

é

ëê

ù

ûú = 2.503mol n

N2

= 50.0g1.00mol

28.0g

é

ëê

ù

ûú = 1.785mol


32

Ptotal

= (3.125 + 2.503 +1.785)RT

Vmix

=

7.413mol(0.08206L - atm

mol - K)(293K)

20.0L= 8.91atm

B) It is reasonable to assume the gases are acting ideally since the pressure of the mixture is not high

enough to produce Z values much different than 1.0.

C) cO

2

=3.125mol

7.413mol= 0.4216 lnc

O2

= ln(0.4216) = -0.8638

cAr

=2.503mol

7.413mol= 0.3377 lnc

Ar= ln(0.3377) = -1.086

cN

2

= 1.000 - (0.3377 + 0.4216) = 0.2407 lncN

2

= ln(0.2407) = -1.424

∆Smix

= -R n1lnc

1+ n

2lnc

2+ n

3lnc

3éë

ùû

= -8.314 J

mol-K3.125mol(-0.8638) + 2.503mol(-1.086) +1.785mol(-1.424)éë

ùû

= +8.314J

K2.699 + 2.718 + 2.542éë

ùû

= 66.17J

K

D) If there are attractive or repulsive interactions among the gas particles, the volumes the gases occupy

at constant P (or pressures, if V constant) will be either lower or higher than expected. This should change

the volume (or pressure) ratio for before and after mixing, so ∆Smix should be affected, but probably only

slightly.

3.15 A) nHe(g)

= 100.0g1.00mol

4.0g

é

ëê

ù

ûú = 25.0molHe n

Ne(g)= 100.0g

1.00mol

20.2g

é

ëê

ù

ûú = 4.95mol Ne

nAr

= 100.0g1.00mol

39.95g

é

ëê

ù

ûú = 2.50mol Ar n

Xe(g)= 100.0g

1.00mol

131.5g

é

ëê

ù

ûú = 0.762mol Xe

ntotal

= (25.00 + 4.95 + 2.50 + 0.762) = 33.26mol

cHe

=25.00mol

33.26mol= 0.752 lnc

He= ln(0.752) = -0.285 c

Ne=

4.95mol

33.26mol= 0.149 lnc

Ne= ln(0.149) = -1.905

cAr

=2.50mol

33.26mol= 0.0752 lnc

Ar= ln(0.0752) = -2.59 c

Xe=

0.762mol

33.26mol= 0.0228 lnc

Xe= ln(0.0228) = -3.78

∆Smix

= ∆Smix

´ ntotal

= -R cHe

lncHe

+ cNe

lncNe

+ cAr

lncAr

+ cXe

lncXe

éë

ùû

= -8.314 J

mol-K´ (33.26mol) 0.752(-0.285) + 0.149(-1.905) + 0.0752(-2.598) + 0.0228(-3.78)é

ëùû

= -276.5J

K´ -0.779 = 215 J / K

B) cHe

= cNe

= cAr

= cXe

=1.00mol

4.00mol= 0.250 lnc

He= ln(0.250) = -1.386

∆Smix

= -R cHe

lncHe

+ cNe

lncNe

+ cAr

lncAr

+ cXe

lncXe

éë

ùû

= -8.314 J

mol-K0.250(-1.386) ´ 4éë

ùû

= -8.314 J

mol-K(-1.386) = 11.52 J

mol-K

Versus ∆Smix

(A) =215

33.26

J

mol-K= 6.46 J

mol-K

So even though the number of moles much smaller, the change in entropy per mol has nearly doubled in

(B) versus (A). So an important factor in the entropy of mixing is the mol fraction of each component.

Analysis: Know three things will happen the “hot” O2(g) mixed with cold Ar(g) that will affect ∆S:

(1) Gases will mix with a positive change in entropy that can be determined using ∆Smix equation. To

calculate ∆Smix will need no. moles each gas, and their mol fractions,

∆Smix

= -R nO

2

lncO

2

+ nAr

lncAr

éëê

ùûú


33

(2) The hotter gas (O2) will transfer heat to the cooler (Ar) gas until the T’s becomes the same. This will

produce an additional positive entropy change for Ar(g) and a negative change for O2(g). You can

determine the ∆S from the temperature change for each gas. To calculate ∆S for each due to T change

must first determine Tfinal for each gas using a –qlost = qgain approach and then use: ∆SO

2

= nCpO

2

lnT

final

Tini,O

2

and ∆SAr

= nCp Ar

lnT

final

Tini,Ar

. (Will need to look up for each gas.)

(3) As the O2 gas changes temperature it will change its volume, as will Ar(g), and the volume changes

will also affect entropy total for each gas. To calculate ∆S for each due to volume change, use:

∆S = nRlnV

2

V1

The three separate entropies can then be added to get the overall entropy change, since the entropy is a

state function, independent of how we get from the initial state to the final state.

Calculation route:

(1) nO

2

= 100.0g1.00mol

32.0g

é

ëê

ù

ûú = 3.125mol n

Ar= 100.0g

1.00mol

39.95g

é

ëê

ù

ûú = 2.503mol

cO

2

=3.125mol

5.628mol= 0.5553 lnc

O2

= ln(0.5553) = -0.5883

cAr

= 1- 0.5553 = 0.4447 lncAr

= ln(0.4447) = -0.8104

∆Smix

(1) = -R nO

2

lncO

2

+ nAr

lncAr

éëê

ùûú

= -8.314 J

mol-K3.125mol(-0.5883) + 2.503mol(-0.8104)éë

ùû

= 32.15 J / K

(2) -qlost byO

2

= qgainby Ar

→ - nO

2

C pO2(T

final- T

ini,O2

)éëê

ùûú

= nAr

C p Ar(Tfinal

- Tini,Ar

)éëê

ùûú

- 3.125mol (29.355J

mol - K(°C))(T

f- 50°C)

é

ëê

ù

ûú = 2.503mol (20.786

J

mol - K(°C))(T

final- 20°C)

é

ëê

ù

ûú

- 91.73Tfinal

+ 4586.7 = 52.03Tfinal

-1040.6

5627.3 = (143.8)Tfinal

Þ Tfinal

= 39.1°C

∆SO

2

= nCpO

2

lnT

final

Tini,O

2

= 3.125mol (29.355 J

mol-K(°C))ln

312.1K

323K

é

ëê

ù

ûú = 91.73 J

K(-0.0342) = -3.14 J / K

∆SAr

= nCp Ar

lnT

f

Ti,Ar

= 2.503mol (20.786 J

mol-K(°C))ln

312.1K

293K

é

ëê

ù

ûú = 52.03 J

K(0.06315) = 3.29 J / K

∆S∆T

(2) = -3.14 + 3.29( ) J / K = +0.15 J / K

(3) Considering gases to be ideal: P V

i

nTi

=P V

f

nTf

®T

f

Ti

=V

f

Vi

then ∆S = nRlnV

final

Vini

= nRlnT

final

Tini

∆SV ,O

2

= nRlnT

final

Tini

= 3.125mol 8.314 J

mol-K( )ln 312K

323K

æ

èç

ö

ø÷ = 25.98 J

K(-0.03465) = -0.900 J / K

∆SV ,Ar

= nArRln

Tfinal

Tini

= 2.503mol 8.314 J

mol-K( )ln 312K

293K

æ

èç

ö

ø÷ = 20.810 J

K(0.06283) = 1.31 J / K

∆SV(3) = -0.900 +1.31( ) J / K = +0.41J / K

∆Soveral l

= ∆Smix

(1) + ∆S∆T

(2) + ∆S∆V

(3) = (32.17 + 0.15 + 0.41) J / K = 32.73 J / K

• In an adiabatic reversible expansion or contraction of an ideal gas, the entropy change for the volume

change is offset by that for the T change, so that ∆S=0. If the T change and volume changes occur

irreversibly, then ∆S≈0. That is definitely the case here, with the ∆Soverall ≈ ∆Smix.

3.17 A) Diagram step-wise pathway:


34

B) Calculation route: ∆Sfus,263K

= nCpH

2O(l)

lnT

1

T2

+ ∆Sfus,273K

+ nCpH

2O(s)

lnT

2

T1

∆Sfus,263K

= 1mol 75.3 J

mol-Kéë

ùûln

273

263

é

ëê

ù

ûú +

∆Hfus,H

2O

273K

é

ë

êê

ù

û

úú

+ 1mol 37.4 J

mol-Kéë

ùûln

263

273

é

ëê

ù

ûú = (2.81 + 22.0 -1.395) J / K = 23.4 J / K

C) The difference in molar heat capacity, ∆Cp is [Cp, final state (or phase) – Cp, initial state (or phase)] of

the substance, which in the diagram in (A) is ∆Cp

= Cp,H

2O(s) - C

p,H

2O(l)é

ëùû.

3.18

A) Diagrams and resulting equations:

∆Hden

(T2) = q

1+ ∆H

den(T

m) + q

3

= Cp,native

(Tm

- T2) + ∆H

den(T

m) + C

p,den(T

2- T

m)

= ∆Hden

(Tm) + C

p,den- C

p,nativeéë

ùû(T

2- T

m)

= ∆Hden

(Tm) + ∆C

p(T

2- T

m)

∆Sden,T

2

= Cp,native

lnT

m

T2

+ ∆Sden,T

m

+ Cp,denatured

lnT

2

Tm

= ∆Sden,T

m

+ Cp,denatured

- C p,native

éëê

ùûúln

T2

Tm

= ∆Sden,T

m

+ ∆Cpln

T2

Tm

B) (a) ∆Sden,340K

=∆H

denaturation

Tm

=640 kJ

mol

340K= 1.882 kJ

K-mol

(b) ∆Hden

(T2) = ∆H

den(T

m) + ∆C

p(T

2- T

m) = 640.1 kJ

mol+ 8.37 kJ

K-mol(310 - 340)K = 389 kJ

mol

∆Sden,310

= ∆Sden,T

m

+ ∆Cpln

T2

Tm

= 1.882 kJ

K-mol+ 8.37 kJ

K-molln

310K

340K

é

ëê

ù

ûú = 1.882 kJ

K-mol+ 8.37 kJ

K-mol(-0.0924) = 1.11 kJ

K-mol

3.19 A) ∆Sden,319K

=∆H

denaturation

Tm

=382 kJ

mol

319K= 1.20 kJ

K-mol

B) ∆S per mol unit = 1.20X 103 J / K

122units= 9.84 J

K-unit

C) ∆SfusH

2O =

6.00X103J

273K= 22.0 J / K ∆S

fusC6H

6

= 10.59X103J

354K= 53.0 J / K

∆SfusCCl

4

= 2.5X103J

250.3K= 9.99 J / K ∆S

fusCH4

= 941J

90.68K= 10.4 J / K

So the ∆Sden per amino acid unit when the protein unfolds is more on the scale of ∆Sfus for CCl4(s) or

CH4(s) where weaker LDF forces maintain the solid structure versus the stronger aromatic forces in

C6H6(s) or H-bonding network in H2O(s).

3.20 A) • Must calculate ∆S at the Tm before you can determine ∆S at a different T.

∆Sden,298K

=∆H

den

Tm

+ ∆Cpln

298K

Tm

=509

348.5

kJ

K-mol+ 6.28

kJ

K-molln

298K

348.5K

é

ëê

ù

ûú

= 1.460 kJ

K-mol+ 6.28 kJ

K-mol(-0.1565) = 1.460 - 0.983( ) kJ

K-mol= 0.477kJ / mol - K


35

B) The ∆S values are very strongly affected by temperature.

C) The ∆S values for the denaturing of the proteins is in kilojoules per mol, while most phase transitions

values are quoted in joules per mol. So the changes for proteins are a thousand times larger than those

for small molecules. This is not surprising since the unfolding involves hundreds of atoms changing

positions. Once put on a per monomer unit scale, as in the previous problem, the values become

joule/mol on the same scale as other “fusion” transitions.

3.21 A) Expect ∆S to be negative, since reaction converts 12 moles of gases (highly disordered state)

to 6 moles of gas and a solid (low disorder state) .

∆Sr

o = 1mol SC

6H

12O

6(s)

o + 6mol SO

2(g)

oéëê

ùûú

- 6mol SCO

2(g)

o + 6mol SH

2O(g)

oéëê

ùûú

= 1mol (209.2) J

mol-K+ 6mol (205.2) J

mol-Kéë

ùû

- 6mol (213.5) J

mol-K+ 6mol(188.8) J

mol-Kéë

ùû

= 1440.4 - 2415éë

ùû

J

K= -974.6 J / K

• So the reaction ∆S° is negative as expected and very large, indicating a major shift in the number of

microstates allowed each molecule.

B) Expect ∆S to be zero, or close to it, since reaction converts 2 moles of gases to 2 moles of gas

maintaining about the same level of disorder.

∆Sr

o = 2mol SNO(g)

oéë

ùû

- 1mol SN

2(g)

o +1mol SO

2(g)

oéëê

ùûú

= 2mol (210.8) J

mol-Kéë

ùû

- 1mol (191.5) J

mol-K+1mol(205.1) J

mol-Kéë

ùû

= 421.5 - 396.6éë

ùû

J

K= +24.9 J / K

• So although not zero, the ∆Sr° is small. A typical reaction ∆Sr° are in hundreds of joules per K.

3.22 A) Expect it will be positive and large since forming 3 moles of gases (high entropy state) from a

solid and liquid (low entropy states). The sign of the ∆H cannot be predicted as easily because ∆H

depends on the changes in chemical bond energies, as well as intermolecular interactions. These would

be very difficult to estimate without specific values. ∆S in contrast depends on the physical states of the

substances and number of their possible microstates, which are not as dependent on the interactions of

the atoms.

B) ∆H

298

o = 2 mol∆Hf ,NH

3(g)

o +1mol∆Hf ,CO

2(g)

oéëê

ùûú

- 1mol∆Hf ,(NH

2)2CO(s)

o +1mol∆Hf ,H

2O(l)

oéëê

ùûú

= 2mol (-46.11) kJ

mol+1mol (-393.5) kJ

moléë

ùû

- 1mol(-333.2) kJ

mol+1mol (-285.8) kJ

moléë

ùû

= 133.3kJ

∆S

r

o = 2mol SNH

3(g)

o +1mol SCO

2(g)

oéëê

ùûú

- 1mol S(NH

2)2CO(s)

o +1mol SH

2O(l)

oéëê

ùûú

= 2mol (192.4) J

mol-K+1mol (231.7) J

mol-Kéë

ùû

- 1mol(104.6) J

mol-K+1mol (69.9) J

mol-Kéë

ùû

= +424 J / K

3.23 A) Balanced reaction: C4H8N2O3(s) + 3 O2(g) → (NH2)2CO(s) + 3 CO2(g) + 2 H2O(l)

B) ∆H298

o = 1 mol∆Hf ,(NH

2)2CO(s)

o + 3mol∆Hf ,CO

2(g)

o + 2mol∆Hf ,H

2O(l)

oéëê

ùûú

- 1mol∆Hf ,glygly(s)

o + 3mol∆Hf ,O

2(g)

oéëê

ùûú

∆H298

o = 1mol(333.2) kJ

mol+ 3mol(-393.5) kJ

mol+ 2mol(-285.8) kJ

moléë

ùû

- 1mol(-745.25) kJ

moléë

ùû

= -673.6kJ

∆Sr

o = 1mol S(NH

2)2CO(s)

o + 3mol SCO

2(g)

o + 2mol SH

2O(l)

oéëê

ùûú

- 1mol Sglycylglycine(s)

o + 3mol SO

2(g)

oéëê

ùûú

∆Sr

o = 1mol (104.6) J

mol-K+ 3mol (213.7) J

mol-K+ 2mol (69.9) J

mol-Kéë

ùû

- 1mol(189) J

mol-K+ 3mol (205.1) J

mol-Kéë

ùû

= 80.3 J / KC)

At 330K, given ∆Cp

=

products

å Cp

-

reactants

å Cp and ∆S

r

o(T2) = ∆S

r

o(T1) + ∆C

pln

T2

T1

æ

èç

ö

ø÷


36

For ∆T = T2

- T1

∆Hr

o(T2) = ∆H

r

o(T1) + ∆C

p´ ∆T

∆Cp

= 1molCp,urea

+ 3molCp,CO

2(g)

+ 2molCp,H

2O(l)

éëê

ùûú

- 1molCp,glycylglycine

+ 3molCp,O

2(g)

éëê

ùûú

∆Cp

= 1mol(92.8) + 3mol(37.1) + 2mol(75.3)éë

ùû

J

mol-K- 1mol(163.6) + 3mol(29.36)éë

ùû

J

mol-K= 103 J / K

∆H

r

o(330K) = ∆Hr

o(298K) + ∆Cp´ ∆T = -673.6kJ + 103

J

K´

1kJ

1000J´ (330 - 298)K

æ

èç

ö

ø÷

= -673.6 + 3.30 = -670.3kJ

∆Sr

o(330K) = ∆Sr

o(298K) + ∆Cpln

330

298= 80.3

J

K+ 103

J

K(0.103)

æ

èç

ö

ø÷ = 90.8

J

K

• So ∆H° is not changed significantly by the increase to 330K, but the ∆Sr° for the reaction shows a 13%

increase.

3.24 A) Reaction I:

Reaction II:

B) Reaction I:

∆Cp(I) = 2molC

p,lactic acid(s)éë

ùû

- 1molCp,C

6H

12O

6(s)

éëê

ùûú

= 2mol(127.6) J

mol-Kéë

ùû

- 1mol(219.2) J

mol-Kéë

ùû

= 36.0 J / K

∆Sr

o(I,310K) = ∆Sr

o(T1) + ∆C

pln

T2

T1

æ

èç

ö

ø÷ = 175 J

K+ 36.0 J

Kln

310K

298K

æ

èç

ö

ø÷ = 175 J

K+ 36.0 J

K(0.0392) = 176.4 J / K

Reaction II: ∆C

p(II) = 2molC

p,CO2(g)

+ 2molCp,C

2H

5OH(l)

éëê

ùûú

- 1molCp,C

6H

12O

6(s)

éëê

ùûú

= 2mol(37.14 +112.3) J

mol-Kéë

ùû

- 1mol (219.2) J

mol-Kéë

ùû

= 79.7 J / K

∆Sr

o(II,310K) = ∆Sr

o(T1) + ∆C

pln

T2

T1

æ

èç

ö

ø÷ = 538.4 J

K+ 79.7 J

Kln

310K

298K

æ

èç

ö

ø÷

= 538.4 J

K+ 79.7 J

K(0.0392) = 541.5 J / K

C) Probably the biggest factor resulting in a minor change in ∆S from the 298K value is the difference in

states of the reactants and products in the two reactions, rather than the T change. Reactions I and II

involve only solids and liquids, whereas the reaction in the previous problem involved gases as well.

3.25 Given:∆Sr

= -62.4J

K= 1mol S

HCl(aq)

o +1mol SAg(s)

oéë

ùû

- 1mol SAgCl(s)

o +1mol Sf ,H

2(g)

oéëê

ùûú

SHCl(aq)

o =-62.4 J

K+ 1mol S

AgCl(s)

o +1 /2mol Sf ,H

2(g)

oéëê

ùûú

-1molSAg(s)

o

1 mol=

=-62.4 J

K+ 96.2 + 0.5(130.7)éë

ùû

J

K- 42.55 J

K

1 mol=

-62.4 J

K+119 J

K

1mol= 56.6

J

K - mol

3.26 Summary:

∆Sr

o(I) = 2mol Slactic acid(s)

oéë

ùû

- 1mol SC

6H

12O

6(s)

oéëê

ùûú

= 2mol (192.1) J

mol-Kéë

ùû

- 1mol (209.2) J

mol-Kéë

ùû

= 384.2 - 209.2éë

ùû

J

K= +175 J / K

∆Sr

o(II) = 2mol SC

2H

5OH(l)

o + 2mol SCO

2(g)

oé

ëêù

ûú- 1mol S

C6H

12O

6(s)

oéëê

ùûú

= 2mol (160) + 2mol (213.8)éë

ùû

J

mol-K- 1mol (209.2)éë

ùû

J

mol-K= 747.6 - 209.2éë

ùû

J

K= +538.4 J / K


37

For A:

w

rev= -nRT ln

V2

V1

æ

èç

ö

ø÷ = -nRT ln

P1

P2

æ

èç

ö

ø÷ = nRT ln

P2

P1

æ

èç

ö

ø÷

= 1.0mol 8.314 J

mol-K(298K)( )ln(10) = 2477.6J(2.303) = 5710J

Since isothermal: q = -w ∆U = q +w = 0

∆H = ∆U + ∆(PV) = ∆U + nR∆T = 0 + 0

∆S = ∆Ssystem

+ ∆Ssurr

=5710 J

298K+ (

-5710 J

298K) = 19.1 -19.1 = 0

For B: w = 0 so q = 0 but

∆S = ∆Ssystem

+ ∆Ssurr

=5710 J

298K+ 0 = 19.1

J

K

• ∆S must be non-zero due to volume change, even though no work done.

3.27 STEP 1:

• Adiabatic process, so q=0 but P1≠ P2. V2= 2V1, T1≠ T2

• Work defined as by wadiab

= nCv∆T , so need to calculate value of T2.

• Cv

= 5 /2R = 2.5(8.314J

mol - K) = 12.47

J

mol - K

Cvln

T2

T1

= -RlnV

2

V1

Þ lnT

2

T1

= -R

Cv

lnV

2

V1

= -

8.314J

mol - K

12.47J

mol - K

ln(2) = -0.667(0.693) = -0.462

T2

T1

= e-0.462 = 0.630 Þ T2

= 0.630(450K) = 283.5K

wadiab

= nCv∆T = 2.50mol(12.47 J

K-mol)(-166.5K) = -5.19X103 J = -5.19kJ ∆U = w = -5.19kJ

∆H = nCp∆T = (n ´ (C

v+ R))∆T = ∆U + nR(∆T)

∆H = -5.19kJ + 2.50mol (8.314 J

mol-K)(-166.5)K

1kJ

1000 J

é

ëê

ù

ûú = -5.19kJ + (-3.46kJ) = -8.65kJ

• Can’t say ∆S = 0 since q = 0, since will have both a volume and T change in the gas which affects

entropy. Must calculate ∆Stotal as:

∆STotal

= nRlnV

2

V1

+ nCvln

T2

T1

= n RlnV

2

V1

+ Cvln

T2

T1

é

ë

êê

ù

û

úú

= 2.50mol (8.314 J

mol-K)ln(2) +12.47 J

mol-Kln

283.5

450

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= 2.50mol (8.314 J

mol-K)(0.693) +12.47 J

mol-K(-0.462)é

ëùû

= 2.50mol (5.76 J

mol-K) + (-5.76 J

mol-K)é

ëùû

= 0.0 J / K

STEP 2:

• Constant V process, so w = 0 so that:

∆U = nCV∆T == nC

V(T

3- T

2) = 2.50mol(12.47

J

mol - K)(450 - 283.5)K = 5.19X103 J = 5.19kJ

∆H = ∆U + nR(∆T) = 5.19kJ + 2.50mol (8.314 J

mol-K)(+166.5)K

1kJ

1000 J

é

ëê

ù

ûú = +8.65kJ

• Only the ∆T contributes to ∆S in this step since volume constant.

∆S∆T

= nCvln

T3

T2

= 2.50mol 12.47 J

mol-Kln

450

283.5

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= 31.18(0.462) J

K= 14.4 J / K

STEP 3:

A (Isothermal)

B (Adiabatic)

w - 5710 J 0

q + 5710 J 0

∆U 0 0

∆H 0 0

∆S 0 19.1 J/K


38

• Isothermal compression process, so ∆U = 0 and ∆H =0

w = nRT4ln

V1

V3

= 2.50mol (8.314 J

mol-K)(450K)ln

V1

2V1

æ

èç

ö

ø÷

é

ë

êê

ù

û

úú

= 21.03 J

K(450K)(-0.693)

1kJ

1000 J

é

ëê

ù

ûú = -6.48kJ

So that: q = -w = +6.48kJ

• Only the ∆V contributes to ∆S in this step, since T constant.

∆S

∆V= nRln

V1

V3

= 2.50mol (8.314 J

mol-K)ln(0.5)é

ëùû

= (20.785 J

K)(-0.693)é

ëùû´

1.0kJ

1000 J= -14.4 J / K

Summary:

STEP 1 STEP 2 STEP 3 TOTAL

w 0 5.19 kJ - 6.48 kJ -1.35 kJ

q - 5.19 kJ 0 6.48 kJ 1.35 kJ

∆U - 5.19 kJ 5.19 kJ 0 0

∆H - 8.65 kJ 8.65 kJ 0 0

∆S 0 14.4 J/K -14.4 J/K 0

3.28 (A): Analysis:

(1) Heat will be transferred, until the same Tf is produced so each block undergoes a ∆S:

∆SA

= nCp,A

lnT

f

TA

and ∆SB

= nCp,B

lnT

f

TB

so that: ∆Stotal

= ∆SA

+ ∆SB

= Cp,A

lnT

f

TA

+ Cp,B

lnT

f

TB

(2) The final temperature has to be determined by calorimetry starting with –qlost = qgain before the ∆S

can be calculated.

(3) Then when the results or Tf is substituted into the ∆S equation, the proof should be clearer.

Calorimetry calculation:

-qlost

= -qBlock A

= qgain

= qBlock A

Þ - nA

Cp,A

(Tf

- TA) = n

BC

p,B(T

f- T

B)

(Tf

- TA) = (T

f- T

B) Þ 2T

f= T

A- T

BÞ T

f=

TA

- TB

2

Calculation of ∆S:

∆Stotal

= ∆SA

+ ∆SB

= lnT

f

TA

+ lnT

f

TB

é

ë

êê

ù

û

úúC

p,M(s)= ln

(TA

+ TB)

2TA

+ ln(T

A+ T

B)

2TB

é

ë

êê

ù

û

úúC

p,M(s)

= ln(T

A+ T

B)

2TA

´(T

A+ T

B)

2TB

é

ë

êê

ù

û

úúC

p,M(s)= ln

(TA

+ TB)2

4TAT

B

é

ë

êê

ù

û

úúC

p,M(s)

B) (a) • Need heat capacity of metal, Fe(s) = 25.10 J/mol-K

• Need T’s in Kelvin for log term:

Set I II III

Block A 373 K 453 K 473 K

Block B 293 K 373 K 873 K

Calculations:

Set I: ∆Stotal

= (25.10 J

mol-K)ln

(373 + 293)2

4(373)(293)

é

ëêê

ù

ûúú

= (25.10 J

mol-K)ln(1.015) = (25.10 J

mol-K)(0.0149) = 0.374 J

mol-K

Set II: ∆Stotal

= (25.10 J

mol-K)ln

(453 + 373)2

4(373)(453)

é

ëêê

ù

ûúú

= (25.10 J

mol-K)ln(1.009) = (25.10 J

mol-K)(0.00924) = 0.232 J

mol-K

Set III: ∆Stotal

= (25.10 J

mol-K)ln

(473 + 873)2

4(473)(853)

é

ëêê

ù

ûúú

= (25.10 J

mol-K)ln(1.115) = (25.10 J

mol-K)(0.1088) = 2.73 J

mol-K

(b) Expected Set III to have the greatest ∆S since the ∆T for each block is the largest of the

combinations. This should result in the greatest change in thermal energy distribution in each block and

the calculations confirm this.

(c) Even though the ∆T for each block is the same in Set II as in Set I, the transfer occurs at higher initial

temperatures. The entropy change will not be as great at the higher temperature, since the molecules are

already in a more disordered state, then when they are at a lower starting temperature.

3.29 A) Processes you would need to consider:


39

• The gas will first cool and then condense, giving off heat.

• The heat will then be gained by the cool liquid bringing it to some higher T.

• Assume no heat loss to or gain from the surroundings: - q lost = q gained.

The key question is - how much heat would be needed to heat liquid up to boiling point (Green box 1)

and would that exceed or match the heat lost when the gas cooled (Red box 1) to the boiling point and

condensed into a liquid? (Red box 2). If the total needed exceeds the total of heat released (Red boxes,

Step 1 and 2), then the re-condensed liquid would have to cool into the liquid phase (Red box 3).

Cool C2H5OH(g) 115 °C → 78.3°C q lost = ?

qlost

= nCp,gas

∆T = 50.0gC2H

5OH ´

1molC2H

5OH

46.0g

æ

èç

ö

ø÷ ´ 65.44 J

mol-K( ) ´ (78.3 -115)K

= (1.09mol) ´ 65.44J

mol-K´ (-36.7K) = -2617J = -2.61kJ

Heat needed to take C2H5OH(l) from 30 °C → 78.3°C q gain = ?

qgain

= nCp,gas

∆T = 100mLC2H

5OH(l) ´

0.789gC2H

5OH

1mL

æ

èç

ö

ø÷ ´

1molC2H

5OH

46.0g

æ

èç

ö

ø÷ ´ 111.5

J

mol-K( ) ´ (78.3 - 30)K

= (1.715mol) ´ 111.5 J

mol-K( ) ´ (48.3K) = 9236J = +9.24kJ

So the gas cooling isn’t enough heat lost. It provides only 2.61 kJ, leaving 6.63 kJ still needed. So now

must consider how much heat lost when gas condenses to liquid at 78.3°C:

Heat lost C2H5OH(g) → C2H5OH(l) @ 78.3°C

qlost

= -∆Hvap

´ nC2H

5OH == (1.09mol) ´

38.6kJ

mol

æ

èç

ö

ø÷ = -42.07kJ

Since the cooler liquid only needs 9.24 kJ total to reach the boiling point, only 15.7% (the extra 6.63 kJ

needed/ 42 kJ) of the gaseous C2H5OH has to condense to bring liquid to the boiling point. The final state

is a combination of gas and liquid states in equilibrium at Tbp, 78.3°C (since liquid cannot be heated

higher than 78.3°C).

B) We then have three entropy changes:

∆S: Cool 50.0 g C2H5OH(g) 115 °C → 78.3°C

∆Scool gas

= nCpln

T2

T1

= (1.09mol) ´65.44J

mol - Kln

351.3K

388K

æ

èç

ö

ø÷ = -7.08 J / K

∆S: Heat lost C2H5OH(g) → C2H5OH(l) @ 78.3°C when 15.7% of moles condensed:

Constants for C2H5OH:

MW = 46.0 g/mol

density liq = 0.789 g/mL

∆Hvap = 38.6 kJ/mol

Tbp = 78.3 °C

Cp(liq) = 111.5 J/mol-K

Cp(g) = 65.44 J/mol-K

1

1

2

1

2

TbpCondense C2H5OH(g)At 78.3°C q lost =?

Cool C2H5OH(l) 78.3°C à T2?

Tbp

Heat C2H5OH(l)30 °C à 78.3°C

q gain = ?

1

2

3

1

C2H5OH(g)

50.0 g,T1= 115°C

C2H5OH(l)

100 mL,T1= 30°C

Open stopcock and mix

Cool C2H5OH(g)115°C à 78.3°C

?


40

∆Scondense

= -∆H

vap

Tbp

´ (Fractionmol condensed) = -

38.6X103 J

mol

351.3K´ (.157(1.09mol))

= -109J / K ´ 0.171 = -18.7 J / K

∆S when heating 100 mL C2H5OH(l) from 30 °C → 78.3°C

∆Sheat liquid

= nCpln

T2

T1

= (1.715mol) ´111.5J

mol - Kln

351.3K

303K

æ

èç

ö

ø÷ = +28.3 J / K

So that:∆Stotal

= ∆Scool gas

+ ∆Scondense

+ ∆Sheat liquid

= (-7.08 + -18.8 + 28.3)J / K = +2.42 J / K

3.30 A) Analysis: For all the ice to melt, must be able to provide enough heat from cooling water to

heat the H2O(s) from -10°C to 0°C and then melt, so that:

Total mass ice = 6.0 cubes ´15.0cm3

1cube´

1.0g

1.0cm3= 90.0g ice

1mol

18.0g

é

ëê

ù

ûú = 5.00mol ice

qmin

= qheat

(-10°C Þ 0°C) + qmelt

= Cp(H

2O(s)) ´ (no.mol ice) ´ ∆T

iceéë

ùû

+ ∆Hfus

´ (no.mol ice)éë

ùû

= 37.4 J

mol-K(°C)´ 5.00mol ´10Ké

ëùû

+ 6.008 kJ

mol´ 5.00molé

ëùû

= 1.87 + 30.04éë

ùûkJ = 31.9kJ

So 31.9 kJ required by the ice to become liquid at 0.0°C. Can the hot water provide this without going

below 0°C? Calculate the ∆T for the hot water when 31.9 kJ is lost.

-3.19X104 J

75.3 J

mol-K(°C)(22.22mol)

= (T2

- 70°C) = -19.1 so T2

= 50.9°C

B) Then the melted ice at 0°C and warm water at 50.9°C will exchange heat

until they come to the same final T, which can be determined by:

qgain (ice)

= -qlost (warm water)

Þ CpH

2O(l)

´ (no.mol ice) ´ (T2

- 0°C)é

ëêù

ûú= - C

pH2O(l)

´ (no.mol water) ´ (T2

- 50.9°C)é

ëêù

ûú

(5.00T2) = -22.22T

2+1131 T

2=

1131°C

27.22= 41.5°C

C) So there will be 3 terms for ice to define its ∆S and only 1 term needed for the warm water:

∆Sice

= nice∆S

fus+ ∆S

heating solid+ ∆S

heating liquidéë

ùû

= (5.00mol)∆H

fus

Tmp

+ Cpsolid

lnT

mp

T1

+ Cp liquid

lnT

2

Tmp

é

ë

êê

ù

û

úú

∆Sice

= 5.006.008X103

273+ (37.4)ln

273

263+ (75.3)ln

314.4

273

é

ëêê

ù

ûúú

J

K= (5.00) 22.0 +1.395 +10.63é

ëùû

J

K= 170 J / K

∆Swarmwater

= nC pH2O(l) ln

T2

T1

= (22.22mol) ´ 75.3J

mol - K´ ln

314.4

343= (22.22)(75.3)(-0.0871)

J

K= -146

J

K

∆Stotal

= 170 + (-146) = +24.0 J / K

1

So the ice will completely

melt, since hot water has

to only cool to 50.9°C.


41

3.31 A) Analysis: First question to ask - Is there enough

heat from steam condensing to heat the

Cu block to 100°C?

qlost

= n(-∆Hvap

)

= (1.0mol) ´ -40.66kJ

1mol

æ

èç

ö

ø÷

= - 40.66kJwould be lost

qCu

(0°C ®100°C) = heat needed = nCp,Cu∆T

= 2000gCu ´1molCu

63.04g

æ

èç

ö

ø÷ ´

24.44J

mol - K

æ

èç

ö

ø÷ ´ (100)K

= (31.48mol) ´ (2444J

mol) = 76937J

= 76.94kJ needed toheat Cuto100°C

Now we know water will undergo two changes: the steam will completely condense to liquid and the

liquid then cools to some T2, losing heat in both changes. The solid Cu will just gain heat going from 0°C

to T2.

-qlost

= - qcondense H

2O(g)

+ qH

2O(l)cooling

éëê

ùûú

= qgain

= nCp,Cu∆T

- -∆Hvap

´ (no.mol H2O)é

ëùû

- (no.mol H2O) ´C

p,H2O(l)

´ (T2

-100)°Céëê

ùûú

= (no.mol Cu) ´ Cp,Cu

´ (T2

- 0)°Céë

ùû

1.0mol(40.66kJ) ´1000 J

1kJ

é

ëê

ù

ûú - 1.0mol ´ 75.3 J

mol-K(°C)´ (T

2-100)°Cé

ëùû

= 31.43mol ´ 24.44 J

mol-K(°C)´ (T

2- 0)°Cé

ëùû

4.066X104 - 75.312T2

- 7531.2 = 769T2

Þ 4.879X104 = 844T2

Þ T2

=4.879X104

844°C = 57.1°C

B) • Must have all T’s in K since will be used in log terms for entropy calculations.

∆Stotal

= ∆Scondense steam

+ ∆SH

2O(l)cools

+ ∆SCuwarms

= -∆H

vap

373K

é

ë

êê

ù

û

úú

+ Cp,H

2O(l)

lnT

2

373K+ C

p,Cu

lnT

2

273K

∆Stotal

= -4.066X104 J

373K

é

ëêê

ù

ûúú

+ 75.312J

mol - K(1.0mol)ln

330K

373K

é

ëê

ù

ûú + 24.44

J

mol - K(31.48mol)ln

330K

273K

é

ëê

ù

ûú

= -109J

K- 75.312(-0.1225)éë

ùû

J

K+ 769(0.1898)éë

ùû

J

K= (-109 - 9.23 +146)

J

K= +27.8

J

K

• Note that the total entropy change is positive, which means the process we have described will occur

spontaneously.

3.32 A) • Isothermal and number of moles constant Assume ideal :P1V

1

n1T

1

=P

2V

2

n2T

2

® P2

=P1V

1

V2

PAr

= 2.14atm ´4.0L

6.5L= 1.321atm P

Ne= 5.312atm ´

2.5L

6.5L= 2.04atm P

total= 3.36atm

B) • Need to calculate moles of each gas and mol fraction

So just condensing the steam does not provide

enough heat energy to bring the Cu block to 100°C.

Therefore the 2nd process, of the liquid cooling

further, must also occur and th final T of Cu will be

below 100°C.

2

1

• T2

Tbp

Condense

H2O(g) at 100°C qvap =-n(∆Hvap)

T

Heat Cu(s)0 °C à ? From

heat lost by H2O

1

2.0 kg Cu(s)@ 0°C

1.0 mol H2O(g) @ 100°C

2.0 kg Cu(s)@ ?°C

H2O(l) @ 100 °C

H2O(g)

Cu(s)

If T of Cu block ≠

100°C then further

cooling must occur for H2O(l) and heating for

Cu(s) until they reach

same final T (T2)

2.0 kg Cu(s)@ Tfinal °C

H2O(l) @ T2l °C

Cool H2O(l)

from 100°C à T2

T Further heat Cu(s) à T2

From heat

lost by H2O(l)

2H2O(l)

Cu(s)

• T2

• Tinitial

• Tinitial

1 2Compare the energy the steam

can lose by condensing at 100°C

to that needed to heat the Cu(s)

to 100°C. if they are equal only

the first process would occur.


42

nAr

=P1V

1

RT=

2.14atm ´ 4.0L

0.08206 L-atm

K-mol´ 293K

= 0.356mol Ar nNe

=P

1V

1

RT=

5.312atm ´ 2.5L

0.08206 L-atm

K-mol´ 293K

= 0.552molNe

cAr

= 0.356 / (0.356 + 0.552) = 0.392 cNe

= 1.0 - 0.392 = 0.608

∆Smix

= -R(cAr

lncAr

+ cNe

lncNe

) = -8.314J

mol - K(0.392ln(0.392) + 0.608ln(0.608))

∆Smix

= -8.314 J

mol-K0.392(-0.936) + 0.608(-0.497)éë

ùû

= -8.314 J

mol-K(-0.367) + (-0.3025)éë

ùû

= 5.56 J

mol-K

C) The two factors, the initial volumes and no. moles of each gas would produce different values for the

reversible work which depends on the ratio of V2/V1 and the value of n for each gas.

3.33 A) Since both liquid and solid C6H6 are in the tube removed from the ice bath, the two states must

be in equilibrium at the freezing point of C6H6.

B) For C6H6 the changes that will affect its entropy are:

(1) All of the C6H6 liquid is cooled from 24/5 °C to 5.6°C, but

(2) Only 73.3 mL (73.3%) freezes into solid at 5.6°C, but the solid does not have time to cool further.

For (1): ∆S for C6H6 liquid calculated from the T-dependence at constant P: ∆S(1)

= nCpln

T2

T1

• Will need to determine how many moles of C6H6 in the tube to determine ∆S.

nC

6H

6

= 100mL0.876g

1.0mL

æ

èç

ö

ø÷

1.0mol

78.0g

æ

èç

ö

ø÷ = 1.123molC

6H

6

∆S(1)

= nCpln

T2

T1

= (1.123mol) ´136J

mol - Kln

278.6K

297.5K

æ

èç

ö

ø÷ = 152.7

J

K(-0.0656) = -10.0

J

K

And for (2): ∆S for C6H6 liquid that froze at 5.6°C would be defined as ∆S for a phase change, but we will

need to determine the ∆S for only the moles that froze:

nC

6H

6(s)

= 0.733(1.123mol) = 0.823molC6H

6(s)

∆SC

6H

6(s)

= -∆H

fus

Tmp

´ (molC6H

6(s)) = -

10.6X103 J

mol

278.5K´ (.823mol) = -

8723.8 J

278.5K= -31.3 J / K

So for C6H6 overall: ∆Stotal,C

6H

6

= ∆SC

6H

6cooling

+ ∆SC

6H

6freezing

= -31.3 -10 = -41.3J

K

C) Although the ice bath T doesn’t change, as long as both phases of water are still present, some ice

must have melted at 0°C, absorbing the heat lost by the benzene cooling and freezing. The ∆S for the

water is then due to this phase change, and we will have to calculate the moles of ice melted by applying

calorimetry.

• Assuming q lost by the benzene = q gain for the water and only the phase change takes place.

-qlost

= - qfusC

6C

6(s)

+ qC

6C

6(l) cooling

éëê

ùûú

= qgain

= nH

2O(s)∆H

fusH2O(s)

• Given that ∆SH

2O(s)

= nH

2O(s)

´∆H

fus,H2O(s)

Tfp

=q

gain

Tfp

then:

Constants for C6H6:

MW = 78.0 g/mol

density liquid = 0.876 g/mL

∆Hfus = 10.6 kJ/mol

Tmp = 5.6 °C

Cp(s) = 118 J/mol-K

Cp(liq) = 136 J/mol-K

Cp(g) = 82.4 J/mol-K

Tube

removed

from ice bathC6H6

cools in

bath.

100 mL

C6H6(l)

20.0°C

C6H6 (l)

26.7 mL

+

C6H6 (s)

T2 = 5.6°C

Ice + water

0.020°C

Ice + water

0.020°C


43

qgain

= -qlost

= - qfreezeC

6H

6(l)

+ qC

6H

6(l)cooling

éëê

ùûú

= - -∆Hfus

(no.mol C6H

6frozen)é

ëùû

+ (no.mol C6H

6)C

p,C6H6

(l)

(5.6 - 24.5)°Cé

ëê

ù

ûú

= 0.823mol ´ (10.6kJ

mol) ´

1000 J

1kJ

é

ëê

ù

ûú - 1.123mol ´136 J

mol-K(°C)´ (-18.9)°Cé

ëùû

= 8716 J + 2887J = 1.160X104 J

Leads to: ∆SH

2O(s)

=q

gain

Tmp

=1.16X104 J

273K= 42.5

J

K

D) We would expect the cooling and freezing of the benzene in the tube to be a spontaneous process,

which means ∆Soverall (= ∆Ssurroundings) will be positive. When it is calculated, we see it is positive and the

process should occur as we have described it.

∆Ssurroundings

= ∆SC

6H

6total

+ ∆SH

2O

= -41.3+ 42.5( ) J / K = 1.2 J / K

E) If all the liquid completely froze and cooled to the bath temperature we should expect that:

(a) The ∆SC6H6 to become more negative, since more heat would be lost.

(b) The ∆Sice to increase, since more ice would melt and become more positive.

(c) The difference between (a) and (b) will still be positive, since this would still be a spontaneous

process.

(d) New values:

For C6H6: The value calculated for the liquid cooling ∆S(2) would stay the same, but the value for the

phase change would change and a third term ∆S(3) for cooling the solid must be added.

New(2):∆SC

6H

6(s)

= -∆H

fus

Tmp

´ (molC6H

6(s)) = -

10.6X103 J

mol

278.5K´ (1.123mol) = -

11.9X103 J

278.5K= -42.7 J / K

New (3):∆S(3)

= nCp,solid

lnT

2

T1

= (1.123mol) ´118 J

mol-Kln

273.1K

278.6K

æ

èç

ö

ø÷ = 132.5

J

K(-0.0196) = -2.60 J / K

Then: ∆Stotal,C

6H

6

= ∆SC

6H

6(l)cooling

+ ∆SC

6H

6freezing

+ ∆SC

6H

6(s)cooling

= -42.7 -10.0 - 2.6 = -55.3 J / K

For H2O(s): We would have to calculate the new value for qgain and ∆S due to melting more ice.

The process is still spontaneous since:

3.34 A) for a reversible expansion or contraction

leads to:

Then:

B) (a)


44

(b) CO2, b = 0.04267 L/mol and CCl4, b = 0.1281 L/mol

So

for gases with larger molar volume corrections, the ∆S is significantly different (about 9.5% higher). In

the smaller volume the attractive forces are creating fewer allowed microstates for the molecules, or we

can say there is less disorder in the initial, smaller volume, because attractive forces are stronger. In the

larger volume, the wider spaces between molecules produces less attraction.

3.35 A) for a reversible expansion or contraction

leads to:

and

B) (a)

(b)

(a) The ∆S values for CO2 and CCl4 as virial gases are lower than the ∆S calculated for the gases as an

ideal gas, which would be the same for CO2 and CCl4. The ∆S values for CO2 is only slightly lower,

indicating the attractive and repulsive forces have not altered the microstates allowed, so that gas is

close to acting ideally. But the value for CCl4 is dramatically lower, indicating the attractive forces must

be influencing the allowed states for the gas particles. The difference of ∆S from the ideal value is being

set by B, the second virial coefficient.

(b) The value of B takes into account both the “a” and “b” terms, so it may be reasonable to neglect them

both for CO2 but not for CCl4, since the Van der Waals gas and Virial gas results are so very different.

(c) The second virial coefficient B is very temperature dependent, whereas the van der Waals are not.


45

PART 4: Free Energy (∆G), Helmholtz energy (∆A) and Phase Equilibrium

4.1 A) (a) T1, T2 and T4 are single phases, only T3 represents a phase transition. The intersection

of gas and liquid lines at T2 has no meaning, since the solid has the lowest µ at that temperature.

(b) At T1 and T2 the solid is most stable and at T4 the gas state.

(c) Gas + solid are in equilibrium at T3 since µsolid = µgas.

B) At no T does the liquid represent the lowest µ, so that it is no transition to the liquid appears for

substance X. It will sublime before it melts.

4.2 A) Given a constant P, a chemical change will always be spontaneous when the entropy of the

change (is positive, remains constant, is negative) and the enthalpy (is positive, remains constant, is

negative) during the process.

The magnitude of ∆G will change with T but it will always be a negative value.

B) When P is increased on a pure substance, at constant T, the molar free energy, G , (increases,

remains constant, decreases).

Since dG

dP

é

ëê

ù

ûúT

= V and the molar volume cannot be negative or zero, then G must increase with P for all

states.

C) When a sample of liquid is converted reversibly to its vapor at its normal boiling point. (q, w, ∆P, ∆V,

∆T, ∆U, ∆H, ∆S, ∆G, or none of these) is equal to zero for the system.

Since P is constant, ∆P =0 and since it is an equilibrium between two states, ∆G = 0.

4.3 A) ∆S = nRlnV

2

V1

= 2.0mol (8.314 J

mol-K)ln

125L

25.0L

é

ëê

ù

ûú = 16.62(1.609) = 26.8 J / K

B) ∆G = ∆H - T ∆S = 0 - T ∆S = -340K(26.8 J

K) = -8564J = -8.56kJ

• ∆H = 0 since isothermal, reversible expansion of an ideal gas.

C) ∆A = ∆U - T ∆S = 0 - T ∆S = -340K(26.8 J

K) = -8564J = -8.56kJ

• ∆U = 0 since isothermal, reversible expansion of an ideal gas.

4.4 A) Isothermal reversible expansion of an ideal gas. ∆U, ∆H

∆U =0 since isothermal, reversible → dq= -dw to keep T the same, ∆H = 0 no bonding or interaction

change, ∆S ≠ 0 since V2 ≠ V1, so ∆G ≠ 0

B) Adiabatic reversible expansion of a non-ideal gas. ∆S

∆U ≠ 0 =dw only, ∆H ≠ 0 since ∆T, ∆S = 0 since ∆S from V2 ≠ V1 compensated by ∆S T1→ T2, and ∆G ≠

0 since have ∆H.

C) Vaporization of liquid water at 80°C and 1 bar pressure. ∆U

∆U = 0 dq =- dw to keep T same, ∆H ≠ 0 since phase change, ∆S ≠ 0 since phase change by ∆S T1→

T2, so ∆G ≠ 0 since have ∆H.

D) Vaporization of liquid water at 100°C and 1 bar pressure. ∆G, ∆U

∆U = 0 dq =- dw to keep T same, ∆H ≠ 0 since phase change, ∆S ≠ 0 since phase change but ∆G = 0

since ∆H = T∆S at normal bp.

E) Reaction between H2 and O2 in a thermally insulated bomb. None

∆U ≠ 0 gas volumes change, ∆H ≠ 0 since chemical reaction, ∆S ≠ 0 since chemical reaction, and ∆G ≠

0 since must be spontaneous or non-spontaneous.

F) Reaction between H2S04 and NaOH in dilute aqueous solution at constant temperature and pressure.

None


46

∆U ≠ 0 no work, but q value, ∆H ≠ 0 since chemical reaction, ∆S ≠ 0 since chemical reaction, and ∆G ≠

0 since must be spontaneous or non-spontaneous.

4.5 A) Need balanced reaction:C6H

6(l) + 7.5O

2(g)® 6CO

2(g) + 3H

2O(l)

• Maximum non-PV work means we need to calculate ∆G:

(a) From table: ∆Hcomb C6H6(l) = -3268 kJ

∆Scomb

o = 3 mol SH

2O(l)

o + 6mol SCO

2(g)

oéëê

ùûú

- 1mol Sf ,C

6H

6(l)

o + 7.5mol SO

2(g)

oéëê

ùûú

= 2(69.9) + 6(213.8)éë

ùû

- 1(173.4) + 7.5(205)éë

ùû

= 1492.5 -1710.9 = -219 J / K

∆G = ∆H - T ∆S = -3268kJ - 298K(-0.219kJ

K)

é

ëê

ù

ûú = -3268kJ + 65.1kJ = -3208kJ

On a per gram basis: -3208kJ

1.0molC6H

6(l)

´1.0molC

6H

6(l)

78.0g= -41.1kJ / g

(b) Since ∆H and ∆S are the same signs there will be a T at which ∆G can become positive, but the

changeover T would be Tchangeover

=∆H

∆S=

-3.203X106 J

-218 J / K= 14,716K so it is not practical to stop reaction by

raising T. The reaction won’t stop (once started) until either all the fuel or the O2(g) is used up.

B) H2(g) +1 /2O

2(g)®H

2O(l) ∆Hcomb = ∆Hf° H2O(l) = -285.8 kJ

∆S

comb

o = 1 mol SH

2O(l)

oéëê

ùûú

- 1mol Sf ,H

2(g)

o + 0.5mol SO

2(g)

oéëê

ùûú

= 1(69.9)éë

ùû

- 1(130.7) + 0.5(205)éë

ùûJ / K = 69.9 - 233.2 = -163 J / K

∆G = ∆H - T ∆S = -285.8kJ - 298K(-0.163kJ

K)

é

ëê

ù

ûú = -285.8kJ + 48.8kJ = -237kJ

On a per gram basis: -237kJ

1.0mol H2(g)

´1.0molH

2(g)

2.0g= -118.5kJ / g

4.6 • Will need to calculate ∆S for the reaction from tabled values

• Can either determine ∆Gf° from tabled values or calculate from ∆H° and ∆S° values

∆Sr

o = 1mol S°NOCl(g)

éë

ùû

- 0.5mol SN

2

o + SO

2

o + SCl

2

oéëê

ùûú

= 261.6éë

ùû

- 0.5 223 + 205 +191.6éë

ùû

= -48.4J / K

∆Gf

o = 66.5kJ or ∆G = ∆H - T ∆S = 51.9kJ - 298K(-48.4J

K)

1kJ

1000J

æ

èç

ö

ø÷

æ

èç

ö

ø÷ = 51.9+14.4 = 66.3kJ

∆U = ∆H - ∆(PV) = ∆H - ∆n(RT) ∆n = 1-1.5 = -0.50 mol

∆U = 51.9kJ - -0.5mol(8.314 J

mol-K)(298K)

1kJ

1000J

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= (51.9 +1.24)kJ = 53.1kJ

∆ A = ∆U - T ∆S = 53.1kJ - 298K(-48.4J

K)

1kJ

1000J

æ

èç

ö

ø÷

æ

èç

ö

ø÷ = (53.1+14.4)kJ = 67.5kJ

C) Both ∆G and ∆A indicate that the reaction is non-spontaneous at all T’s since the difference between

∆H (or ∆U) and T∆S terms will always be positive.

4.7 A) ∆H

298

o = 1 mol∆Hf ,C

5H

6(l)

o + 2mol∆Hf ,HI(g)

oéëê

ùûú

- 1mol∆Hf ,C

5H

8(l)

o +1mol∆Hf ,I

2(s)

oéëê

ùûú

= 1mol (134.3) kJ

mol+ 2mol (26.1) kJ

moléë

ùû

- 1mol (58.2) kJ

mol+ 0é

ëùû

= (186.5 - 58.2)kJ = 128.3kJ

∆S298

o = 1 mol Sf ,C

5H

6(l)

o + 2mol Sf ,HI(g)

oéëê

ùûú

- 1mol Sf ,C

5H

8(l)

o +1mol Sf ,I

2(s)

oéëê

ùûú

= 1mol (274.1) J

K-mol+ 2mol (206) J

K-moléë

ùû

- 1mol (291.3) J

K-mol+1mol(116 J

K-mol)é

ëùû

= (686 - 407) J

K-mol= 279 J / K

H2(g) has a much higher fuel

value. It is nearly 3 times

greater than that of C6H6.


47

∆G298

o = ∆H° - T ∆S° = 128.3kJ - 298K(279J

K)

1kJ

1000J

æ

èç

ö

ø÷

æ

èç

ö

ø÷ = 128.3- 83.1 = 45.2kJ

B) ∆n = 2 and ∆U298

o = 128.3kJ - 2mol(8.314 J

mol-K)(298K)

1kJ

1000J

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= (128.3 + 4.95)kJ = 133.2kJ

∆ A298

o = ∆Uo - T ∆So = 133.2kJ - 298K(279J

K)

1kJ

1000J

æ

èç

ö

ø÷

æ

èç

ö

ø÷ = (133.2- 83.1)kJ = 50.1kJ

C) There will be a changeover T since the signs of ∆H and ∆S are the same.

Tchangeover

(∆G) =∆H

r

∆Sr

=128.3kJ

0.279kJ / K= 460K(187°C) and Tchangeover,(∆ A) =

∆Ur

∆Sr

=133.2kJ

0.279kJ / K= 477K(204°C)

So the T needed for ∆G to change sign is not the same value as for ∆A. This is because ∆H ≠ ∆U.

4.8 A) ∆H

298

o = 6 mol∆Hf ,N

2(g)

o +12mol∆Hf ,CO(g)

o +10mol∆Hf ,H

2O(g)

o + 7 mol∆Hf ,O

2(g)

oéëê

ùûú

- 4mol∆Hf ,C

3H

5N

3O

9(l)

oéëê

ùûú

= 0 +12(-110.5) +10 (-241.8) + 0éë

ùûkJ - 4(-370)é

ëùûkJ = (-3081 +1480)kJ = -1601kJ

∆S

298

o = 6 mol SN

2(g)

o +12mol SCO(g)

o +10mol SH

2O(g)

o + 7 mol SO

2(g)

oéëê

ùûú

- 4mol SC

3H

5N

3O

9(l)

oéëê

ùûú

= 6(191.6) +12(198) +10(189) + 7(205)éë

ùû

J

K- 4 (545)éë

ùû

J

K= (6.851 - 2.180) kJ

K= +4.67kJ /K

∆G298

o = ∆H298

o - T ∆S298

o = -1601kJ - 298K(4.67 kJ

K)( ) = -1601-1392 = -2993kJ

B) ∆n = 35 and ∆U298

o = -1601kJ - 35mol(8.314 J

mol-K)(298K)

1kJ

1000J

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= (-1601 - 86.7)kJ = -1688kJ

∆ A298

o = ∆Uo - T ∆So = -1688kJ - 298K(4.67kJ

K)

æ

èç

ö

ø÷ = (1688-1392)kJ = -3080kJ

C) % extra work =3080 - 2993

2993´100 = 2.9%

• So would get about 3.0% more work if chemical reaction was under constant volume conditions

4.9 A) Since the transformation starts at 13.2°C, that is when the two forms must be in equilibrium:

µa

= µb and can apply: ∆µ = ∆G = 0 =∆H – T∆S at 13.2°C (286.4K) so that:

∆Sa®b

=∆H

a®b

Ttr

=2.238 kJ

mol

286.4K= 7.81X10-3 kJ

mol-K= 7.81 J / mol - K

B) Since then signs of ∆H and ∆S are the same, T will definitely affect the sign of ∆G or

∆µ = µb

- µa.

∆Ga®b

= µb

- µa

= ∆Ha®b

- T ∆Sa®b

= 2.232 kJ

mol- 273.15K(0.00781 kJ

K-mol)( ) = 2.232- 2.133 = +0.105kJ

C) At T = 273K, if we assume ∆H and ∆S constant, then can apply: ∆µ = ∆G = ∆H – T∆S

Since the ∆µ is positive, µβ is larger than µα, which means µα has the lower value and is the

most stable state. Therefore Sn will spontaneously transform to the alpha form at 13.2°C.

D) We can apply the pressure dependence of dG (dµ) for solids to each phase and then get the difference

to define how a change in P will affect transition:

Since µa

= dGa

= VadP and µ

b= dG

b= V

bdP thenµ

b- µ

a= V

b- V

a( )dP

• To calculate the molar volumes, assume the density of beta form is equal to the 13.2°C value, AW Sn =

118.7 g/mol then:

Vb

=AW Sn

db

=118.7 g

mol

7.31g

cm3

1000cm3

1.0L

æ

è

çç

ö

ø

÷÷

= 0.01624L

mol V

a=

AW Sn

da

=118.7 g

mol

5.77g

cm3

1000cm3

1.0L

æ

è

çç

ö

ø

÷÷

= 0.02057L

mol


48

∆Ga®b

= µb

- µa

= 0.01624 - 0.02057( ) L

mol´ (10atm) ´

101.3 J

1L - atm

æ

èç

ö

ø÷ = -4.39

J

mol

Since the ∆µ is negative, µβ is lower than µα, making the beta form the most stable state at the

higher P at 15°C.

4.10 A) (a) For CH4(g) since it is a gas, G(P2) = G(P

1) + RT

P1

P2

òdP

PÞ G(P

2) - G(P

1) = ∆G = nRT ln

P2

P1

• Assuming CH4 is acting as an ideal gas

∆G = nRT lnP

2

P1

= 2.0mol(8.314 J

mol-K)(293K)ln 100é

ëùû

= (4872 J)(4.605) = 22,436 J = 22.4kJ

(b) For C2H6(g): All the same values apply, so ∆G = 22.4 kJ

(c) • Need to apply G(P2) - G(P

1) = V(P

2- P

1) as the result of dG = VdP for liquids

• Will need to calculate molar volume from the density of the liquid

Vliquid

=MW C

6H

6

dC6H

6(l)

=78.0 g

mol

0.879g

cm3

1000cm3

1.0L

æ

è

çç

ö

ø

÷÷

= 0.0887L

mol

G(P2) - G(P

1) = V(P

2- P

1) = 0.0887

L

mol200 - 2( )atm

101.3 J

1L - atm

æ

èç

ö

ø÷ = 1780J = 1.78 kJ

B) The ∆G for CH4 and C2H6 is the same since both are gases (so the same equation applies) and we have

the same number of moles of each. As long as gases are ideal, the change in ∆G with P is independent of

chemical identity and only depends on the number of moles of gas.

C) C6H6 is a liquid and cannot substitute Vgas for ∆Vm. We will need actual density of liquid so chemical

identity will affect the ∆G with P.

4.11 A) ∆P = P2

- P1

= 1.0atm - 2000m14.5psi

10.06m

æ

èç

ö

ø÷

51.7torr

1psi

æ

èç

ö

ø÷

1.0atm

760torr

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= 1 -196 = -195atm

B) (a) Vliquid

=MW C

57H

110O

6

dliquid

=891.5 g

mol

0.950g

cm3

1000cm3

1.0L

æ

è

çç

ö

ø

÷÷

= 0.9384L

mol

∆G = V ∆P = 0.9384L

mol-195atm( ) 101.3 J

1L - atm

æ

èç

ö

ø÷ = -1.86X104 J = -18.6kJ

B) (b) We need to apply the Clapeyron equation for the solid → liquid phase transition:

dP =∆H

fus

∆Vfus

dT

TÞ

P1

P2

ò dP =∆H

fus

∆Vfus

T1

T2

òdT

TÞ P

2- P

1=∆H

fus

∆Vfus

lnT

2

T1

æ

èç

ö

ø÷ @

∆Hfus

∆Vfus

´∆T

T1

Þ ∆P =∆H

fus

∆Vfus

´∆T

fp

Tfp

o

• Assuming ∆Hfus, ∆Vfus constant from T1 → T2

• Making the series approximation: lnT

2

T1

æ

èç

ö

ø÷ = ln

T1

+ ∆T

T1

æ

èç

ö

ø÷ = ln 1 +

∆T

T1

æ

èç

ö

ø÷ »

∆T

T1

• Will need to calculate the change in molar volume for transition:

Vfus

= Vliq

- Vsolid

=MW(g / mol)

dliq

(g / L)-

MW(g / mol)

dsolid

(g / L)= MW

dsolid

- dliq

dliq

´ dsolid

é

ë

êê

ù

û

úú

= 891.5g

mol

950 - 856

950(856)

é

ëê

ù

ûúL

g= 0.1052

L

mol

Then with ∆P = -195 atm

• ∆Hfus must be in joule/mol = 2.033 X 105 J/mol, will need conversion factor 101.3 J = 1 L-atm

• Then can use 101.3 J = 1 L-atm to introduce pressure units

• Tfp° = 0°C at 1.0 atm pressure, but must be in K

Rearrange to solve for ∆T :

∆Tfp

=∆V

fusT

fp

o

∆Hfus

é

ë

êê

ù

û

úú∆P =

0.1052 L

mol(344K)

2.033X105 J

mol

101.3 J

1L - atm

æ

èç

ö

ø÷ (-195atm) = -3.52K


49

So the Tmp decreased by 3.5°C when P decreased. The Ttr from alpha to beta form, although also

decreased, would not have the same ∆T since the ∆V and ∆Htr would be different values.

C) ∆Sa®b

=∆H

a®b

Ttr

=145 kJ

mol

327K= 0.443

kJ

K - mol= 443

J

K - mol ∆S

mp=∆H

fus

Tmp

=203.3 kJ

mol

347K= 587

J

K - mol

The ∆S° for the α → β transition in the fat involves about two-thirds of ∆S° for the melting, so the α → β

transition must involve a significant increase in disorder or freedom of movement for the chains.

(a) The α → β transition in Sn results in a ∆S° 4.67 kJ/K-mol, about 10 times the size of the ∆S° in the

fat. The transition in the fat involves a shift of the chains and a small compression, whereas many more

atoms have to change position in Sn, so it is consistent.

4.12 A) ∆Hden

= qtr

= 544kJ

mol ∆G

den= 0 so ∆S

den=∆H

den

Tm

=544 kJ

mol

344.5K= 1.58

kJ

K - mol

B) Calculate new values for ∆Hden and ∆Sden at new T:

∆H(T2) = ∆H(T

1) + ∆C

p∆T Þ ∆H

den(310K) = ∆H

den(344.5K) + ∆C

p(310 - 344.5)K

= 544 kJ

mol+ 6.28 kJ

K-mol(-34.4K) = (544 - 216) kJ

mol= 328 kJ

mol

∆Sden

(T2) = ∆S

den(T

1) + ∆C

p,denln

T2

T1

= 1.58 kJ

K-mol+ 6.28 kJ

K-molln

310K

344.5K

é

ëê

ù

ûú

= 1.58 + (6.28(-0.1055))kJ

K - mol= 1.58 - 0.663( ) kJ

K - mol

1000J

1kJ

æ

èç

ö

ø÷ = 917

J

K - mol

∆G310

o = ∆H310

o - T ∆S310

o = 328kJ - 310K(0.917 kJ

K)( ) = (328- 284)kJ = +43.1kJ

• Since ∆G is positive, the lysozyme will not spontaneously unfold at 37°C.

C) If used the ∆Hden and ∆Sden from (A) and only changed T, then:

∆G310

o = ∆H310

o - T ∆S310

o = 544kJ - 310K(1.58kJ

K)

æ

èç

ö

ø÷ = (544- 490)kJ = +54.0kJ

% error =54.0 - 43.6

43.6´100 = 23.9%

• As we saw in the calculations for protein folding in PART 2, because the ∆Cp values are in

kilojoules/ mol-K, they are too large to neglect.

D) ∆Tm

=∆V

denT

m

o

∆Hden

é

ë

êê

ù

û

úú∆P =

4.58 cm3

mol(344.5K)

5.44X105 J

mol

1L

1000cm3

æ

èç

ö

ø÷

101.3 J

1L - atm

æ

èç

ö

ø÷ (10atm) =

1577.8(1013)

5.44X108K = 0.003K

• So the 10 atm pressure increase would have no effect on the transition temperature, Tm.

4.13 A) From tabled values:

∆S

tr= S°

S(s),monoclinic- S°

S(s),rhombic= (32.6 - 31.8) J

K= 0.80 J / K - mol

B) • For ∆H and ∆S for the transition at 95.5°C (368.5 K), need to apply Kirchhoff’s Law:

∆Cp

= Cp,monoclinic

- Cp,rhombic( ) = (23.6 - 22.64) J

K-mol= 0.96 J

K-mol

∆H368.5K

= ∆H298K

+ 0.96 J

K-mol(70.5)K = 330 + 67.7( ) J

mol= 398 J / mol

Since: ∆G

368.5K= 0 then ∆S(368.5K) =

∆H368.5

T1

=398 J / mol

368.5K

é

ëê

ù

ûú = 1.08

J

K - mol

• To make rhombic form stable at 100 °C, the transition T will have to increase by about 5K (°C)

• We will need to calculate ∆P from the Clapeyron equation, to produce a ∆T of 5.0 K

• Also need to determine ∆Va®b

= ∆Vb,monoclinic

- ∆Va ,rhombic

from the respective densities.

∆Vtr

= Vb

- Va

=MW(g / mol)

db(g / L)

-MW(g / mol)

da(g / L)

= MWd

a- d

b

dad

b

é

ë

êê

ù

û

úú

= 8(32.0)g

mol

270 - 207

270(207)

é

ëê

ù

ûúL

g= 0.02885

L

mol

So ∆Gden would have been positive, but

overestimated by about 24%.


50

dP

dT=∆S

(a®b)

∆V(a®b)

=∆H

(a®b)

T ∆V(a®b)

»∆P

∆T becomes: ∆P =

∆S(a®b)

∆V(a®b)

∆T =1.08 J

K-mol(5K)

0.02885 L

mol

1L - atm

101.3 J

æ

èç

ö

ø÷ = 1.71atm

So that:∆P = 1.71 atm = (P2

-1.0atm) Þ P2

= 2.71atm

4.14 Transition: ∆Htr(-). ∆Str(-)

Because the signs of ∆H and ∆S are the same, the sign of ∆G depends

on the temperature at which the phase change takes place. At low

temperatures the 1st term dominates, and the conversion favors the

beta form. When T in increases, the 2nd term gets larger, until it

dominates. The changeover temperature will mark the highest

temperature at which the folded state would be spontaneously

produced, or favored [See diagram]. At this temperature, the two

states are equally probable and ∆G would equal zero. As T gets larger,

the resulting ∆G(+) means the alpha state is favored.

Tchangeover

=∆H

∆S=

-209 kJ

mol

-0.554 kJ

K-mol

= 377K(104°C)

4.15 We need to apply the Clapeyron equation since solid → liquid phase transition for mercury, Hg:

dP =∆H

fus

∆Vfus

dT

TÞ

P1

P2

ò dP =∆H

fus

∆Vfus

T1

T2

òdT

TÞ P

2- P

1=∆H

fus

∆Vfus

lnT

2

T1

æ

èç

ö

ø÷ @

∆Hfus

∆Vfus

´∆T

T1

• Assuming ∆Hfus, ∆Vfus constant from T1 → T2

• Make the series approximation: lnT

2

T1

æ

èç

ö

ø÷ = ln

T1

+ ∆T

T1

æ

èç

ö

ø÷ = ln 1 +

∆T

T1

æ

èç

ö

ø÷ »

∆T

T1

For Vfus

= Vliq

- Vsolid

=AW(g / mol)

dliq

(g / L)-

AW(g / mol)

dsolid

(g / L)= AW

dsolid

- dliq

dliq

´ dsolid

é

ë

êê

ù

û

úú

Then the change in molar volume with fusion is:

Vfus

,Hg = 200.6g

mol

14.193X103 -13.690X103

(14.193X103) ´ (13.690X103)

é

ëêê

ù

ûúú

L

g=

200.6 ´503

1.943X108= 5.193X10-4 L

mol

• Densities must be converted to g/L to produce units of L/mol

• For final calculation with P2 = 100 atm

• ∆Hfus must be in joule/mol

• Then can use 101.3 J = 1 L-atm to introduce pressure units

• T1 = mp at 1.0 atm pressure = listed Tfp, but must be in K

P2

- P1

=∆H

fus

∆Vfus

´∆T

fp

Tfp

orearrangesto:

∆Vfus

Tfp

o

∆Hfus

P2

- P1( ) = ∆T

fp

∆Tfp

=∆V

fusT

fp

o

∆Hfus

P2

- P1( ) =

5.193X10-4 L

mol(234.2K)

2.292X103 J

mol

101.3J

1L - atm

æ

èç

ö

ø÷ (100 -1)atm =

12.04

22.62K = 0.53K

So the melting point increases Tfp @ 100 atm = -38.9°C + 0.53°C = -38.4°C (234.7 K)

B) Increase of ∆T

T1

´100 = % increase =0.53°C

38.9´100 = 1.4%

C) If Tfp° substituted as -38.9°C instead of the degrees Kelvin value then

∆Tfp

==5.193X10-4 L

mol(-38.9°C)

2.292X103 J

mol

101.3J

1L - atm

æ

èç

ö

ø÷ (100 -1)atm =

-2.000

22.62°C = -0.0884°C


51

The ∆T would have been calculated as negative (-0.0884°C), saying the melting point was lowered. That

can’t be true since when P increases µ increases, and the intersection of µsolid with µliquid should have been

shifted up, not down. Furthermore the magnitude of ∆T is only 16.4% of what it should be,

0.0884

0.54´100 = 16.4%.

4.16 A) The premise is the ∆P from the skate will cause a change the melting point of H2O(s). Given ∆Hfus

= 6010 J/mol, d H2O(l) = 997 kg/m3 and d H2O(s) = 997 kg/m3 at 0°C.

• The Clapeyron equation would apply to the solid liquid phase

equilibrium (consult map).

Need P2

= P1

+∆H

fus

∆Vm,fus

lnT

2

T1

é

ë

êê

ù

û

úú

• Will need to calculate the ∆Vm for the transition, from respective

densities:

∆Vm,fus

= Vm,H

2O(l)

- Vm,H

2O(s)

= MWH

2O

dH

2O(s)

- dH

2O(l)

dH

2O(l)

dH

2O(s)

é

ë

êê

ù

û

úú

= 18.0g

mol

920 g

L- 997 g

L

997g

L(920

g

L)

é

ë

êê

ù

û

úú

= 18.0 g

mol[-8.395X10-5 L

g] = -1.51X10-3 L

mol

Calculating P2:

P2

= P1

+∆H

fus

∆Vm,fus

lnT

2

T1

é

ë

êê

ù

û

úú

= 1.0atm +6010 J

mol

-1.51X10-3 L

mol

1L - atm

101.3 J

æ

èç

ö

ø÷ ln

268

273

æ

èç

ö

ø÷

é

ë

êê

ù

û

úú

= 1.0atm + (726.3)atm = 727atm

If used the approximation for log, then you would get a different value:

P2

= P1

+∆H

fus

∆Vm,fus

∆T

T1

é

ë

êê

ù

û

úú

= 1.0atm +6010 J

mol

-1.51X10-3 L

mol

1L - atm

101.3 J

æ

èç

ö

ø÷

-5.0K

273K

æ

èç

ö

ø÷

é

ë

êê

ù

û

úú

= 1.0atm + (718.5)atm = 719.5atm

The value for the P needed is significantly lower so the approximation is not valid.

B) What P can skate provide means must apply some physics (equations given in problem):

• Pressure defined as force/area and force would be: F = m(kg) ´ g(m

s2) and area must be in m2.

Force = 85.0kg ´ (9.807m

s2) = 833.6

kg - m

s2 Area = l ´ blade width = 0.25m ´ (2.5X10-4m) = 6.25X10-5m2

P(pascals) =833.6 kg-m

s2

6.25X10-5m2= 1.334X107Pa

1atm

101,325Pa

æ

èç

ö

ø÷ = 131.6 = 132atm

4.17 To answer question, need to go back to the original Clapeyron equation:

dP

dT=∆S

(a®b)

∆V(a®b)

=∆H

tr(a®b)

T ∆V(a®b)

ÞdT

dP=

T ∆V(a®b)

∆Htr(a®b)

given ∆µ = 0 at T for transition.

• Tbp must be in Kelvin

• Will need 101.3 J = 1 L-atm conversion

dT

dP=

Tbp∆V

(liq®gas)

∆Htr(liq®gas)

=T

bpV

gas- V

liquidéë

ùû

∆Hvap

=373K 30.20 - 0.0188é

ëùû

L

mol

40.66X103 J

mol

101.3 J

1L - atm

æ

èç

ö

ø÷ =

(373K)(30.18)

401.4atm= 28.0

K

atm

B) The Tbp will increase if the pressure is increased (as expected, see “Key Points”).

C) Whenever T is used in an equation, it must be in Kelvin, since it is coming from the kinetic energy

contribution. Only ∆T can be in °C, since the amount of heat in a ∆T in °C is the same as that in the ∆T in

K.

D) Can see a remarkable difference, since a change in P of nearly 100 atm changes the melting point by

only about 0.5 K, whereas a 1.0 atm change will increase the boiling point by 28 K. The two factors that

could contribute would be that:

Converting densities:

• 1m3 = 1000 L

• 1 kg = 1000 g, so that:

dH

2O(s)

=920kg ´

1000g

1kg

1m3 ´1000L

1m3

= 920g

L

So a DECREASE in molar volume

occurs for the phase change.

So skate does not apply enough

pressure to melt the ice and

hypothesis is incorrect.


52

(1) The difference in using ∆Hvap versus ∆Hfus. Since ∆Hvap is always larger than ∆Hfus, this should actually

make dT/dP smaller, not larger as we see here.

(2) The difference between the molar volumes of gas and liquid is always much larger than that of the

solid and the liquid. This is the primary reason why the effect of a pressure increase is so much larger

when the transition involves a gas state.

4.18 A) Because boiling occurs when the vapor pressure of the liquid equals the external pressure,

lowering the P from 760 torr at sea level to 380 torr will cause the boiling point T to be lower than 100°C,

which is when the vapor pressure of water equals 1.0 atm.

B) The Clausius Clapeyron equation would apply, not the Clapeyron, since the approximation that ∆Vm ≈

Vm,gas can be made for a transition involving a gas.

C) lnP

2

P1

= -∆H

vap

R

1

T2

-1

Tbp

o

é

ë

êê

ù

û

úúÞ ln

380torr

760torr

é

ëê

ù

ûú =

-(40.66) kJ

mol

8.314J

mol-K

1000J

1kJ

æ

èç

ö

ø÷

1

T2

-1

373K

é

ë

êê

ù

û

úú

-0.693 = -4890.51

T2

-1

373K

é

ë

êê

ù

û

úú

=-4890.5

T2

+13.11Þ T2

=-4890.5

-13.80K = 354.4K(81.3°C)

4.19 Could apply Clausius - Clapeyron equation (since involving the liquid to gas transition) or use the

modified Clapeyron equation derived in Problem 4.7, knowing 10 psi = P2-P1.

• Pressure will have to converted to atm: 10psi51.7torr

1psi

æ

èç

ö

ø÷

1atm

760torr

æ

èç

ö

ø÷ = 0.680atm

Using the modified equation:dP

dT=∆S

(a®b)

∆V(a®b)

=∆H

tr(a®b)

T ∆V(a®b)

ÞdT

dP=

T ∆V(a®b)

∆Htr(a®b)

»∆T

∆P

then ∆Tbp

=T

bp

o ∆Vliq®gas

∆Hvap

∆P =373K(30.18 L

mol)

4.066X103 J

mol

101.3 J

1L - atm

æ

èç

ö

ø÷

é

ë

êê

ù

û

úú(0.680atm) = 19.1°C and T

2= 100 + ∆T = 119°C

• Using Clausius - Clapeyron equation yields T2= 388 K = 115°C so modified equation valid.

4.20 • The situation describes a phase equilibrium between C6H6(l) and C6H6(g) at the normal boiling

point (P = 1 atm) and at a lower temperature.

A) The Clausius Clapeyron equation relates vapor pressure, ∆Hvap and T’s for the liquid gas

equilibrium.

• Let T1 = Tbp° so that P1 = 1.0 atm and , then P2 = vapor pressure in atm at T2.

lnP2

= lnP1

-∆H

vap

R

1

T2

-1

T1

é

ë

êê

ù

û

úú

= ln(1) -∆H

vap

R

T1

- T2

T1T

2

é

ë

êê

ù

û

úú , P

2= 0.10bar

1.0atm

1.01bar

é

ëê

ù

ûú = 0.0991atm

ln(0.0991) = 0 -∆H

vap

8.314 J

mol-K

353 - 273

353(273)

1

K

é

ëê

ù

ûú = -

∆Hvap

(5.80X10-4)

8.314 J

mol

Þ ∆Hvap

= 3.31X104 J

mol= 33.1 kJ

mol

B) Since ∆G = 0 for an equilibrium, ∆Sbp = ∆Hvap/Tbp° ∆Svap

=∆H

vap

Tbp

=3.31X104 J

mol

353K= 93.8

J

mol - K

4.21 A) To calculate work, decide whether reversible or irreversible work then need ∆Vvap:

• The expansion occurs against a constant P in an open system, as described, so treat as irreversible

work, w = -P∆Vvap

• ∆Vvap = Vgas-Vliquid ≈ Vgas and if ideal gas, Vgas

=nRT

P

w = -P∆V = -nRT = -1.0mol(8.314 J

mol-K)(373K) = -3101J = -31.0kJ

q = ∆Hvap

kJ

mol( ) ´ (no.mol) = 40.6 kJ

mol( ) ´ (1.0mol) = 40.6kJ

∆U = q +w = 40.6kJ - 31.0kJ = 9.6kJ

• Since ∆Gvap = 0 at 100°C, then:


53

∆Svap

=∆H

vap

Tbp

=40.6 kJ

mol

373K= 0.1088

kJ

mol - K

1000 J

1kJ

æ

èç

ö

ø÷ (1.0mol) = 108.8

J

K

B) Apply P change to ∆Gvap(P1) where since P2 = 10 at, ∆Gvap ≠ 0.

∆G°(P2) = ∆G°(P

1) + n

gasRT ln

P2

P1

= 0 + (1mol)(8.314 J

mol-K)(373K)ln

10atm

1atm

æ

èç

ö

ø÷ = 3101J(2.303) = 7140J = 7.14kJ

• Because ∆Gvap has become positive, the conversion would be non-spontaneous at the higher P meaning

the liquid would be the stable state at higher P’s.

C) When P is lowered, the log of the fraction (P2/P1) will be negative, so ∆Gvap becomes negative and this

indicates a spontaneous conversion to gas at the lower P (since the gas has the lower chemical potential).

∆G°(P2) = ∆G°(P

1) + n

gasRT ln

P2

P1

= 0 + ngas

RT lnP

2

P1

∆G°(P2) = n

gasRT ln

P2

P1

= (1mol)(8.314 J

mol-K)(373K)ln 0.50( ) = 3101J(-0.693) = -2150J = -2.15kJ

4.22 A) Reminder: Using a catalyst will not change the value of ∆G since ∆G is a state function and as

long as the initial and final states stay the same, ∆G does not change.

• Also assuming the state of pyruvate and acetaldehyde will not significantly affect ∆Gr° so can use the

solid values.

∆G

r

o = 1mol∆Gf ,acetaldehyde

o +1mol∆Gf ,CO

2(g)

oéëê

ùûú

- 1mol∆Gf ,pyruvate

oéë

ùû

= -463.4 + (-394.4)éë

ùûkJ - -133.3é

ëùûkJ = -724kJ

B) Le Chatelier’s Principle would say that an increase in pressure will cause the reaction to try to contract

and favor the side with the least number of moles of gas. Since the reactant side has fewer moles of gas,

an increase in P to 100 atm should cause the ratio of pyruvate to acetaldehyde to increase and the

amount of CO2(g) to decrease.

C) • Apply the Clapeyron equation: Convert dG = VdP to d(∆Gr) = ∆V

rdP where ∆V

r=∆n

gasRT

P

∆G°(P1)

∆G°(P2)

ò d(∆Gr

o) =P2

P2

ò ∆Vrd(P) = ∆n

gasRT

P2

P2

òd(P)

PÞ ∆G°(P

2) - ∆G°(P

1) = ∆n

gasRT ln

P2

P1

∆G°(P2) = ∆G°(P

1) + ∆n

gasRT ln

P2

P1

= -724kJ + (1mol)8.314 J

mol-K(298K)ln(100)é

ëùû

= -724kJ + 2477.6(4.605)éë

ùûkJ = (-724 +11.4)kJ = -712.6kJ

The increase in P to 100 atm has made the value of ∆G is less negative (spontaneous), meaning less CO2

will be made at the higher P and the ratio of pyruvate to acetaldehyde to increase,.

4.23 A) • To define the vapor pressure need to use the Clausius Clapeyron equation and let P1 = 1.0

atm, T1 = Tbp°:

lnP = lnPo -

∆Hvap

R

1

T-

1

Tbp

o

é

ë

êê

ù

û

úú

= 0 -23.8X103 J

mol

8.314 J

mol-K

1

300K-

1

272K

é

ëê

ù

ûú

= -2863 -3.30X10-4éë

ùû

= 0.9438 Þ P = e0.9438 = 2.57atm

B) ∆G

vap(P

2) = ∆G

vap(P

1) + RT ln

P2

P1

= 0 + 8.314 J

mol-K(300K)ln(2.57)é

ëùû

= 2497(0.9439)éë

ùûkJ = 2354J = 2.35kJ

• ∆G is positive for the transition from liquid to gas at 2.57 atm, so the liquid is the most stable state and

is why you see liquid in the lighter.

4.24 Given P1 = 4.583 torr, T1 = 0°C and P2 = ? when T2 = -10.5°C


54

lnP

2

P1

= -∆H

sub

R

1

T2

-1

T1

é

ë

êê

ù

û

úú

=-(40.66 + 6.00) kJ

mol

8.314 J

mol-K

1000J

1kJ

æ

èç

ö

ø÷

1

262.6K-

1

273K

é

ëê

ù

ûú

= 5612.2(-1.467X10-4) = -0.8233 ÞP

2

P1

= e-0.8233 = 0.439 Þ P2

= 0.4399 4.583torréë

ùû

= 2.01torr

• So must reduce P to about 2 torr on the solid when T = -10.5°C to spontaneously sublime the solid.

4.25 Apply the Clausius Clapeyron equation to each liquid, letting P1 = 1.0 atm and T1= 298K

lnP = lnPo -∆H

vap

R

1

T-

1

Tbp

o

é

ë

êê

ù

û

úú

= 0 -∆H

vap

R

Tbp

o - T

T(Tbp

o )

é

ë

êê

ù

û

úú

For ethanol: lnP = -

39.32X103 J

mol

8.314 J

mol-K

351.65 - 298

298(351.65)

é

ëê

ù

ûú1

K= - 4729.4 5.12X10-4é

ëùû

= -2.42 Þ

P = e-2.42 = 0.0888atm = 67.5torr

For acetone: lnP = -

30.25X103 J

mol

8.314 J

mol-K

329.65 - 298

298(329.65)

é

ëê

ù

ûú1

K= -3638.4 3.22X10-4é

ëùû

= -1.17 Þ

P = e-1.17 = 0.310atm = 235torr

For n-hexane: lnP = -

28.85X103 J

mol

8.314 J

mol-K

490.9 - 298

298(490.9)

é

ëê

ù

ûú1

K= -3470 1.32X10-4é

ëùû

= -4.576 Þ

P = e-4.58 = 0.0103atm = 7.83torr

The vapor pressures are so different it should be very easy to identify what the liquid is from its vapor

pressure.

4.26 A) Apply the Clausius Clapeyron equation:

lnP

2

P1

= -∆H

vap

R

T1

- T2

T1T

2

é

ë

êê

ù

û

úú

= ln(40) -8000 J

mol

8.314 J

mol-K

300 - 350

300(350)

é

ëê

ù

ûú1

K

= 3.689 - 962.2(- 4.752X10-4) = 3.689 + 0.458 = 4.147 Þ P2

= e4.157 = 63.3torr

B) Instead of assuming ∆Hvap is constant, calculate the new value at 350 K:

∆Hvap

(T2) = ∆H

vap(T

1) + ∆C

p,liq®gas∆T = 8000 J

mol+ (35.0 - 67.0) J

mol-K(350 - 300)K

= 8000J

mol- (1600)

J

mol= 6400

J

mol

Recalculate P2 with the new value of ∆Hvap:

lnP2

= lnP1

-∆H

vap

R

1

T2

-1

T1

é

ë

êê

ù

û

úú

= ln(40) -6400 J

mol

8.314 J

mol-K

300 - 350

300(350)

é

ëê

ù

ûú1

K

= 3.689 - 769.8(- 4.752X10-4) = 3.689 + 0.3658 = 4.054 Þ P2

= e4.054 = 57.7torr

C) The vapor pressure decreases by about 10% when the new ∆Cp term adjusts the ∆Hvap so it is

significant. The effect is more pronounced because of the low ∆Hvap value. If it were in ten’s of kilojoules,

which is more normal, the change in ∆Hvap would have been much smaller and could be neglected.

4.27 A) lnP2

= lnP1

-∆H

vap

R

1

T2

-1

T1

é

ë

êê

ù

û

úú

Þ y = mx + b Let P1 = 1.0 atm, then ln P1 = 0 and T1 = Tbp°

Then lnP2

= -∆H

vap

R

1

T2

-1

Tbp

o

é

ë

êê

ù

û

úú

= -∆H

vap

RT2

+∆H

vap

RTbp

o and lnP

2= y

1

T2

= x -∆H

vap

R= m

∆Hvap

RTbp

o= b

B) Plotting ln P versus 1/T produces:(a) ∆Hvap from: slope = -∆H

vap

RÞ -R ´ (slope) = ∆H

vap and


55

(b) Tfp° from the y-intercept given: y - intercept,b =-1 ´ (slope)

Tbp

oÞ T

bp

o =-1 ´ (slope)

b.

(c) The units of the vapor pressure must be in atm (so that ln P1 = 0) and T in degrees Kelvin.

C) ∆Hvap

= -R ´ slope = -8.314 J

mol-K(-5420.7K) = 45.1 kJ / mol

and Tbp

o =-1 ´ (slope)

b=

+(5420.7)

14.646K-1= 370K

D) • Since the tabled value is 47.0 kJ/mol the calculation underestimates the ∆H by 4.0%

• This calculated Tbp value equals the tabled value.

E) Pc

= 52.62bar1.01atm

1.00bar

æ

èç

ö

ø÷ = 53.15atm then lnP

2= slope

1

T2

æ

èç

ö

ø÷ + b Þ

lnP2

- b

slope=

1

T2

Þ T2

=slope

lnP2

- b So

that: Tc

=slope

lnPc

- b=

-5420.7K

ln(53.15) -14.646=

-5420.7

-10.673= 509K

• Since the tabled value is 537K, the calculation underestimates the Tc by 5.2%. The assumption made in

the original derivation that ∆Hvap stays constant from T1 → T2 is not likely to be correct going from Tbp° to

Tc since the T range is large.

4.28 A) Use the given information to calculate the ∆Hvap from the Clausius Clapeyron equation:

lnP

2

P1

= -∆H

vap

R

T1

- T2

T1T

2

é

ë

êê

ù

û

úúÞ ln

100torr

40torr

é

ëê

ù

ûú =

-∆Hvap

8.314 J

mol-K

250.65 - 268.05

250.65(268.05)

é

ëê

ù

ûúK

0.916(8.314 J

mol-K)

-2.590X10-4 K= -∆H

vapÞ ∆H

vap= 2.94X104 J

mol= 29.4kJ / mol

For the normal boiling point let P1 = 760 torr and T1 = Tbp°

lnP

2

P1

= -∆H

vap

R

1

T2

-1

T1

é

ë

êê

ù

û

úú

Þ ln100

760

æ

èç

ö

ø÷ = -

29.4X103 J

mol

8.314 J

mol-K

1

268K-

1

Tbp

o

é

ë

êê

ù

û

úúÞ -2.028 = -3536.2 3.73X10-3 -

1

Tbp

o

é

ë

êê

ù

û

úú

-2.028 = -13.195 +3536.2

Tbp

oÞ T

bp

o =3536.2

11.167K = 316.7K (43.7°C)

B) Can determine ∆Hvap at 268K using:∆Hvap

(T2) = ∆H

vap(T

1) +[C

p,gas- C

p,liquid]∆T

• Since phase transition is liquid → gas then ∆Cp = (Cp,gas – Cp,liquid)

Calculating the adjusted value for ∆Hvap:

∆Hvap

(T2) = 29.4

kJ

mol+ [45.4 - 73.7] J

mol-K(268 - 250)K

1kJ

1000J

æ

èç

ö

ø÷

æ

èç

ö

ø÷ = 29.4

kJ

mol+ -0.545( ) kJ

mol= 28.85

kJ

mol

lnP

2

P1

= -∆H

vap

R

1

T2

-1

T1

é

ë

êê

ù

û

úú

Þ ln100

760

æ

èç

ö

ø÷ = -

28.85X103 J

mol

8.314 J

mol-K

1

268K-

1

Tbp

o

é

ë

êê

ù

û

úúÞ -2.028 = -3470 3.73X10-3 -

1

Tbp

o

é

ë

êê

ù

û

úú

-2.028 = -12.94 +3470

Tbp

oÞ T

bp

o =3470

10.92K = 317.9K (44.8°C)

C) The calculated boiling point is closer to the true value (46.3°C) when the ∆Cp term is included.

The % error is reduced from 5.6% to 3.2% so it may be worth the extra effort.

4.29 A) Starting point, the triple point T is where the vapor pressure of the solid is equal to that of the

liquid so that since µsolid = µgas and µliquid = µgas, then µsolid = µgas = µliquid

Setting ln P2 equal for the Clausius Clapeyron equations for the solid → gas and liquid → gas

Start with: lnP2

= lnPliquid

-∆H

vap

R

1

Ttriple

-1

Tliquid

é

ë

êê

ù

û

úú

= lnPsolid

-∆H

sub

R

1

Ttriple

-1

Tsolid

é

ë

êê

ù

û

úú

Multiply through the T terms: lnPliquid

-∆H

vap

R

1

Ttriple

é

ë

êê

ù

û

úú

+∆H

vap

RTliquid

= lnPsolid

-∆H

sub

R

1

Ttriple

é

ë

êê

ù

û

úú

+∆H

sub

RTsolid


56

Then “collect like terms”:

lnPliquid

- lnPsolid

= +∆H

vap

R

1

Ttriple

é

ë

êê

ù

û

úú

-∆H

vap

RTliquid

-∆H

sub

R

1

Ttriple

é

ë

êê

ù

û

úú

+∆H

sub

RTsolid

Leads to:

R lnP

liquid

Psolid

é

ë

êê

ù

û

úú

= ∆Hvap

- ∆Hsub( ) 1

Ttriple

é

ë

êê

ù

û

úú

+∆H

sub

Tsolid

-∆H

vap

Tliquid

R lnP

liquid

Psolid

é

ë

êê

ù

û

úú

-∆H

sub

Tsolid

+∆H

vap

Tliquid

= ∆Hvap

- ∆Hsub( ) 1

Ttriple

é

ë

êê

ù

û

úú

→ Ttriple

=∆H

vap- ∆H

sub( )

R lnP

liquid

Psolid

é

ë

êê

ù

û

úú

-∆H

sub

Tsolid

+∆H

vap

Tliquid

4.29 B) (a) For ∆Hvap need to use vapor pressures for liquid:

Given T1 = 356.5 K, P1 = 1.1744 kPa, and T2= 464.85 K, P2 = 53.774 kPa

• Can leave pressure units of kPa, since will be calculating with a pressure ratio of P2/P1

lnP

2

P1

= -∆H

vap

R

T1

- T2

T1T

2

é

ë

êê

ù

û

úú

Þ ln53.774kPa

1.1744kPa

é

ëê

ù

ûú =

-∆Hvap

8.314 J

mol-K

356.5 - 464.85

356.5(464.85)

é

ëê

ù

ûúK

3.824(8.314 J

mol-K)

-1.467X10-4 K= -∆H

vapÞ ∆H

vap= 4.863X104 J

mol= 48.6kJ / mol

For ∆Hsub will need vapor pressures for solid:

Given T1 = 333.4 K, P1 = 0.2448 kPa, and T2=349.86 K, P2 = 0.7893 kPa

lnP

2

P1

= -∆H

sub

R

T1

- T2

T1T

2

é

ë

êê

ù

û

úú

Þ ln0.7893kPa

0.2448kPa

é

ëê

ù

ûú =

-∆Hsub

8.314 J

mol-K

333.4 - 349.9

333.4(349.9)

é

ëê

ù

ûúK

1.171(8.314 J

mol-K)

-1.414X10-4 K= -∆H

vapÞ ∆H

vap= 6.883X104 J

mol= 68.8kJ / mol

(b) ∆Hsub

= ∆Hvap

+ ∆Hfus

Þ ∆Hfus

= ∆Hsub

- ∆Hvap

= 68.8 - 48.6( )kJ / mol = 20.2kJ / mol

(c) To calculate the normal boiling point we need P1 to be 1.0 atm = 101.3 kPa so that T1 = Tbp° and

choose one of the liquid values for P2 and T2. The most useful form of the Clausius Clapeyron to use

would be: lnP

2

P1

= -∆H

vap

R

1

T2

-1

T1

é

ë

êê

ù

û

úú

Þ lnP

2

P°= -

∆Hvap

R

1

T2

-1

Tbp

o

é

ë

êê

ù

û

úú

• You can substitute and solve for Tbp°, or rearrange the equation to solve for Tbp°, whichever you think is

easiest.

lnP

2

P°= -

∆Hvap

R

1

T2

-1

Tbp

o

é

ë

êê

ù

û

úú

Þ ln53.774kPa

101.3kPa

é

ëê

ù

ûú = -

48.6X103 J

mol

8.314J

mol-K

1

464.85-

1

Tbp

o

é

ë

êê

ù

û

úú

ln(0.5308) =-5845.6

464.85+

5845.6

Tbp

oÞ -0.6333 = -12.575 +

5845.6

Tbp

o

12.575 -0.6333( ) =5845.6

Tbp

oÞ T

bp

o =5845.6K

11.94= 489.5K (216.4°C)

(d) For the triple point T:

Ttriple

=∆H

vap- ∆H

sub( )

R lnP

liquid

Psolid

é

ë

êê

ù

û

úú

-∆H

sub

Tsolid

+∆H

vap

Tliquid

=(48.6 - 68.8)X103 J

mol

8.314J

mol-Kln

1.1744

0.7893

æ

èç

ö

ø÷

æ

èç

ö

ø÷ -

68.8X103

349.9

J

mol-K+

48.6X103

356.5

J

mol-K

Ttriple

=-20.2X103

(3.303) -196.65 +136.3K =

-20.2X103

-57.0K = 354K (81.0°C)

4.30 A) (1) (a) A (b) F, H (c) C (2) (a) D (b) E (c) G (3) B (4) I (5) C

B) Region (1): D Region (2): C Region (3): E Region (4): H Region (5): K


57

C)

Point G Point H Point I Point J

X exists as gas only

Two phases exist:

X(l) X(g)

Two phases exist: X(l) X(g)

But dliquid = dgas so the meniscus is not visible

Single phase X(g) in supercritical region

4.31 A) Region would be P ranges 5.12 → 72 atm, and T range is

unlimited, as shown in “A” box.

B) T must be above 31.2°C and P to be in supercritical region,

“B” on diagram.

C) Must be at the Triple point: P = 5.11 atm T = -56.4 °C, point

“C” on diagram.

4.32 A) Phase rule: F = C – P + 2

Points P F Phases in equilibrium Single phase present

A 1 2 S (rhombic)

B 2 1 S (rhombic)+ gas

C 3 0 S (rhombic), S (monoclinic) + Liquid

D 2 1 S (rhombic) + S (monoclinic)

E 1 2 S (monoclinic)

F 2 1 S (monoclinic) + gas

G 2 1 S (rhombic) + S (monoclinic)

H 2 1 S (rhombic)+ Liquid

I I 2 Liquid

J 2 1 S (monoclinic) + Liquid

K 3 0 S (monoclinic) + Liquid + gas

L 2 1 Liquid + gas

B) Starting with an equilibrium between the two solid phases (rhombic, monoclinic) and must change

both P and T to maintain equilibrium (saying there is only one degree of freedom).

C) Increasing T but keeping P constant, D → J → L.

Point D Observe all solid, but would be equilibrium of the two crystalline forms,

monoclinic and rhombic.

Point J Increased T, only some of the solid will have melted to produce liquid, and have equilibrium between monoclinic solid form and liquid.

Point L All solid would be melted and would have liquid in equilibrium with gas.

D) Decreasing P holding T constant, A → D → E → F.

At Point A At Point D

µ(monoclinic) > µ (rhombic)

So only rhombic form stable since it has lower µ value.

µ (monoclinic) = µ (rhombic)

µ (monoclinic) decreased with P decrease so both crystalline forms are stable and coexist.

At Point E At Point F

µ(monoclinic) < µ (rhombic)

Decreased P has decreased µ monoclinic further so that complete conversion from rhombic form has occurred.

µ (monoclinic) = µ (gas)

µ (monoclinic) has decreased with P so that gas state has same potential as the solid. The monoclinic form produces its vapor pressure.

E) Point C at 95.4 °C meets the criteria of a triple point in that there are three phases in equilibrium, with

the two crystalline forms in equilibrium with the gas phase, so that µ(monoclinic) = µ(rhombic) = µ(gas).

It is also the lowest temperature at which the rhombic form spontaneously converts to the monoclinic

Super-critical fluid

Gas

Liquid

Solid

T(°C) à31.1-56.4-78.5

Pre

ssure

(atm

) à

73.0

5.11

1.00

65.0 atm

80.0 atm

B

A

C


58

form, so it is considered the transition temperature. But 119°C would be considered the traditional triple

point because µ(solid, monoclinic) = µ(liquid) = µ(gas).

F) Point L at 445°C is when µ(liquid) = µ(gas) at 1.0 atm so it would be considered the normal boiling

point for sulfur.

G) The normal melting point would be the T at which µ(liquid) = µ(solid) when P = 1.0 atm and that is

Point J on the diagram.

4.33 A) Answers:

Shifts (1) (2) (3) (4) (5)

A) isobaric isothermal Neither isobaric isobaric

B) No change No change Liquid appears in the tube

No change Both solid and liquid in tube

(6) (7) (8) (9) (10)

A) Neither Neither isothermal Neither isobaric

B) Liquid disappears, only solid in tube

Solid + Liquid + gas in tube

Solid + liquid only in tube

Liquid + gas, but no solid in tube

No change

C) Boundaries will appear in shifts (3), (5), (7), (8) and (9)

4.34 A) Sketch of diagram should look something like that given below:

4.35 A) False

Don’t need equal amounts of the two phases, just that the chemical potentials are the same. The key

factors were ∆Sm and ∆Vm, as you can see from the map.

B) True

If heating ∆Str would be (+) but if cooling ∆Str would be (-). ∆Gtr would be 0, but not ∆Str since there is

no instance where ∆Htr ÷ T =0.

C) False

The Clapeyron equation applies to the solid liquid transition, not Clausius Clapeyron. The Clausius

Clapeyron only applies to a phase transition involving a gas as the final or initial state.


59

4.36 A) This equilibrium line is described by the Clapeyron equation where dP/dT = group of constants,

so it is a linear equation. Whether the slope is positive or negative depends on the sign of ∆Vm = Vm,liquid -

Vm,solid which in turn is set by the relative densities of the two states.

B) Since Vm,gas >> Vm,liquid , ∆Vm = Vm,liquid - Vm,solid is always positive so the slopes are positive. The

lines follow the Clausius Clapeyron equation, so the equilibrium line is a log function, and the slope is not

equal to a group of constants and changes value as T changes.

C) The chemical potentials must be equal.

D) (a) It is the triple point. (b) The chemical potentials of the 3 phases must be equal.

(c) Because you cannot change either P or T and maintain the three way equilibrium. Even if you changed

them both, you could not replicate the equilibrium so there is only one possible combination of P and T

for the triple point.

4.37 A) (a) It is also consistently greater than 1.0, like Z.

(b) Yes, the fugacity coefficient is less than 1.0 for CO2 at the pressures shown. The crossover to the

values greater than 1.0 can’t be seen on the fugacity graph since it does not go to as high enough

pressures, but it would likely show the same behavior.

B) The trend for N2 and H2 similar since they are both small molecules with very weak attractive forces

between the molecules. However, both CO2 and NH3 are larger molecules with polar bonds. NH3 has a

permanent dipole, resulting in strong attractive forces. (The H-bonding may also play a role, but that

requires more organization between the gaseous molecules.) CO2 does not have a permanent dipole, but

has temporary dipoles and a shape that would allow the molecules to cluster. So it has the potential for

attractive forces being stronger than the repulsive forces even at low pressures.

C) ∆µ would be positive when Φ is greater than 1.0 since ln Φ would be positive. Similarly, ∆µ would be

negative when Φ is less than 1.0 since ln Φ would be negative.

4.38 A) f = f ´ P = (e-0.03379)(20.0atm) = 0.9668(20.0atm) = 19.3atm which is a 3.5% decrease

And f = f ´ P = (e-0.3378)(200atm) = 0.7133(200atm) = 143.7atm which is a 28.2% decrease

B) For 20 atm: µreal

- µideal

= RT lnf = 8.314J

mol - K(423K)(-0.03389) = -119

J

mol

For 200 atm: µreal

- µideal

= RT lnf = 8.314J

mol - K(423K)(-0.3378) = -1188

J

mol

4.39 A) In the Van der Waals result: For the Virial equation result:

lnf =bP

RT-

aP

(RT)2Þ

L

mol(atm)

L - atm

mol - K(K )

-

L2 - atm

mol2(atm)

L2 - atm2

mol2 - K2

æ

èçç

ö

ø÷÷(K2 )

lnf =BP

RT=

L

mol(atm)

L - atm

mol - K(K )

• So no units for ln Φ if: R = 0.08206 L-atm/K-mol,

b in L/mol, P in atm, and T in Kelvin

• So no units on ln Φ if: R = 0.08206

L-atm/K-mol, B in L/mol. P in atm,

and T in Kelvin.

B) Results of calculations:

Van der Waals P ideal = 10 atm

T(°C) ln ϕ ϕ Fugacity,ϕP ∆µ = RTln ϕ

25 -0.05222 0.9491 9.49 -129.5 J/mol

150 -0.02230 0.9780 9.78 -78.4 J/mol

300 -0.00978 0.9903 9.90 -46.6 J/mol

Virial Equation

T(°C) ln ϕ ϕ Fugacity,ϕP ∆µ = RTln ϕ

25 -0.10831 0.8973 8.97 -268.5 J/mol

150 -0.02886 0.9716 9.72 -101.5 J/mol

300 -0.00970 0.9904 9.90 -46.2 J/mol

C) (a) Both equations produce the same values at 300 K, and very close values for the fugacity at 150°C,

but not ∆µ. But at 25°C the result for the virial equation is twice that of the Van der Waals value for ∆µ


60

and the fugacities are not that close. They definitely both agree on the sign of ln Φ and the ∆µ value at all

three temperatures. At all three temperatures, µreal is less than µideal.

(b) ∆µ is decreasing as T is increased indicating that at some higher T, the repulsive forces will be

countered balancing the attractive forces (discussed earlier as the Boyle T).

(c) As T increases, NH3 is becoming

more like an ideal gas.

4.40 A) If B is negative, then the

result of the integral will be

negative, meaning Φ will be less

than 1.0 and the effective pressure

will be less than the expected ideal

gas pressure. When it becomes

positive (350K and above) the real

pressure will exceed the ideal gas

pressure.

B) Calculated results: (see table)

C) (a) The ∆µ values are getting

smaller as the P is lowered. (See

table)

(b)

(c) The T regions where the attractive and repulsive forces dominate definitely change with P.

T P = 100 atm P = 20 atm P = 2.0 atm

200 K Attractive forces Attractive forces Attractive forces

273 K Attractive forces Attractive forces Neither dominating

350 K Attractive forces Neither dominating Neither dominating

450 K Repulsive forces Repulsive forces Neither dominating

550 K Repulsive forces Repulsive forces Neither dominating

(d) Yes, the behavior observed with the P decrease fits the figure shown. However, depending on the

temperature, when pressure is constant, the µreal could also be greater than, equal to or less than µideal,

for a range of temperatures. So a plot of µ versus T(K) would look somewhat the same as the µ versus P

plot, except the µreal > µideal values will appear at low temperatures, whereas they should appear at

higher pressures as in the figure shown. Increasing T increases kinetic energy and makes it harder for the

molecules to cluster.

P µreal > µideal µreal < µideal µreal = µideal

100 350 → 550K 200, 273 K None listed

20 450 → 550 K 200, 273 K ≈ 350 K

2 None listed 200 K ≈ 273 → 550K


61

PART 5: PHASE DIAGRAMS FOR MIXTURES, IDEAL AND REAL SOLUTIONS AND COLLIGATIVE

PROPERTIES

5.1 Since the gases are at the same P, T the molar ratio matches the volume ratio, so that:

A) V

N2

Vtotal

=n

N2

ntotal

= cN

2

=50L

100L= 0.50 and c

O2

=50L

100L= 0.50

(a)

∆Gmix

= RT cN

2

lncN

2

+ cO

2

lncO

2

éëê

ùûú

= 2 8.314J

mol - K(295K)

æ

èç

ö

ø÷ 0.50ln(0.50)éë

ùû

= 2 2.453kJ

mol

æ

èç

ö

ø÷ (-0.693) = -1.70kJ

(b) ∆Smix

= -∆G

mix

T= -

(-1700J

mol)

295K= 5.76

J

mol - K

B) If the P of the mixture deceases, but the composition doesn’t change, than the ∆Gmix and ∆Smix will not

change.

5.2 The molar ratio is set by the volume ratio, so that:

A) V

CO2

Vtotal

=n

CO2

ntotal

= cCO

2

=4.0L

44.0L= 0.0909 and c

O2

= cHe

=20L

44L= 0.4545

B) ntot

=P

totV

tot

RT=

1.0atm(44.0L)

0.08206L - atm

mol - K(300K)

= 1.80mol

∆Gmix

= ntot

RT cCO

2

lncCO

2

+ cO

2

lncO

2

+ +cHe

lncHe

éëê

ùûú

= 1.80mol 8.314J

mol - K(298K)

æ

èç

ö

ø÷ 0.0909ln(0.0909) + 2 0.4545ln(0.4545)( )éë

ùû

= 1.80mol 2.4776kJ

mol

æ

èç

ö

ø÷ (-0.9338) = -4.16kJ

5.3 A) Start with knowing you need to add “x” liters of pure O2 to the 100 L to change the ratio:

21.0 + x

100 + x=

32

100Þ 68x = 110L Þ x = V

O2

needed =110

68L = 16.18LO

2

B) (a) The ∆Gmix increases up to a 50/50 mixture and then decreases. The 32% mixture of O2 is closer to

that maximum, so we should see that ∆Gmixx becomes more negative than ∆Gmix for air.

(b)

∆Gmix

,air = RT cN

2

lncN

2

+ cO

2

lncO

2

éëê

ùûú

= 8.314J

mol - K(298K)

æ

èç

ö

ø÷ 0.21ln(0.21) + 0.79ln(0.79)éë

ùû

= 2477.5J

mol(-0.514) = -1273 J / mol

∆Gmix

,Nitrox I = RT cN

2

lncN

2

+ cO

2

lncO

2

éëê

ùûú

= 8.314J

mol - K(298K)

æ

èç

ö

ø÷ 0.32ln(0.32) + 0.68ln(0.68)éë

ùû

= 2477.5J

mol(-0.627) = -1553 J / mol

So the ∆Gmix for Nitrox I is more negative by 280 J/mol than ∆Gmix for air.

5.4 A) Need mol fraction of O2 to define pressure ratio: P

O2

Ptotal

=n

O2

ntotal

. Choose 100 g of Trimix:

• Must calculate moles of other gases to get ntotal


62

26.0gO2(g)

1molO2

32.0g

é

ë

êê

ù

û

úú

= 0.8125molO2(g) 17.0gHe(g)

1mol He

4.0g

é

ëê

ù

ûú = 4.25molHe(g)

57.0gN2(g)

1mol N2

28.0g

é

ë

êê

ù

û

úú

= 2.036molN2(g) n

total= 0.8125 + 4.25 + 0.2868 = 7.098mol

cO

2

=0.8125mol

7.098mol= 0.1145 P

total=

PO

2

cO

2

=120kPa

0.1145= 1048.0kPa

1.0atm

101.3kPa

é

ëê

ù

ûú = 10.35atm

B) • To get the partial pressures and will need the mole fractions of He and N2.

cHe

=4.25mol

7.098mol= 0.5988 c

N2

=2.036mol

7.098mol= 0.2868

PHe

= cHe

Ptotal

= 0.5988(1048kPa) = 628.6kPa1.0atm

101.3kPa

é

ëê

ù

ûú = 6.21atm

PN

2

= cN

2

Ptotal

= 0.2868(1048kPa) = 300.6kPa1.0atm

101.3kPa

é

ëê

ù

ûú = 2.97atm

C)

∆Gmix

Trimix = RT cO

2

lncO

2

+ cHe

lncHe

+ +cN

2

lncN

2

éëê

ùûú

= 8.314J

mol - K(298K)

æ

èç

ö

ø÷ 0.1145ln(0.1145) + 0.5988ln(0.5988) + 0.2968ln(0.2968)éë

ùû

= 2.4776kJ

mol

æ

èç

ö

ø÷ (-0.913) = -2.26kJ

5.5 A) Since both gases are at the same Pand T, the molar ratio will equal the volume ratio.

∆Gmix

= RT cN

2

lncN

2

+ cXe

lncXe

éëê

ùûú

= 8.314J

mol - K(298K)

æ

èç

ö

ø÷ 0.667ln(0.667) + 0.333ln(0.333)éë

ùû

= 2477.5J

mol(-0.643) = -1593 J / mol = -1.59kJ / mol

∆Smix

= -∆G

mix

T= -

(-1593J

mol)

298K= 5.35

J

mol - K

B) (a) Because the moles of gas on each side changes, the ∆Gmix and ∆Smix will have to change in value.

• Need to calculate the moles of each gas to get mol fractions, using initial conditions

• Assume ideal gas behavior.

nN

2

=P

1V

1

RT=

4.0atm(3.0L)

0.08206L - atm

mol - K(298K)

= 0.4907mol N2 n

Xe=

P1V

1

RT=

2.0atm(2.0L)

0.08206L - atm

mol - K(298K)

= 0.1636mol Xe

Then: cN

2

=0.4907mol

0.6543mol= 0.7499 c

Xe= 1 - 0.7499 = 0.2501

∆Gmix

= RT cN

2

lncN

2

+ cXe

lncXe

éëê

ùûú

= 8.314J

mol - K(298K)

æ

èç

ö

ø÷ 0.750ln(0.750) + 0.250ln(0.250)éë

ùû

= 2477.5J

mol(-0.562) = -1393 J / mol = -1.39kJ / mol

∆Smix

= -∆G

mix

T= -

(-1393J

mol)

298K= 4.67

J

mol - K

• So both ∆Gmix and ∆Smix decrease

(b) ∆Gmix will not be the same total ∆G, as in (A), since there will be a pressure change for each gas,

which will produce new values for the ∆G of each gas.


63

Since ∆G is a state function we can make the changes step-wise and calculate each change

separately: ∆Gtotal

= ∆Gmix

+ ∆GN

2,∆P

+ ∆GXe,∆P

and G(P2) = G(P

1) + RT

P1

P2

òdP

PÞ ∆G = nRT ln

P2

P1

• Will need to calculate final pressure (P2) for both gases and then new ∆G value.

P2

=n

TOTRT

mix

Vmix

=

0.6543mol (0.08206L - atm

mol - K)(298K)

5.0L= 3.20 atm

∆GN

2

= nRT lnP

2

P1

= 0.4907mol 8.314J

mol - K(298K)

é

ëê

ù

ûú ln

3.20

4.0

é

ëê

ù

ûú = -271 J

∆GXe

= nRT lnP

2

P1

= 0.1636mol 8.314J

mol - K(298K)

é

ëê

ù

ûú ln

3.20

2.0

é

ëê

ù

ûú = 190 J

∆Gtotal

= ∆Gmix

+ ∆GN

2,∆P

+ ∆GXe,∆P

= 0.6543mol(-1393J

mol)

é

ëê

ù

ûú + (-271J) + (190) = -992 J

• Since the total ∆G is still negative, the gases will spontaneously mix and change pressure.

5.6 A) Since ∆Gmix at any T depends only on mole fractions of the solutes, the most negative value for

∆Gmix occurs when chexane

= cheptane

= 0.500 and the molar ratio is 1:1.

B) Molar mass C6H14 = 86.18 g/mol and C7H16 = 100.21 g/mol then for a molar ratio is 1:1, then:

%massC6H

14=

86.18g

(100.21 - 86.14)g´100 = 46.2% %massC

7H

16= 100 - 46.2 = 51.76%

C) Molar volumes: C6H14: C7H16:

Vm,C

6H

14

=MW

d=

86.18g

mol

655g

L

= 0.1316L

mol= 131.6

mL

mol V

m,C7H

16

=MW

d=

100.21g

mol

685g

L

= 0.1463L

mol= 146.3

mL

mol

So for a 1:1 mixture: Vtotal = 131.6 + 146.3 = 278 mL since volumes additive

• To adjust to a total of 250 mL for the 1:1 mixture , set a proportion:

VC

6H

14

needed

131.6mL=

250mL

278mLÞV

C6H

14

needed = 131.6mL250mL

278mL

é

ëê

ù

ûú = 118.3mLC

6H

14

VC

7H

16

needed = 146.3mL250mL

278mL

é

ëê

ù

ûú = 131.7mLC

7H

16

5.7 A) Water is the solvent and at χB = ≈ 0.30 there are 7 water molecules for every 3 CH3OH

molecules. The contraction occurs because the methyl group of CH3OH cannot H-bond so it is disrupting

water’s organization. The water molecules have multiple H-bonding sites and small size, so is highly

structured and CH3OH is causing that structure to partially collapse on itself.

B) (a) Pure molar volumes: VA* =

18.01g

mol

0.9982g

cm3

= 18.04cm3

mol ; V

B* =

32.04g

mol

0.7872g

cm3

= 40.703cm3

mol

(b) From graph: ∆VA

= -0.35cm3

mol and ∆V

B= -2.0

cm3

mol so that V

A= 18.04 + (-2.0) = 16.0

cm3

mol and

VB

= 40.703 + (-0.40) = 40.3cm3

mol Both agree with open circle values on second graph.

C) (a) Approximately 0.90 - 0.95 cm3/mol from the graph at χ CH3OH = 0.70

(b) From tangent: ∆VA

= -2.0cm3

mol and ∆V

B= -0.40

cm3

mol so that:


64

(c) VA

= 18.04 + (-2.0) = 16.0cm3

mol and V

B= 40.703 + (-0.40) = 40.3

cm3

mol.

Which agree well with red circle values on the second graph.

5.8 A) (a) The solution is contracting since negative molar volume changes are observed throughout

all the mole fractions measured.

(b) Expect that since both water (H2O) and isopropanol (CH3CHOHCH3) have H-bonding as intermolecular

forces, the A···B interactions would be similar in strength to A···A and B···B interactions and the two

substances should form a mixture. The ∆Gmix overall should then be negative so that mixing is

spontaneous.

(c) Since the isopropanol has only one OH group centered between bulky methyl groups, the H-bonding

network in water is being disrupted by the alcohol molecules (as in the CH3OH and C2H5OH mixtures

shown previously). Then as the mole fraction of the alcohol increases, the water molecules between the

alcohol molecules allow them to be closer together than they would be in pure isopropanol.

B) (a) Need to calculate pure molar volume for isopropanol since: Videal

= cH

2OV *

H2O

+cisop

V *isop

V *isop

=

60.096g

mol

0.7804g

cm3

= 77.01cm3

mol then V

ideal= 0.6971(18.054

cm3

mol) + 0.3029(77.01

cm3

mol) = 35.71

cm3

mol

(b) Vsoln

= Videal

+ ∆Vmix

= 35.71 + (-0.9842)cm3

mol= 34.93

cm3

mol

(c) Visop

=V

soln- c

H2OV

H2O

cisop

=

34.93cm3

mol- 0.6971(17.154)

cm3

mol

0.3029= 75.83

cm3

mol

Since the pure value is 77.01, this is a significant contraction in partial molar volume ≈ 1.5%.

5.9 A) V *H

2O

=

18.02g

mol

0.9982g

cm3

= 18.05cm3

mol and V *

ethanol=

46.07g

mol

0.7893g

cm3

= 58.37cm3

mol

B) (a) Only from about 0.0 to 0.1 (b) No mol fraction at which it exceeds its pure molar volume

C) (a) Expected A···B would be H-bonding since both water and ethanol can H-bond.

(b) No, we would expect H-bonding throughout.

D) The partial molar volumes show that water’s network has

expanded, and ethanol’s collapsed into much lower volume, so it is

likely the waters are staying connected, but expanding the network

to include the CH3CH2OH molecule, as shown in the figure. But as

the number of molecules of ethanol increases, the original water

network cannot accommodate replacing a water molecule with 4

binding sites with ethanol that only has two H-bonding sites.

Consequently, the H-bonding network of water collapses and

decreases the partial molar volume.

5.10 A) • First calculate moles of ethanol needed, then use pure molar volume to get volume needed.

50.00mLH2O

1mol H2O

18.05mL

é

ë

êê

ù

û

úú

= 2.77mol H2O and

nH

2O

ntotal

= cH

2O

= 0.600 =2.770

2.770 + nethanol

So that:

nethanol

=(2.770 -1.662)

2.770= 1.847mol and V

C2H

5OH

= 1.847mol58.37mL

1molC2H

5OH

é

ë

êê

ù

û

úú

= 107.8mLC2H

5OH

B) Since volumes add together, if ideal:Vsoln

= nH

2OV

H2O

+ nC

2H

5OH

VC

2H

5OH

= 50.0 +107.8 = 157.7mL

• Then using the actual partial molar volumes, calculate true volume of solution:


65

Vsoln

= nH

2OV

H2O

+ nC

2H

5OH

VC

2H

5OH

= 2.770mol(17.4mL

mol) +1.847mol(56.5

mL

mol) = 48.2 +104.4 = 152.6mL

So the difference is -5.14 mL = ∆Vmix which will be the total excess volume observed.

Per mole the value would be: ∆V

mix

ntotal

=-5.14cm3

(2.77 +1.845)mol= -1.11

cm3

mol

C) Yes, the graph shows a molar excess volume close to -1.1 cm3/mol at χC2H6O = 0.400.

D) Yes, it should be visible. Since the volume contraction is about 5.1 mL, the total volume should be one

marking lower than what you expected for an ideal solution.

5.11 A) • Use density and total weight of solution to get actual volume

Vsoln

=total wt.

density=

(64.084 + 43.65)g

0.7851g / cm3= 137.22cm

3

B) (a) V *CH

3OH

=

32.04g

mol

0.7872g

cm3

= 40.70cm3

mol and V *

(CH3)2CHOH

= 77.01cm3

mol

(b) 64.084g1molCH

3OH

32.04g

é

ë

êê

ù

û

úú

= 2.000molCH3OH 43.65g

1mol (CH3)2CHOH

60.096g

é

ë

êê

ù

û

úú

= 0.7263mol(CH3)2CHOH

Then: Videal

= nCH

3OH

VCH

3OH

* + nisop

Visop

* = 2.000mol(40.70cm3

mol) + 0.7263mol(77.01

cm3

mol) = 137.33cm

3

C) (a) dideal

=total wt.

Videal

=(64.084 + 43.65)g

137.33cm3= 0.7845g /cm

3

• Actual density is slightly greater at 0.7851 g/cm3

(b) Solution has contracted, since there is more mass per unit volume than expected.

(c) The solution is very near ideal at this composition since the contraction is very small. Both molecules

have LDF interactions between the methyl groups in the pure state as well as one OH group for H-

bonding so forces very similar in both so the solution is very close to ideal. The different shapes of the

molecules is probably causing the “normal” H-bonding and LDF networks in the pure liquids to be

disrupted slightly are the two alcohols are mixed.

5.12 A) (a) H-bonding (b) Dipolar forces

(c) Dipolar forces, unless the waters molecules H-bond to the O in the C=O bond.

B) (a) It is contracting, so Vsoln < Videal at all mole fractions of acetone.

(b) cacetone

= 0.400so there are 6 molecules of H2O for every 4 molecules of (CH3)2CO (or 3:2)

C) (a) water (b) At about 0.20

(c) That the acetone···water interactions are stronger than the forces in pure

water, so energy is released. It is consistent with some form of a H-bonded

complex cluster (such as that in figure on the right) occurring between water and

acetone when the mole fraction of acetone is low.

D) (a) acetone

(b) That the acetone···water interactions are weaker than the forces in pure

acetone so acetone largely interacts with itself in the solution arrangements. This is

consistent since if you need more water molecules than acetone to form the cluster and at high mole

fraction acetone, there are too few waters to form the cluster.

5.13 A) (a) Dipolar forces in pure chloroform (b) Dipolar forces in pure acetone

B) (a) It is less than the ideal volume up to about cacetone

» 0.7 (contracts), then the volume of solution

becomes greater than ideal (expands).


66

(b) The maximum occurs near cacetone

» 0.3, where there are about 7 chloroform molecules for every 3

acetone molecules.

C) (a) When the ratio is about 1:1 or cacetone

» 0.5 .

(b) The chloroform-acetone interaction is stronger than those in either pure chloroform or acetone.

D) (a) Since H-bonding interactions are stronger than dipolar forces, energy would be released and

produce the exothermic ∆Hmix values, if they are being formed in the solution. The maximum loss should

appear at a molar ratio of 1:1 and the data indicates that.

(b) It would be very unusual since H’s bonded to carbon do not participate in H-bonding. Only H’s bonded

to O or N are generally involved, so this would be a unique situation.

5.14 A) The maximum would be when cA

= cB

= 0.5, so that:

∆Gmix

= RT cAlnc

A+ c

Blnc

Béë

ùû

= 8.314J

mol - K(298K) 2 ´ (0.5ln(0.5))é

ëùû

= -1861J

mol and

T ∆Sideal

= -∆Gideal

= 1861J

mol

B) Rearrange definition of GE: GE = ∆Gsoln

- ∆Gideal

Þ ∆Gsoln

= ∆Gideal

+ GE , then:

Mixture I (CHCl3/C6H14) III ((CH3)2CO / CHCl3) IV (C2H5OH/ C6H14)

Estimated GE (J/mol) 250 -500 ≈ 1500

∆Gsoln (J/mol) 250 – 1861 = -1611 -500 – 1861 = -2361 1500 – 1861 = -361

(b) In all cases, the ∆Gsoln is negative, so a solution will form. But the C2H5OH/ C6H14 is the least

spontaneous and at -0.4 kJ/mol of mixture, is much less than the ideal value of 1.86 kJ/mol.

C) Considering each comparison:

Compare: Trends in HE Forces between A···B Trends in SE

(a)

I

CHCl3/

C6H14

and

IV

C2H5OH/

C6H14

Both mixtures show

endothermic values

for HE.

In I: The maximum

is lower at 750 J/mol

at cCHCl

3

= 0.50.

In IV: The maximum

occurs earlier at

cC

2H

5OH

= 0.25 and is

about 1000 J/mol

In I: The HE indicates A···B is

weaker, so the interaction

between them likely dipole-

induced dipole, so energy has

added to overcome the dipolar

forces in CHCl3

In IV: The endothermic HE

indicates the interaction A···B is

weaker than the H-bonding in

C2H5OH. So more energy has to

be added to overcome the H-

bonding and which is consistent

with the higher value of HE.

In I: The SE is positive,

disorder increases with mole

fraction of CHCl3 and the

maximum occurs at 1:1

composition, as in ideal

solution. So molecules are

staying independent.

In IV the situation is very

different: The SE is first

positive, (disorder increases) at

low mole fraction C2H5OH, but

then it becomes negative

indicating some H-bonding

network is returning as the

number of C2H5OH molecules

increases.

(b)

II

(CH3)2CO/

CH3OH

and

III

(CH3)2CO/

CHCl3

In II, HE is

endothermic all mole

fractions of (CH3)2CO.

In III, HE is

exothermic at all

mole fractions of

(CH3)2CO.

In II: The endothermic HE with

1:1 as the maximum suggest no

clustering is occurring, so the

interaction for CH3OH-(CH3)2CO

are likely just dipolar. (as seen

in V)

In III: Pure (CH3)2CO forms a

complex with CHCl3 as discussed

in 5.12. So although the C=O

can H-bond with the H on CHCl3,

it does not H-bond with the OH

group in CH3OH.

In II: The SE is positive all

mole fractions of acetone, the

solvent and solute interactions

are being disrupted as

expected and no clustering

occurring.

In III: The SE is negative, and

maximum order occurs at 1:1

composition, which is

consistent with a 1:1 complex

formation.


67

The two mixtures clearly represent opposite situations for acetone as the solute, which is

why the properties of the mixture can be measured but not easily predicted.

(c)

V

C2H5OH/

CHCl3

and

VI

C2H5OH/

H2O

In V, HE is

endothermic at low

mole fractions and

exothermic at high

mole fractions of

C2H5OH.

In VI, HE also shows

two regions. It is

exothermic at up to

about 0.5 mole

fraction of C2H5OH

and close to zero

from 0.5 to 1.0 mole

fraction.

In V: Pure C2H5OH has H-

bonding and CHCl3 has dipolar

forces. The endothermic HE

suggests that when CHCl3 is the

solvent, the C2H5OH-CHCl3

interactions are weaker, as

would be expected since dipolar

interactions are likely. But when

C2H5OH outnumbers CHCl3 some

clustering may occur producing

the exothermic HE.

In VI: Both C2H5OH and H2O

have H-bonding forces in their

pure states. The exothermic HE

indicates that some clustering

occurs when H2O outnumbers

C2H5OH, but they mix with no

extra energy needed or released

when the C2H5OH to H2O ratio

exceeds 1:1 where HE is zero.

In V: The excess entropy

shows two regions. Initially the

entropy increases as C2H5OH

added to CHCl3, consistent with

no clustering, but then changes

to negative values at higher

mole fractions of C2H5OH, This

is consistent with some

clustering when C2H5OH

outnumbers CHCl3.

In VI: The excess entropy is

negative at all mole fraction of

C2H5OH so more ordering is

occurring in the solution than is

the pure states. This would be

consistent with clustering

occurring between C2H5OH and

H2O.

5.15 A) For the acetone, (CH3)2CO /H2O mixture: As the temperature is increased, HE becomes less

endothermic at low mole fraction of acetone and more exothermic at higher mole fractions. The transition

occurs at lower values of mole fractions at higher temperature.

For the acetonitrile, CH3CN /H2O mixture: As the temperature is increased, HE becomes more

endothermic. Both changes are consistent with a positive term being added to HE as the T is increased.

B) Kirchhoff’s Law:∆H(T2) = ∆H(T

1) +

T1

T2

òCpdT

C) The heat capacity of the mixture at each composition from: HE(T2) = HE(T

1) + C

mix∆T .

5.16 A) The P and T dependence of G (or ∆G): dG

dP

é

ëê

ù

ûúT

= -V and d(G /T)

dT

é

ëê

ù

ûú = -

H

T 2

B) It changes the units on the righthand and lefthand sides to K-1 which can be easier to apply or plot.

C) The conversion factor 101.3 J/L-atm is needed for VE.

D) V E

RT=

-0.15 cm3

mol

8.314 J

mol-K(298K)

´101.3 J

1L - atm

æ

èç

ö

ø÷ ´

1.0L

1000cm3

æ

èç

ö

ø÷ = -6.13X10-6 atm-1 and

d(GE / RT)

dT

é

ëêê

ù

ûúú

=HE

RT 2=

-1900 J

mol

8.314 J

mol-K(298K)2

= -0.0214K-1

Since the volumes of liquids are generally not influenced by pressure changes, it is not surprising that the

T dependence of GE is much larger than the P dependence.

5.17 A) Both molecules are weakly polar and would have dipolar forces so that A···B ≈ A···A ≈ B···B.

B) Using:Ptotal

= cC

2H

5ClP

C2H

5Cl

* + cC

2H

5BrP

C2H

5Br

* nC

2H

5Cl

= 100.0g1.00mol

92.57g

é

ëê

ù

ûú = 1.080molC

2H

5Cl

nC

2H

5Br

= 100.0g1.00mol

137.02g

é

ëê

ù

ûú = 0.7298molC

2H

5Br c

C2H

5Cl

=1.080mol

1.810mol= 0.597 c

C2H

5Br

= 1.00 - 0.597 = 0.403

PC

2H

5Cl

* = 16.84kPa1L - atm

101.3 J

æ

èç

ö

ø÷

1.00atm

760torr

æ

èç

ö

ø÷ = 126.3torr


68

Ptotal

= cC

2H

5ClP

C2H

5Cl

* + cC

2H

5BrP

C2H

5Br

* = 0.597(126.3) + 0.403(52.34) = 75.4 + 21.1= 96.5 torr

C) yC

2H

5Br(vapor)

=P

C2H

5Br(g)

Ptotal

=21.1torr

96.5torr= 0.219

D)

∆Gmix

= RT cC

2H

5Cl

lncC

2H

5Cl

+ cC

2H

5Br

lncC

2H

5Br

éëê

ùûú

= 8.314 J

mol-K(298K) 0.579ln(0.579) + 0.403ln(0.403)é

ëùû

= 2.519kJ

mol(-0.674) = -1.70

kJ

mol

5.18 A) Both molecules have the same basic shape (tetrahedral) and similar size so that the contact

area (which affects LDF) should be about the same. CHCl3 is weakly polar, whereas CCl4 is non-polar, but

expect A···B as an LDF would be very close in strength to the forces with the pure components.

B) • Use the liquid composition and Raoult’s Law to get partial pressure for each component, Ptotal and

then calculate the mole fractions in vapor. Given: PCHCl

3

* = 26.54kPa, PCCl

4

* = 15.27kPa

Since mole ratio in liquid is 2 mols CHCl3:1 mol CCl4 so that cCHCl

3

= 0.667 and cCCl

4

= 0.333, so that:

Ptotal

= cCHCl

3

PCHCl

3

* + cCCl

4

PCCl

4

* = 0.667(26.54kPa) + 0.333(15.27kPa) = 17.70 + 5.09= 22.79kPa

Then: yCHCl

3

=P

CHCl3(g)

Ptotal

=17.70kPa

22.79kPa= 0.777 and y

CCl4

= 1.0 - 0.777 = 0.223

C) The molar ratio in vapor is 0.777 CHCl3:0.223 CCl4 = 3.5:1.0 or 7 mol CHCl3 for every 2 mol CCl4

which is much higher than the 2:1 ratio in the liquid. So the vapor is much “richer” in CHCl3.

D) For the mixture to boil, Ptotal must equal atmospheric pressure, so the atmospheric pressure would

have to be lowered to 22.79 kPa or 171 torr.

5.19 A)

Ptotal

= chexane

Phexane

* + coctane

Poctane

* == Poctane

* + chexane

(Phexane

* - Poctane

* )

666torr = (354) + chexane

(1836 - 354) Þ chexane

=(666 - 354)torr

1482torr= 0.210

B) yhexane

=385.6torr

665.3torr= 0.580 and y

octane= 1.0 - 0.580 = 0.420

C) (a) For the solution phase:

∆Gmix

(liquid) = RT chexane

lnchexane

+ coctane

lncoctane

éë

ùû

= 8.314J

mol - K(373K)

æ

èç

ö

ø÷ 0.210ln(0.210) + 0.790ln(0.790)éë

ùû

= 3101J

mol(-0.328 + -0.186) = -1594 J / mol = -1.54kJ / mol

(b) For the vapor phase:

∆Gmix

(vapor) = RT yhexane

lnyhexane

+ yoctane

lnyoctane

éë

ùû

= 8.314J

mol - K(373K)

æ

èç

ö

ø÷ 0.580ln(0.580) + 0.420ln(0.420)éë

ùû

= 3101J

mol(-0.316 + -0.364) = -2111 J / mol = -2.11kJ / mol

∆Gmix (liquid) and ∆Gmix (vapor) should not be equal since the mole fractions of each component are

different.

5.20 A) Start with Ptotal

=P

A

*PB

*

PA

* + (PB

* - PA

*)yA

, then multiply through by denominator:

Ptotal

PA

* + (PB

* - PA

*)yA

éë

ùû

= PA

*PB

* Þ PA

*Ptotal

+ PB

*yAP

total- P

A

*yAP

total= P

A

*PB

*

PA

*Ptotal

- PA

*PB

* = PA

*yAP

total- P

B

*yAP

total= y

AP

totalP

A

* - PB

*( ) yields: P

A

*Ptotal

- PA

*PB

*

Ptotal

PA

* - PB

*( )= y

A

B) (a) Rearranging: Ptotal

= cacetone

Pacetone

* + cCCl

4

PCCl

4

* gives:


69

PCCl

4

* =P

total- c

acetoneP

acetone

*

cCCl

4

=8.76 - 0.520(11.24)( )atm

0.480= 6.08atm

(b) yA

=P

A

*Ptotal

- PA

*PB

*

Ptotal

PA

* - PB

*( )=

11.24(8.76)atm2 -11.24(6.08)atm2

8.76atm 11.24 - 6.08( )atm= 0.667

(c) Actual versus calculated (ideal value): %difference =0.667 - 0.631

0.631´100 = 5.7%

(d) gacetone

=P

acetone,obs

Pacetone,ideal

=0.631

0.667= 0.946

5.21 A) (a) Raoult’s law (b) PB (c) 7 (and possibly 10) (d) (1), (4)

(e) (1) (and possibly 2) (f) Ptotal, ideal (g) Actual Ptotal (h) PB

*

B) (a) and (d) • Since there are negative deviations, A···B is stronger and the activity coefficients for

both A and B must less than 1.0

5.22 A) Since negative deviations seen, GE and HE must be exothermic values.

B) It is likely there would be a maximum azeotrope in the T-xy diagram for this mixture since the

deviations are negative.

C) Component A, since the mole fraction in the vapor at this point is 0.717 from estimated values of PA

and Ptotal from the graph (or because PA

* > PB

*).

• Reading the values of Ptotal and PA from the graph: yA

=P

A

Ptotal

=190torr

265torr= 0.717

D) KH,B

< PB

* since negative deviations produce an intersection below the pure vapor pressure.

5.23 A) Diethylether (CH3CH2OCH2CH3) has dipolar forces, acetone (CH3OCH3) also has dipolar forces

B) (a) Table of results:

(b) The plot of the data

shows positive deviations

from Raoult’s Law, so

the mixture is not acting

as an ideal mixture.

C) (c)

gacetone

=P

acetone,obs

Pacetone,ideal

=168torr

141.5torr= 1.19

gethylether

=P

ethylether ,obs

Pethylether,ideal

=391torr

323torr= 1.21

5.24 A) The pure vapor pressure of both liquids increase as T is

increased. This behavior should then be true of all liquid mixtures.

B) Yes the basic shape is retained and the curves come together at

the same mole fraction (≈ 0.90) at each temperature.

C) (a) P = 64 kPa, 303.15 K: cdiethylether

= 0.30 ydiethylether

= 0.58

(b) P= 36 kPa, 293.15K: cdiethylether


= 0.40

(c) P= 20 kPa, 273.15K: cdiethylether


= 0.65

D) For two reasons:


70

(1) The difference between the pure vapor pressures is decreasing. [As can be seen on the graph at mol

fractions 0 and 1.0].

(2) As T is increased the vapor phase becomes “richer” in the more volatile component and the curves

are separated more. [As can be seen by the difference between the values in (D)].

5.25 A) Raoult’s Law diagram: B) Answers:

(a) Positive deviations

(b) CHCl3 shows much greater positive deviations than

ethanol. That the Ethanol - CHCl3 forces are much weaker

than the pure ethanol forces, or CHCl3 forces, since CHCl3

escapes much more easily from the mixture than what we

would expect (ideal value) if it experiences the same

forces.

(c) The activity coefficient (ϒ) is always greater than 1.0

since Pobs, CHCl3 > Pideal, CHCl3 at all mole fractions.

(d) The value for ϒ C2H5OH would be close to 1.0 from χ

CHCl3 = 0 → 0.4 but will be greater than 1.0 at higher

mole fractions CHCl3.

5.26 A) Yes, positive deviations indicate the A···B interactions are weaker than what the pure liquids

have and that should indicate a positive value for GE, so it is consistent.

B) The log of the activity coefficient for ethanol starts out above 1.0, but then drops to lower values

indicating the interactions of ethanol and CHCl3 molecules start out very weak. It takes added energy to

replace the ethanol H-bonds with the dipolar ethanol-CHCl3 interactions and HE is endothermic. However,

as ethanol’s mole fraction increases ethanol can reestablish H-bonding and the HE becomes exothermic.

C) The activity coefficient for CHCl3 quickly increases and doesn’t show a region where it is close to 1.0,

where the log ϒ would be zero. That correlates to the deviations seen on the Raoult’s Law diagram at

nearly all mole fractions.

5.27 A) Yes, it has an azeotrope. The plot shows a minimum Tbp for the mixture, below either of the

boiling points of the pure components. If the deviations from Raoult’s Law are positive, then a minimum

boiling temperature will be observed so the T-xy plot shows the expected behavior.

B) Because the highest vapor pressure will produce the lowest boiling point for the mixture.

C) Yes, it does. The equality occurs at ≈ 0.85 mol fraction CHCl3.

D) (a) χ CHCl3 = 0.40, y CHCl3 = 0.60.

E) Apply the lever rule: Relative amt. solute in liquid phase

Relative amt. solute in vapor phase=

nB,liq

nB,vapor

=L

vapor

Lliquid

»0.15

0.20» 0.75

5.28 A) Positive deviations and would expect only LDF forces between acetone and cyclohexane, since

acetone has dipolar forces but cyclohexane, being non-polar, has only LDF forces.

B) Yes there will an azeotrope and it will be a minimum Tbp. It will occur at χ acetone = 0.75.

C) Results:

Point A B C D Only liquid mixture

present (single phase)

χ acetone = 0.30

Liquid + vapor

phases in equilibrium

χ acetone = 0.12

y acetone = 0.52

Vapor in equilibrium with a

drop (small amount) of liquid

χ acetone = 0.30

y acetone = 0.30

Vapor only

(single phase)

y acetone =

0.30

D) Relative amt. solute in liquid phase

Relative amt. solute in vapor phase=

nB,liq

nB,vapor

=L

vapor

Lliquid

»0.20

0.18» 1.1


71

5.29 A) Raoult’s law Diagram on the right.

B) Since both molecules are polar, A···B should be dipolar

forces. A···B must be a slightly weaker interaction since the

positive deviations are small but do occur.

C) (a) Plot the partial pressure of iodoethane versus mole

fraction iodoethane for the Henry’s Law region (first 5

values).

Then use the equation for the line to get the value of y where it crosses the line at mol fraction C2H5I

equal to 1.0, so that KH = 449 torr for C2H5I.

(b) Calculate the mole fraction of C4H8O2 in the Henry’s

Law region for ethylacetate (χ C2H5I = 0.65 → 0.0) and

plot versus partial pressure of ethyl acetate. Then use

the same approach as in (a) to get KH= 337 torr for

C4H8O2.

D) Activity coefficients: Compare the observed partial

pressure to the ideal value

χ C2H5I = 0.2353 χ C2H5I = 0.5473 χ C2H5I = 0.8253

g C2H

5I =

105.4

83.2= 1.27 g C

2H

5I =

213.3

193.4= 1.10 g C

2H

5I =

296.9

291.7= 1.02

g C4H

8O

2=

220.8

214.2= 1.04 g C

4H

8O

2=

144.2

126.8= 1.14 g C

4H

8O

2=

66.6

45.94= 1.36

5.30 A) Positive deviations must occur that produce a minimum boiling temperature for the azeotrope.

B) nC

3H

7Br

= 79.0g1.0molC

3H

7Br

123.0g

æ

èç

ö

ø÷ = 0.642molC

3H

7Br n

CH3OH

= 21.0g1.0molCH

3OH

32.0g

æ

èç

ö

ø÷ = 0.656molCH

3OH

so cCH

3OH

=0.656

0.656 + 0.642= 0.505

5.31 A) (a) Yes, the deviations are consistent since the excess properties HE and GE had indicated the

A···B forces were stronger than those in pure acetone or CHCl3. There is nothing to indicate in the

Raoult’s law diagram that a complex is being formed. We need properties like the excess thermodynamic

properties to substantiate that.

(b) The activity coefficient, measured by the separation of the ideal partial pressure and actual partial

pressure seems to occur at χ CHCl3 = 0.60, where it is approximately 0.78.

(c) It is approximately 21 kPa.

B) (a) Yes, there should be a maximum boiling point since the deviations are negative from Raoult’s Law.

(b) χ CHCl3 = 0.60 (c) χ CHCl3 = 0.25, y CHCl3 = 0.20

5.32 A) (a) Since the ln ϒ values are positive, the activity coefficients are greater than 1.0, so positive

deviations will occur and Ptotal, actual > Ptotal, ideal.


72

(b) KH for ethanol > PC

2H

5OH

* because of the positive deviations.

(c) Pure C2H5OH would have H-bonding forces whereas DIPE, (CH3)2COC(CH3)2, will have dipolar forces, but

not H-bonding. The A···B forces will then be largely dipolar, which is weaker than the H-bonding between

C2H5OH molecules.

B) (a) Yes, it will. (The relationship is illustrated in Problem 5.27)

(b) It will be a minimum Tbp since the deviations are positive (opposite behavior than the P-xy graph).

(c) It will occur at cC

2H

5OH

= 0.40

C) (a) Worked out answers:

(b) Yes the sign for GE mixture

is positive throughout all mole fractions. This is what we would expect, that overall energy must be added

to sever the stronger forces in the pure components to form the weaker interaction between ethanol and

DIPE based on the thermodynamics of the interactions, and since the observed deviations are positive.

5.33 A) • Need to look up ∆Hvap and normal Tbp for Br2 liquid.

lnP2

= lnP1

-∆H

vap

R

1

T2

-1

T1

é

ë

êê

ù

û

úúÞ lnP

298= ln(1.0) -

∆Hvap

R

1

Tbp

*-

1

T298

é

ë

êê

ù

û

úú T

bp

* = 332.4K ∆Hvap

= 29.45kJ

mol

lnP298

* = 0 -

2.945X104 J

mol

8.314J

mol - K

1

332.4-

1

298.2

é

ëê

ù

ûúK

-1 = -1.403 and P298

* = e-1.403 = 0.246atm760torr

1atm

æ

èç

ö

ø÷ = 187torr

Once we have the pure vapor pressure at 25°C, we can

calculate the expected ideal vapor pressure for each mole

fraction using Raoult’s Law.

B) The deviations are positive as the table below proves: mol

fraction Br2 Observed P Br2

(torr) Calc'd ideal P Br2(torr)

Difference Pobs - Pideal

0.00394 1.50 0.74 0.76

0.00420 1.60 0.79 0.81 0.00599 2.39 1.12 1.27 0.01020 4.27 1.91 2.36 0.01300 5.43 2.43 3.00 0.02360 9.57 4.41 5.16 0.02380 9.83 4.45 5.38

0.02500 10.27 4.68 5.60 C) The equation for the Henry’s law plot shown above produces: KH = 414 torr when x =1.0.

The calculated KH is greater than the pure vapor pressure (187 torr) for Br2 which is consistent with

positive deviations from Raoult’s Law.

5.34 A) • Must first calculate ideal values for the partial pressure and then compare the observed

partial pressure to the ideal value


73

B) Positive deviations are occurring since the activity coefficients are greater than 1.0.

C) (a) Plot shown on right:

(b) The plot shows is very similar to that for di-isopropyl ether

mixed with ethanol in Problem 5.32. Since both are

ether/alcohol mixtures, we could expect that the disruptions

caused by the formation of A···B would be very similar,

since the H-bonding of the alcohol is most affected.

(c) Since the logs of the activity coefficients are nearly the

same at Χ = 0.50 we would expect they both contribute about

one-hallf of the GE for the mixture.

D) Calculated values for (a) and (b):

(c) The two components will spontaneously mix at all mole fractions. The least spontaneous is a small

amount of methanol is being added to the ether.

5.35 A) No azeotrope is visible so it is called a “zeotropic” diagram.

B)

Point A Point B Point C Point D Only vapor mixture

present (single phase) y hexane ≈ 0.45

Liquid + vapor phases in

equilibrium χ hexane = 0.20 y hexane = 0.75

Liquid + vapor phases in

equilibrium χ hexane = 0.30 y hexane = 0.30

Liquid mixture only

(single phase) χ hexane = 0.68

C) The vapor pressure of hexane is much higher than that of m-xylene, indicating it has much weaker

forces to overcome, so that unless it has a very small mole fraction it is always going to produce a higher

proportion of the total pressure over the solution.

D) It would not be very easy to predict what the ideal boiling point should be for each mixture, to gauge

the deviation, since both components are volatile and contribute to the total vapor pressure, so the P-xy

graphs are generally used for that determination.

5.36 PA

= cAP

A

* ÞP

A

PA

*= c

AÞ (1.0 - 0.0045) = c

A=

nA

nA

+ nB

= 0.9955


74

nA

= nC

2H

5OH

= 98.0g1.0molC

2H

5OH

46.0g

æ

èç

ö

ø÷ = 2.13molC

2H

5OH

nC

2H

5OH

nC

2H

5OH

+ nX

= 0.9955 Þ2.13

2.13 + nX

= 0.9955 Þ nX

= 9.19X10-3 mol X

MW X =2.0g X

9.19X10-3mol X= 217.5 = 218g / mol

5.37 • Can apply iB

=∆P

obs

∆Pi=1.0

if calculate ∆P when i=1.0:∆P = cBP

A

*

nBa(NO

3)2

= 40.0g1.0mol Ba(NO

3)2

261.34g

æ

èç

ö

ø÷ = 0.153molBa(NO

3)2

cBa(NO

3)2

=n

Ba(NO3)2

nH

2O

+ nBa(NO

3)2

=0.153

55.555 + 0.153= 2.75X10-3 ∆P

i=1.0= 2.75X10-3(55.324torr) = 0.152torr

iBa(NO

3)2

=∆P

obs

∆Pi=1.0

=(55.324 - 54.909)torr

0.152torr= 2.73

B) %dissociation =actual i value

i valuefromformula´100 =

2.73

3.00´100 = 91.0%

5.38 • Must calculate the mole fraction of NaCl needed in solution. ∆P

PA

*= c

B=

20.0torr

233.7torr= 0.08558

• Must calculate i factor from % dissociation: actual i factor =97.5%

100%(2.0) = 1.95

Then apply i factor in mol fraction: cNaCl

=(i ´ n

NaCl added)

(i ´ nNaCl added

) + nH

2O

=(1.95n

NaCl added)

(1.95nNaCl added

) + nH

2O

= 0.08558

Leads to: 0.1669(nNaCl added

) + 4.754 = 1.95(nNaCl added

) Þ nNaCl added

= 2.666molNaCl1.0mol

58.44g

æ

èç

ö

ø÷ = 156gNaCl

5.39 ∆PC

6H

6

= cBP

C6H

6

* Þ cB

=14torr

400torr= 0.0035 n

C6H

6

= 500.0g1.0molC

6H

6

78.0g

æ

èç

ö

ø÷ = 6.41molC

6H

6

0.0035 =n

B

nC

6H

6

+ nB

=n

B

6.41 + nB

Þ 0.0035(6.41 + nB) = n

BÞ n

B= 0.0225 MW B =

19.0g

0.0225mol= 844

g

mol

5.40 A) nC

6H

18O

6

= 70.0g1.0mol

182.1g

æ

èç

ö

ø÷ = 0.384molC

6H

18O

6 n

H2O

= 30.0g1.0mol

18.02g

æ

èç

ö

ø÷ = 1.667mol H

2O

cC

3H

18O

6

=n

C3H

18O

6

nH

2O

+ nC

3H

18O

6

=0.387

1.667 + 0.387= 0.187

B) (a)∆Tfp

= -R(T

fp,H2O

* )2

∆Hfus,H

2O

cB

= -8.314 J

mol-K(273.15K)2

6.008X103 J

mol

(0.187) = -103.3(0.187)K = -19.3K = -19.3°C

(b) ∆Tfp

= -Kf ,H

2O(m

B) = -1.86

°C - kg

mol

0.384mol

0.030kg

æ

èç

ö

ø÷ = -23.8°C

C) The values don’t agree, with the approximation producing the larger value. It is likely that the

approximation cB

»n

B

nA

doesn’t apply since n

B

nA

=0.384mol

1.667mol= .230, not 0.187.


75

5.41 A) ∆Tfp

= -Kf ,C

6H

12

(mB) leads to: m

B=∆T

fp

Kf ,C

6H

12

=(-2.3 - 6.4)°C

-20.86°C-kg

mol

= 0.431m

nB

= 0.4307mol

kg0.250kgC

6H

12( ) = 0.1077mol MW B =15.0g

0.1077mol= 139

g

mol

B)∆Tbp

= Kb,C

6H

12

(mB) = 2.79 K(°C)-kg

mol(0.4037m) = 1.13°C T

bp,solution= T

bp

* + ∆Tbp

= 80.74 +1.13 = 81.87°C

5.42 nI2

= 30.0g1.0mol I

2

258.0g

æ

èç

ö

ø÷ = 0.1163mol I

2 ∆T

bp= K

b,CHCl3

(mB) = 3.88

°C - kg

mol

0.1163mol I2

0.200L1.479kg

1.0L

æ

èç

ö

ø÷

æ

è

ççççç

ö

ø

÷÷÷÷÷

= 1.53°C

Tbp,solution

= Tbp

* + ∆Tbp

= 61.2 +1.53 = 62.7°C

5.43 A) Since the constants depend on different properties of A, they would not be equal. The value for

Kf for A depends on the normal freezing point of pure A, the ∆Hfus and its molecular weight, while Kb

depends on the normal boiling point of pure A, its ∆Hvap and also the molecular weight of A.

B) The Kf is always larger than Kb for the same solvent since although the boiling point is a higher

temperature than the freezing point, the division by the much larger ∆Hvap should makes the Kb a smaller

value.

5.44 Kf ,A

=R(T

fp,A

* )2

∆Hfus,A

1000

MWA

æ

èç

ö

ø÷ =

8.314 J

mol-K(273.15K)2

1.319X104 J

mol

88.06 g

mol

1000g

kg

æ

è

çç

ö

ø

÷÷

=7.018X107K - kg

1.319X107mol= 5.32

K(°C) - kg

mol

5.45 Kf

=R(T

fp,H2O

* )2

∆Hfus,H

2O

1000

MWA

æ

èç

ö

ø÷ Þ ∆H

fus,A=

R(Tfp,A

* )2

Kf ,A

1000

MWA

æ

èç

ö

ø÷

∆Hfus,A

=8.314 J

mol-K(158.85K)2

1.79K(°C)-kg

mol

74.0 g

mol

1000g

kg

æ

è

çç

ö

ø

÷÷

=1.552X107 J

1.74X103mol= 8673

J

mol= 8.67

kJ

mol

5.46 A) wt.H2O = 1000mLsoln

1.044gsoln

1.0mL

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

- 3.0mol60.06g

1.0molCO(NH2)2

æ

èç

ö

ø÷

é

ë

êê

ù

û

úú

= 1044 -180.2 = 863.8gH2O

∆Tbp

= Kb,H

2O(m

urea) = 0.512

K(°C) - kg

mol

3.0molurea

0.8638kg

æ

èç

ö

ø÷ = 1.78°C ∆T

bpsolution = 101.8°C

B) P =n

BRT

Vsoln

=3.0mol(0.08206 L-atm

mol-K(298K))

1.0L= 73.4atm

C) If urea forms dimers then the effective number of particles decreases which would produce lower

values for the boiling point elevation and in osmosis. The most noticeable change of a small decrease in

moles will be the change in the osmotic pressure, i.e. if molarity lowered to 2.89 mol/L osmotic pressure

becomes 70.7 atm versus a boiling point of 101.5 °C.

5.47 A) Results:

Salt solution MW

(g/mol)

∆Tfp, obs m

(mol/kg)

∆Tfp when

i = 1.0

Calc’d i No.ions

salt

% dev

in i

14.0% NaCl 58.5 -9.94 °C 2.785 -2.544 1.92 2 4.0%

14.0% MgSO4 120.4 -2.86 °C 1.352 -2.515 1.14 2 43%

14.0% BaCl2 208.4 -3.92 °C 0.7811 -1.453 2.70 3 10%

14.0% (NH4)2SO4 132.1 -4.07 °C 1.232 -2.291 1.78 3 40.7%

14.0% AgNO3 169.9 -2.55 °C 0.958 -1.782 1.43 2 28.5%

B) Smallest % dev → largest % dev: NaCl < BaCl2 < AgNO3 <(NH4)2SO4 < MgSO4

Sample Calculations:


76

msalt

=14.0g

MWsalt

( g

mol) ´ 0.086kgH

2O

∆Tfp(i = 1.0) = -1.86

°C - kg

mol(m

salt) i calc 'd =

∆Tfp,obs

∆Tfp, i = 1

%deviation =(no.ions salt) - (i calc 'd)

(no.ions salt)´100

5.48 A) Msalt

=

14.0g1mol BaCl

2

208.4g

æ

èç

ö

ø÷

100g1.0mLsoln

1.1342g

æ

èç

ö

ø÷

1.0L soln

1000mL

æ

èç

ö

ø÷

=0.06718mol BaCl

2

0.08817L soln= 0.7619M

P = i ´n

BRT

Vsoln

é

ë

êê

ù

û

úú

= 2.700.7619mol(0.08206 L-atm

mol-K(298K))

1.0L

é

ë

êê

ù

û

úú

= 2.70[18.63]atm = 50.31atm

B) If van’t Hoff factor = 3.00, P = 3.0[18.63]atm = 55.89atm

Overestimation: %error =55.89 - 50.31

50.31

é

ëê

ù

ûú ´100 = 11.1%

5.49 Msalt

=

14.0g1mol MgSO

4

120.4g

æ

èç

ö

ø÷

100g1.0mLsoln

1.148g

æ

èç

ö

ø÷

1.0L soln

1000mL

æ

èç

ö

ø÷

=0.1163mol MgSO

4

0.08711L soln= 1.335MMgSO

4

PMgSO

4

= Pglucose

Then: i ´n

MgSO4

Vsoln

é

ë

êê

ù

û

úúRT =

nglucose

Vsoln

é

ë

êê

ù

û

úúRT = 1.14

1.335mol

1.0L

é

ëê

ù

ûú =

nglucose

Vsoln

é

ë

êê

ù

û

úú

= 1.522Mglucose

5.50 A) PNaCl

= Psucrose

Þ Csucrose

(mol

L) = i ´ C

NaCl(mol

L) so that C

sucrose(mol

L) = 2.0 ´ 0.15 mol

L= 0.30M

B) • Must convert % in D5W to molarity and compare to 0.30M

%(w /v) =5.00gC

6H

12O

6

100mLsolnÞ

5.00gC6H

12O

6

0.100Lsoln

1.0molC6H

12O

6

180g

æ

èç

ö

ø÷ = 0.278MC

6H

12O

6

Since the D5W concentration of solute is lower than 0.30 M, water would flow into the cell because cH

2Ois

lower there than in the D5W solution.

C) The factor would need to be: i =0.278M

0.150M= 1.85

D) iNaCl

=∆T

obs

∆Ti =1

=-0.59°C

-1.86 °C

molal(0.172m)

= 1.83 so the van’t Hoff factor can be below 2.0 even in low

concentrations of solute. Applying i = 1.85 in (A), makes Csucrose = 0.278M, not 0.30 M.

5.51 A) • Need molarity of mannitol

%(w /w) =5.00gC

6H

14O

6

100gsolnÞ

5.00gC6H

14O

6

1.0molC6H

14O

6

180g

æ

èç

ö

ø÷

100gsoln1.00mL

1.06gsoln

æ

èç

ö

ø÷

1.00L

1000mL

æ

èç

ö

ø÷

= 1.164MC6H

14O

6

Pmannitol

=1.164mol(0.08206 L-atm

mol-K(310K))

1.0L

é

ë

êê

ù

û

úú

= 29.60atm

B) Know 5%, D5W, is a 0.278M solution of glucose (from Problem 5.52(b)), so:

P =0.278mol(0.08206 L-atm

mol-K(310K))

1.0L

é

ë

êê

ù

û

úú

= 7.07atm ∆P = Pmannitol

- PD5W

= 29.6atm - 7.07atm = 22.5atm

5.52 A) Derivation: (• Need to subtract equations to get ratio of mol fractions)


77

∆P = Pmannitol

- Pvitreous

Þ ∆P =RT

Vm,H

20

lncH

2O,mannitol

é

ë

êê

ù

û

úú

-RT

Vm,H

20

lncH

2O,vitreous

é

ë

êê

ù

û

úú

=RT

Vm,H

20

lnc

H2O,mannitol

cH

2O,vitreous

é

ë

êê

ù

û

úú

(Pmannitol

- Pvitreous

)Vm,H

20

RT= ln

cH

2O,mannitol

cH

2O,vitreous

é

ë

êê

ù

û

úúÞ

cH

2O,mannitol

cH

2O,vitreous

é

ë

êê

ù

û

úú

= exp(P

mannitol- P

vitreous)V

m,H20

RT

æ

è

çç

ö

ø

÷÷

B) Given VH

2O

L

mol

æ

èç

ö

ø÷ =

MW g

mol( )d@310K g

L

æ

èç

ö

ø÷

=18.02 g

mol( )993.3 g

L

æ

èç

ö

ø÷

= 0.01814L

mol

cH

2O,mannitol

cH

2O,vitreous

é

ë

êê

ù

û

úú

= exp(22.5atm)(0.01814 L

mol)

0.08206 L-atm

K-mol(310K)

æ

èç

ö

ø÷ = exp(0.01604) = 1.016

C) A mole fraction difference of 2% produces a 22.5 atm change in pressure, so even much smaller

changes can be measured easily from osmotic pressure.

5.53 A) P =massB,g

MWB

æ

èç

ö

ø÷

RT

Vsoln

Þ MWB

=massB,g

P

æ

èç

ö

ø÷

RT

Vsoln

=0.150g(0.08314 L-bar

mol-K)(298K)

(0.515bar)(0.100L)= 72.1g / mol

B) Only the mole fraction of A is involved in the original derivation so that no properties specific to water

are included in the final equations. Therefore the solvent can be any liquid as long as the membrane is

permeable to only that liquid.

5.54 A) MWcreatine

=mass,g

P

æ

èç

ö

ø÷

RT

Vsoln

=0.100g(0.08206 L-atm

mol-K)(298K)

(0.0171atm)(1.00L)= 142.9g / mol

B) Given PH

2O

* (37°C) = 47.067torr , creatine a non-volatile solute, then:∆P = ccreatine

PH

2O

*

ccreatine

=n

creatine

nH

2O

+ ncreatine

=6.99X10-4

55.56 + 6.99X10-4= 1.26X10-5 ∆P = 1.26X10-5(47.067torr) = 5.93X10-4 torr

This vapor pressure difference is too small to measure accurately, so that applying Raoult’s law for a

molecular weight determination won’t be practical.

C) Blood: 20 mg/L → P =20.0X10-3g

142.9 g

mol

æ

èç

ö

ø÷0.08206 L-atm

mol-K(298K)

1.0L= 3.42X10-3atm

760torr

1atm

æ

èç

ö

ø÷ = 2.60torr

Urine: 750 mg/L → P =0.750g

142.9 g

mol

æ

èç

ö

ø÷0.08206 L-atm

mol-K(298K)

1.0L= 0.128atm

760torr

1atm

æ

èç

ö

ø÷ = 97.5torr

The osmotic pressures could easily be measured accurately, in either torr or kPa, for these

concentrations.

5.55 A) n

B

Vsoln

= Cpolymer

mol

L( ) =

P

RT=

10torr1.0atm

760torr

æ

èç

ö

ø÷

(0.08206 L-atm

mol-K(298K))

= 5.17X10-4M

B) 5.17X10-4mol

1.0L

1.0LH2O

1.0kgH2O

æ

èç

ö

ø÷

2.10X105g polymer

1.0mol polymer

æ

èç

ö

ø÷ = 108.6g polymer per kg

5.56 A) A linear equation can be developed after substituting nB = mass B / MW B, so that:

P =mass

MW

RT

Vsoln

é

ë

êê

ù

û

úú then y = P and x =

massB,g

Vsoln

(L) then slope =

RT

MWB

, b(y - intercept) = 0

MWpolymer

=RT

slope=

8.314 kPa-L

mol-K(317K)

6.90X10-3 kPa-L

g

= 381,965g

mol= 3.82X105 g

mol


78

B)

MWpolymer

=RT

slope=

0.08206 atm-L

mol-K(310K)

6.90X10-4 atm-L

g

= 381,965g

mol= 3.82X104 g

mol

5.57 • Change pressures to atm from torr before plotting data.

Data: Plot:

MWpolymer

=RT

slope=

0.08206 atm-L

mol-K(310K)

4.74X10-4 atm-L

g

= 53,670g

mol= 5.37X104 g

mol

5.58 A) P = -RT

VH

2O

lncH

2O and V

H2O

L

mol

æ

èç

ö

ø÷ =

MW g

mol( )d

298

g

L( )

=18.02 g

mol( )997.04 g

L

æ

èç

ö

ø÷

= 0.01807L

mol

cH

2O

=n

H2O

nH

2O

+ nglu

=55.56

55.56 + 0.300= 0.9946 so that lnc

H2O

= ln(0.9946) = -5.386X10-3

(a) P = -0.08206 L-atm

mol-K(298K)

0.01807 L

mol

(-5.386X10-3) = 7.288 = 7.29atm

(b) After substituting: ln(1-ΧB) = - ΧB gives: P =RT

VH

2O

cC

6H

12O

6( ) and c

C6H

12O

6

= 1.0 - 0.9946 = 0.0054

P = -0.08206 L-atm

mol-K(298K)

0.01807 L

mol

(0.0054) = 7.311 = 7.31atm

(c) After assuming cB

=n

B

nA

: PVsoln

= nBRT → P =

0.08206 L-atm

mol-K(298K)

1.0154L(0.300mol) = 7.22atm

B) The approximation that cB

=n

B

nA

produces the most significant change.

5.59 • Get the concentration of solute from first property, then use for the second colligative property

Ccompd

mol

L( ) =

P

RT=

99.0kPa

(8.314 L-kPa

mol-K(288K))

= 0.04135M 0.04135mol compd

1.0Lsolution®

0.04135mol compd

?kg solvent= m

• Assume dsoln = dsolvent and total weight solution = wt. C6H12, since solute not known:

wt.C6H

12,kg = 1.0LC

6H

12

779kg

1.0m3

é

ëê

ù

ûú

1.0m3

1000L

é

ëêê

ù

ûúú

= 0.779kg

y = 0.00690x - 0.00034

R² = 0.99186

0.000

0.020

0.040

0.060

0.080

0.100

0.00 5.00 10.00 15.00

Osm

oti

c p

ressu

re (

kP

a)

Concentration (g/L)

Polystyrene in Cyclohexane


79

∆T

fp= -K

f ,C6H

12

(mB) = -20.2

°C

molal

0.4135mol

0.779kg

æ

èç

ö

ø÷ = -1.07°C

5.60 A) • Need to determine the i factor for NaCl by comparing osmosis data.

• Must have concentration units in molarity before comparing osmotic pressures

Totalwt.Mannitol(aq)

= 1000g + 0.2164mol182.2g

1.0mol

æ

èç

ö

ø÷ = 1039.4g V

soln= 1039.4g

1.0mL

1.012g

æ

èç

ö

ø÷

1.0L

1000mL

æ

èç

ö

ø÷ = 1.039L

Totalwt.NaCl(aq)

= 1000g + 0.1165mol58.44g

1.0mol

æ

èç

ö

ø÷ = 1005g V

NaClsoln= 1005g

1.0mL

1.005g

æ

èç

ö

ø÷

1.0L

1000mL

æ

èç

ö

ø÷ = 1.001L

Then: Cmannitol

=0.2164mol

1.027L=

P

RT= iC

NaCl

0.1165mol

1.001L

é

ëê

ù

ûú →

Cmannitol

CNaCl

= i =0.2107M

0.1163M= 1.81 for NaCl

∆Tfp

= -Kf ,H

2O(i ´ m

NaCl) = -1.86

°C

molal1.81(0.1165m)( ) = -0.393°C

B) Yes, the two solutions would have the same freezing point, within a hundredth of a degree, if the i

factor is applied to the NaCl solution.

∆Tfp,mannitol = -K

f ,H2O(m

NaCl) = -1.86

°C

molal0.2164m)( ) = -0.402°C

5.61 A) nalbumin

= 40.0g1.0mol Albumin

69,000g

æ

èç

ö

ø÷ = 5.80X10-3 mol

nglobulin

= 20.0g1.0mol Albumin

160,000g

æ

èç

ö

ø÷ = 1.25X10-3 mol

Pglobulin

=0.08206 L-atm

mol-K(310K)

1.0L(1.25X10-3M) = 0.00318atm

760torr

1.0atm

æ

èç

ö

ø÷ = 2.42torr

B) P

blood ions

Palbumin

+ Pglobulin

=7.07atm

(0.01475 + 0.00318)atm= 395 times greater.

C) We can assume each value is independent of any other, since as a colligative property it doesn’t

matter what the solute particle is, but how many there are. Therefore, they should always add to each

other (like partial pressures in a mixture of gases.)

5.62 Pneeded

=(1.0mol ions)(0.08206 L-atm

mol-K)(300K)

1.0L= 24.6atm

5.63 Pneeded =(0.1802mol ions)(0.08206 L-atm

mol-K)(298K)

1.0L= 4.41atm

5.64 • Must calculate the mole fraction of water in solution for lowered vapor pressure, assuming ideal.

nC

3H

8O

3

= 60.0g1.0molC

3H

8O

3

92.09g

æ

èç

ö

ø÷ = 0.6515molC

3H

8O

3 n

H2O

= 40.0g1.0mol H

2O

18.02g

æ

èç

ö

ø÷ = 2.22mol H

2O

cH

2O

=n

H2O

nH

2O

+ nglycerol

=2.222

2.222 + 0.6515= 0.7731 and g

A=

PA,obs

PA,ideal

=P

A,obs

cAP

A

*=

33.93torr

0.7731(47.12torr)= 0.9053

Tfp,soln

= Tfp,C

6H

12

* + ∆Tfp

= -6.47 -1.07 = -7.54°C

Palbumin =0.08206 L-atm

mol-K(310K)

1.0L(5.80X10-3M) = 0.0145atm

760torr

1.0atm

æ

èç

ö

ø÷ = 11.2torr

P


80

B) aH

2O

= gH

2Oc

H2O

= 0.9053(.7731) = 0.6999 So the activity of water is below the limit for slowing down

enzymatic reaction, but above that for enhancing possible metal oxidation reaction rates.

5.65 A) (a) P

obs

PH

2O

*= g

H2O given P

H2O

* = 9.514kPa then: SolutionI :gH

2O

=9.239

cH

2OP

H2O

*=

9.239

(.9790)(9.514)= 0.973

SolutionII :gH

2O

=9.239

cH

2OP

H2O

*=

9.239

(.9585)(9.514)= 0.958

(b) No, it is decreasing with increasing solute concentration as would be expected.

B) Because the value of γ is less than 1.0, the ln(γ) is negative and we would expect HE for the mixing to

be exothermic. It also indicates the raffinose ··· water interactions are stronger than in water, indicating

some clustering of water and the saccharide is likely occurring.

C) (a) m =208kg

1.0m3

1000g

1.0kg

æ

èç

ö

ø÷

1.0m3

1000L

æ

èç

ö

ø÷ =

208g

1.0LH2O

1.0LH2O

1.0kgH2O

æ

èç

ö

ø÷

1.0mol

504.4g

æ

èç

ö

ø÷ = 0.4026m

(b) Given that it is likely that γ for water will be less than 1.0, the vapor pressure would not be

lowered, because of solute, as much as it would be for an ideal solution. Therefore the freezing point

depression observed for the solution would be less than -0.75°C.

5.66 A) gH

2O

=9.239

cH

2OP

H2O

*=

6.877

(0.813)(9.514)= 0.889 B)∆T

fp,obs= g

A∆T

fp,A= 0.889(-19.3°C) = -17.2°C

C) Quite a significant impact. The freezing point obtained with the correction for the activity of water is 2°

higher than the value calculated with the first equation, and 6.6°C higher than calculated from the

simplified equation.

5.67 A) Use: gH

2O

=P

obs

Pg =1.0

and Pg =1.0

= cH

2OP

H2O

* nH

2O

= 500g1.0molH

2O

18.02g

æ

èç

ö

ø÷ = 27.78molH

2O

nC

6H

14O

6

= 105.56g1.0mol

182.2g

æ

èç

ö

ø÷ = 0.5794molC

6H

14O

6 c

H2O

=n

H2O

nH

2O

+ nmann

=27.78

28.357= 0.9797

Pg =1.0

= cH

2OP

H2O

* = 0.9797(42.2torr) = 41.34torr gH

2O

=40.51torr

41.34torr= 0.980

B) ∆Tfp

= -gH

2OK

fm

BÞ T

fp,soln- 0°C = -g

H2OK

fm

B= -0.980(1.86

°C

molal)

0.5794mol

0.500kg

æ

èç

ö

ø÷ = -2.11°C

C) Pobs

= gH

2OP

g =1.0= g

H2O

nBRT

Vsoln

é

ë

êê

ù

û

úú Vsoln,mL =

Totalwt.,g

d(g / mL)

æ

èç

ö

ø÷ =

605.6g

1.06 g

mL

= 571mL

Pobs

= 0.1980.5794mol(0.08206 L-atm

mol-K)(308K)

0.571L

é

ë

êê

ù

û

úú

= 0.198(25.6atm) = 25.1atm

D) Not a significant difference (only about 2%) when gH

2O since P

g =1.0= 25.6atm.

5.68 A) nH

2O

= 40.0g1.0mol

18.02g

æ

èç

ö

ø÷ = 2.22molH

2O

nC

3H

8O

3

= 60.0g1.0mol

92.09g

æ

èç

ö

ø÷ = 0.6515molC

3H

8O

3

cH

2O

=n

H2O

nH

2O

+ nlactose

=2.22

2.874= 0.7725

gH

2O

=P

obs

Pg =1.0

=0.0434atm

0.7725(0.0620atm)= 0.906

B) Ratio =n

H2O

nlactose

=2.22

0.6515= 3.41 • So about seven waters for every two lactose molecules

Would expect most of the water molecules to form H-bonded clusters around lactose molecules (since it’s

a triol) and not be “free” unbound water molecules.


81

C) The actual value of clactose

is 0.2275, but if calculated from the vapor pressure lowering a higher value

results: ∆PA

= cBP

A

* Þ cB

=∆P

A

PA

*=

0.0186

0.0620= 0.30 so the solution does illustrate this effect.

.

5.69 A) nlactose

= 20.0g1.0molC

12H

22O

11

342.0g

æ

èç

ö

ø÷ = 0.234molC

12H

22O

11

∆Tfp,g

A=1.0

= -Kfm

BÞ T

fp,so ln- 0°C = -(1.86

°C

molal)

0.234mol

0.920kg

æ

èç

ö

ø÷ = -0.473°C

gH

2O

=∆T

fp,obs

∆Tfp,g =1.0

=-0.500

-0.473= 1.06

B) The value for activity coefficient for water is greater than 1.0 instead of being less than 1.0 as it was in

the other problems.

C) Yes, even though the solutes are non-volatile, it is the strength of the A···B interactions and the

possible clustering and grouping of molecules, as we saw earlier, that can affect vapor pressure and

produce both values of activity coefficient that are higher than one or less than one. The colligative

properties should also reflect what is going on between the molecules in solution.

5.70 A)

Solution ∆Tfp (°C) ∆Tfp,g =1.0

B) No. “free” water

molecules vs. pure water

20.0 % ethylene glycol -7.93 -7.49 1.06 Increased

20.0 % 1-propanol -7.76 -7.74 1.00 Stayed the same

20.0 % urea -7.00 -7.74 0.904 Decreased

Sample Calculation:

∆Tfp,g

A=1.0

= -Kfm

B= -(1.86

°C

molal) 4.028m( ) = -7.49°C g

H2O

=∆T

fp,obs

∆Tfp,g =1.0

=-7.93

-7.49= 1.06

B) All interactions between water and the solutes are H-bonding, but urea has a flat planar structure with

many H-bonding sites compared to ethylene glycol or propanol, creating clustering within the solution.

Ethylene glycol appears to increase the number of free water molecules (as did lactose, another “polyol”)

so it may be disrupting the normal H-bonding arrangements in water.

C) Pobs

= gA(P

g =1.0) = g

AC

BRTé

ëùû

= 0.904 3.506(0.08206 L-atm

mol-K)(298K)é

ëùû

= 0.904(85.72atm) = 77.66atm

%overestimate =85.72 - 77.66

77.66´100 = 10.4% Yes, this is a significant difference.

5.71 Must change units on KH: KH

= 1.4X10-5 mol

m3 - Pa

1.0m3

1000L

æ

èç

ö

ø÷

1.013X105 Pa

1.0atm

æ

èç

ö

ø÷ = 1.418X10-3 mol

L - atm

So that PAr(g)

(atm) =C

Ar

mol

L( )

KH,Ar inH

2O

mol

L-atm( )=

0.010 mol

L

1.418X10-3 mol

L-atm

= 7.05atm

5.72 A) • Assume 1.0 L of solution = 1.0L of water (55.56 mol)

4.17X10-5 =n

CCl2F2

nH

2O

+ nCCl

2F2

=x

55.56mol + xÞ 4.17X10-5(55.56 + x) = 0.00232 + 4.17X10-5x = x

nCCl

2F2

= x = 2.32X10-3mol then MCCl

2F2

= 2.32X10-3 mol

L in the saturated solution.

B) KH,CCl

2F2inH

2O

mol

L-atm( ) =

CCCl

2F2

mol

L( )

PCCl

2F2(g)

(atm)=

2.32X10-3 mol

L

1.0atm= 2.32X10-3 mol

L - atm

5.73 A) Cheptachlor

=0.056mg

1.0L

1.0g

1000mg

æ

èç

ö

ø÷

1.0mol

393.35g

æ

èç

ö

ø÷ = 1.42X10-7 M

gH

2O


82

KH

= 82.6Pa - m3

mol

1000L

1.0m3

æ

èç

ö

ø÷

1.0atm

1.013X105 Pa

æ

èç

ö

ø÷ = 0.8154

L - atm

mol

Pgas

= Cgas

mol

L

æ

èç

ö

ø÷ K

H

L - atm

mol

æ

èç

ö

ø÷ = 1.42X10-7(0.8154) = 1.15X10-7atm

nheptachlor

V=

P

RT=

1.15X10-7atm

0.08206 L-atm

mol-K(308K)

=4.58X10-9mol

1.0L

6.02X1023molecules

1.0mol

æ

èç

ö

ø÷

1.0L

1000cm3

æ

èç

ö

ø÷ = 2.76X1012 molecules / cm3

C) If KH is greater, the pressure of the gaseous molecules over the solution would also be greater, if the

concentration of the dissolved gas were the same.

D) K

H,seawater

KH,water

=P

gas,seawater/ C

gas

Pgas,water

/ Cgas

then P

gas,seawater

Pgas,water

=n

gas,seawater

ngas,water

=No.molecules / cm3,seawater

No.molecules / cm3,water

No.molecules / cm3,seawater = 2.76X1012 211.6

82.6

æ

èç

ö

ø÷ = 7.06X1012

5.74 A) For lncB

=∆H

fus,B

R

1

Tmp,B

*-

1

T

é

ë

êê

ù

û

úú need ∆Hfus,B in J/mol: ∆H

fus,B=

4450cal

1.0mol

4.184J

1.0cal

æ

èç

ö

ø÷ =

1.862X104 J

1.0mol

lncB

=1.862X104 J

mol

8.314 J

mol-K

1

373K-

1

293K

é

ëê

ù

ûú = 2239.5K(-7.32X10-4)K-1 = -1.64 c

B= e-1.64 = 0.194

Assume 1.0 L solution, so 1.0 L benzene: nC

6H

6

= 1.0L1000mL

1.0L

æ

èç

ö

ø÷

0.876mL

1.0mL

æ

èç

ö

ø÷

1.0moL

78.0g

æ

èç

ö

ø÷ = 11.23mol

0.194 =n

C6H

6

nC

6H

6

+ nC

14H

10

=x

11.23 + xÞ 2.179 = 0.806x x = n

C6H

6

=2.179

0.806= 2.703

MC

14H

10

=x

1.0Lsoln= 2.70M

B) Can set a proportion:

1.0gC14

H10

2.0mLC6H

6

=x gC

14H

10

1000mLC6H

6

Þ x = 500gC14

H10

1.0molC14

H10

178.2g

æ

èç

ö

ø÷ = 2.80molC

14H

10per LC

6H

6

So the result is very close (within 4.0%) which could be due to round off error. Consequently the values

agree and the equation has predicted the solubility correctly.

5.75 A) lncB

=

1.816X104 J

mol

8.314J

mol - K

1

325.7K-

1

293K

é

ëê

ù

ûú = 2184K(-3.43X10-4)K -1 = -0.749 c

B= e-0.749 = 0.473

B) 20.57% by mass → 20.57 g C6H4Cl2, 79.43 g C6H14

nC

6H

4Cl

2

= 20.57g1.0mol

147.0g

æ

èç

ö

ø÷ = 0.1399molC

6H

4Cl

2 n

C6H

14

= 79.43g1.0molC

6H

14

86.0g

æ

èç

ö

ø÷ = 0.9236molC

6H

14

cB

=n

C6H

4Cl

2

nC

6H

14

+ nC

6H

4Cl

2

=0.1399

0.9236 +0.1399= 0.132

C) p-Dichlorobenzene is a non-polar molecule (since Cl’s opposite each other on the ring) with strong LDF

forces, largely due to the aromatic ring. Hexane is also non-polar, but its LDF forces would be much

weaker, so that A···B forces are weaker than B···B and they don’t mix as well and do not form an ideal

solution.

5.76 nC

10H

8

= 1.0g1.0mol

128.0g

æ

èç

ö

ø÷ = 7.813X10-3 molC

10H

8

nC

6H

6

= 1.0g1.0mol

78.0g

æ

èç

ö

ø÷ = 0.01282molC

6H

6


83

cB

=n

C10

H8

nC

10H

8

+ nC

6H

6

=0.007813

0.007813 +0.01282= 0.3786 ln(0.3786) =

148.95J

g

8.314J

mol - K

128.0g

1.0mol B

é

ëê

ù

ûú

1

353.15K-

1

T

é

ëê

ù

ûú

Then: -0.9712 = 2293.2K (2.832X10-3) -1

T

é

ëê

ù

ûú Þ -7.465 =

-2293.2

TÞ T = 307.2K = 34.0°C


84

PART 6: FREE ENERGY (∆G), CHEMICAL EQUILIBRIUM AND ELECTROCHEMISTRY

6.1 A) Dissociation reaction:

(NH4)2CO3(s) 2 NH3(g) + HCl(g) so:Kp

= PHCl(g)

PNH

3(g)( )

2

PTotal

= 3x = 124torr1atm

760torr

æ

èç

ö

ø÷ = 0.1632atm,x = 0.0544atm

so Kp

= (2x)2(x) = 4x3 = 4(0.0544)3 = 6.46X10-4

B) ∆Gr

o = -RT lnKp

= -8.314J

mol - K(303K)ln(6.46X10-4) = -18.50

kJ

mol

(1mol)∆Gf ,(NH

4)2CO

3

o = (1mol)∆Gf ,HCl

o + (2mol)∆Gf ,NH

3

oéëê

ùûú

- ∆Gr

oéë

ùû

= (-95.30) + 2(-16.45)éë

ùû

- (-18.50)kJ

= (-128.2) +18.5éë

ùûkJ Þ ∆G

f ,(NH4)2CO

3

o = -109.7kJ

mol

6.2 A)

∆Gr

o = å∆Gf

o

,products- å∆G

f

o

,reactants= 4mol∆G

f

o,Ag(s) + 4mol∆Gf

o,O2(g)é

ëùû

- 2mol∆Gf

o,Ag2O(s)é

ëùû

= 0 + 0éë

ùû

- 2mol(-11.3kJ

mol)

é

ëê

ù

ûú = 22.6kJ

(b) Kp

= PO

2

Since heterogeneous equilibrium, activities of pure solids and liquids taken as 1.0.

(c) K should be less than 1.0, since ∆G(+) and reaction should be weak.

B) lnKp

= -∆G°

RT= -

22.6X103 J

molO2

8.314 J

mol-K(298K)

= -9.12 gives: PO

2

= KP

= e-9.12 = 1.09X10-4 atm

C)

lnKT

2

= lnKT1

+∆H

R

1

T1

-1

T2

é

ë

êê

ù

û

úúÞ ln(1.0) = -9.12 +

-31.0X103 J

molO2

8.314 J

mol-K

æ

è

çç

ö

ø

÷÷

1

298K-

1

T2

é

ë

êê

ù

û

úú

Þ 9.12 = -12.512 +3728.7

T2

Þ T2

=3728.7

21.63K = 172K (-101°C)

6.3 A) Dissociation reaction: NO2(g) → NO(g) + ½ O2(g) and Kp

=P

NO(g)P

O2(g)( )

1/2

PNO

2(g)

B) Have to define the equilibrium pressures and relate to Ptotal, so best to use the iCe table approach to

organize information.

Given the equilibrium ratio between NO and NO2, can solve for the equilibrium P

of NO2 as

@700 K: x

PNO2,eq

= 0.872 Þ P NO2,eq

=x

0.872= 1.15x

Then get value of “x” from Ptotal equation: Ptotal

= 1.0atm = 2.65x Þ x =1

2.65= 0.377atm

So P NO2 = 1.15(0.377) atm = 0.4336 atm, P O2 = 0.5(0.377) = 0.189 atm at equilibrium at 700 K

@ 700 K Kp

=P

NO(g)P

O2(g)( )

1/2

PNO

2(g)

=0.377 0.189( )

1/2

0.4336= 0.378 ∆G°

700= -8.314

J

mol - K(700K)ln(0.378) = +5.68kJ

@ 800 K: x

PNO2,eq

= 2.50 Þ P NO2,eq

=x

2.50= 0.400x

Then get value of “x” from Ptotal equation: Ptotal

= 1.0atm = 1.90x Þ x =1

1.90= 0.526atm


85

So P NO2 = 0.400(0.526) atm = 0.2105 atm, P O2 = 0.5(0.526)=0.283 atm at equilibrium at 800K

Kp

=P

NO(g)P

O2(g)( )

1/2

PNO

2(g)

=0.526 0.283( )

1/2

0.2104= 1.33 ∆G

800

o = -RT lnKp

= -8.314J

mol - K(800K)ln(1.33) = -1.90kJ

C) Expect that if K when T , then ∆H°(+). Would expect ∆S°(+) as well since total number moles of

gas greater on product side than reactant side. That means ∆G° can change sign with T, since signs of

∆H° and ∆S° are the same. The reaction should become spontaneous at high T’s.

D) ∆Hr

o = 1mol∆Hf ,NO(g)

o +1

2mol∆H

f ,O2(g)

oé

ëê

ù

ûú - 1mol∆H

f ,NO2(g)

oéëê

ùûú

= 90.25kJ + 0éë

ùû

- 33.18kJéë

ùû

= 57.1kJ

∆Sr

o = 1mol SNO(g)

o +1

2mol S

O2(g)

oé

ëê

ù

ûú - 1mol S

NO2(g)

oéëê

ùûú

= 210.8J

K+ 0.50(205)

J

K

é

ëê

ù

ûú - 240.1

J

K

é

ëê

ù

ûú = 73.2

J

K

The changeover T for ∆G° to change sign would be: T =∆H

r

o

∆Sr

o=

57.1X103 J

73.2 J

K

= 780K

The experimental data correlates to the calculated value, since ∆G° is negative at 800 K, but not at 700

K.

6.4 A) Trial 1: KP

=P

HI( )2

PH

2( ) P

I2

( )=

0.9367( )2

(0.27618)(0.06438)= 49.3 Trial 2:

KP

=P

HI( )2

PH

2( ) P

I2

( )=

0.7176( )2

(0.10027)(0.10306)= 49.8

(a) Average KP = 49.6

(b) KP

=0.9367bar( )

21.013atm

1.00bar( )2

(0.27618) 1.013atm

1.00bar

æ

èçç

ö

ø÷÷ (0.06438) 1.013atm

1.00bar

æ

èçç

ö

ø÷÷

= 49.3

Doesn’t make a difference in value of KP since the conversion factors cancel, since ∆n gases equals zero

in this reaction. If the ∆n ≠ 0 there would be a difference in the calculated value, but likely very small.

B) ∆G° = -RT lnKp

= -8.314J

mol - K(731K)ln(49.6) = -23,726 J = -23.7kJ

C) ∆H(T

2) = ∆H(T

1) + ∆C

p∆T = -9.4kJ + (-7.1 J

K)(731- 298)K = -12.47kJ

∆S(T2) = ∆S(T

1) + ∆C

pln

T2

T1

= 21.8J

K+ (-7.1

J

K)ln

731K

298K

æ

èç

ö

ø÷ = 21.8

J

K+ (-6.37

J

K) = 15.4

J

K

∆G731

o = ∆H731

o - T ∆S731

o = -9.4kJ - (731K)(15.4J

K) ´

1kJ

1000 J

æ

èç

ö

ø÷ = (-12.47 -11.3) = -23.8kJ

• So the values agree for ∆G° at 731 K.

6.5 A) Will need to determine initial pressure for PCl5 using ideal gas law

P =nRT

V=

5.00g

208.24g / mol

é

ëê

ù

ûú0.08206 L-atm

mol-K(523K)

2.0L= 0.515atm

KP

=P

PCl3

PCl

2

PPCl

5

= 1.05 =x2

0.515 - x Need to apply quadratic formula to

solve for “x”: x =-b ± b2 - 4ac

2a=

-1.05 ± (1.05)2 - 4(1)(-0.5408)

2=

-1.05 ±1.807

2= +0.379

PPCl

5

= (0.515 - 0.379)atm = 0.136atm


86

B) (a) PTotal

= (0.515 - 0.379) + 2(0.379)éë

ùûatm = 0.894atm %PCl

5=

0.136

0.894´100 = 15.2%

(b) Expect that when K >1.0 will have more products than reactants and that is the case here, with the

PCl5 only making up 15.2% of the final mixture.

C) mass =PV

RT(MW) =

(0.136atm)2.0L

0.08206 L-atm

mol-K(523K)

208.24g

1mol PCl5

é

ë

êê

ù

û

úú

= 1.32gPCl5

6.6 A) • Q vs. K situation, compute initial of Q from data:

KC

= KP

for this reaction, since ∆n = 0, KC

=n

iso-B/V

tot

nB

/Vtot

=n

iso-B

nB

=wt.

iso-B/ MW

iso-B

wt.B

/ MWB

=wt.

iso-B

wt.B

• So for this reaction, the weight ratio equals the mole ratio since the isomers have the same MW’s.

• Therefore Q =wt.

iso-B

wt.B

=4.0g

16.0g= 0.25 > K = 0.106

The system has too many products to be at equilibrium, product weight must decrease and reactant

weight increase to achieve equilibrium.

B) K = 0.106 =wt.

iso-B

wt.B

=4.0g - x

16.0g + xÞ1.106x = 2.304 Þ x = 2.08g is the weight converted:

Wt. borneal at equilibrium = 18.08 g and Wt. iso-borneal at equilibrium=1.82 g

C) ∆G = ∆G° + RT lnQ = (-RT lnK) + RT lnQ = RT ln

Q

K

æ

èç

ö

ø÷ = (8.314 J

mol-K(503K)ln

0.25

0.106

æ

èç

ö

ø÷

= -4182J(0.858) = -3588 J = -3.59kJ

6.7 A) KC

=CH

3CO

2C

5H

10éë

ùû

CH3CO

2Hé

ëùû

C5H

10éë

ùû

• Need to determine molarities, cannot use moles in Kc

CH3CO

2Hé

ëùû0

=0.0010mol

0.845L= 1.183X10-3M C

5H

10éë

ùû0

=0.00645mol

0.845L= 7.633X10-3M

At equilibrium:

CH3CO

2C

5H

11éë

ùûeq

=7.84X10-4 mol

0.845L= 9.278X10-4M = "x"

Allows calculation of all other equilibrium calculations (see box):

KC

=CH

3CO

2C

5H

10éë

ùû

CH3CO

2Hé

ëùû

C5H

10éë

ùû

=0.000928

(0.006702)(0.0002552)= 542

B) %conversion =x

[CH3CO

2H]

0

´100 =9.28X10-4M

1.18X10-3M´100 = 78.6%

For K > 1.0, we would expect a high % conversion. The 78.6% conversion means there will be more

products than the limiting reactant left at equilibrium, which correlates to a strong reaction.

C) ∆G° = -RT lnKc

= -8.314 J

mol-K(298K)ln(542) = -15596 J = -15.6kJ

6.8 A) Given stoichiometry (see table) and Kp

= PHg( )

2

PO

2

At 420°C: Ptotal

= 3x = 50kPa1.0atm

101.3kPa

é

ëê

ù

ûú = 0.494atm x =

0.494atm

3= 0.165atm

Kp,693K

= PHg( )

2

PO

2

= (0.330)2(0.165) = 0.01797 = 0.0180

At 450°C: Ptotal

= 3x = 108kPa1.0atm

101.3kPa

é

ëê

ù

ûú = 1.066atm x =

1.066atm

3= 0.355atm

Kp,723K

= PHg

2PO

2

= (0.710)2(0.355) = 0.1790 = 0.179


87

B) At 420°C: KC ,693K

=K

P

(RT)3=

0.0180

0.08206 L-atm

mol-K(693K)( )

3=

0.0180

1.839X105= 9.79X10-8

At 450°C: KC ,723K

=K

P

(RT)3=

0.179

0.08206 L-atm

mol-K(723K)( )

3=

0.179

2.088X105= 8.59X10-7

C) There would be some difference, since the log terms would be different. It’s mostly due to rounding off

error - since only about 5%:

lnK

P,693K

KP,723K

é

ë

êê

ù

û

úú

= ln0.0180

0.179

é

ëê

ù

ûú = ln(0.1005) = -2.30 ln

KC,693K

KC,723K

é

ë

êê

ù

û

úú

= ln9.79X10-8

8.59X10-7

é

ëêê

ù

ûúú

= ln(0.1140) = -2.17

D) ∆H values from: lnK

T2

KT1

é

ë

êê

ù

û

úú

= -∆H

R

1

T1

-1

T2

é

ë

êê

ù

û

úú

= -∆H

R

(T2

- T1)

T1T

2

é

ë

êê

ù

û

úú

Kp’s: -2.30 = -∆H

r

8.314 J

mol-K

30K

(693)(723)K2

é

ëê

ù

ûú Þ ∆H

r=

2.30(8.314 J

mol-K)

(5.99X10-5 1

K)

1kJ

1000J

æ

èç

ö

ø÷ = 319kJ

Kp’s: -2.17 = -∆H

r

8.314 J

mol-K

30K

(693)(723)K2

é

ëê

ù

ûú Þ ∆H

r=

2.17(8.314 J

mol-K)

(5.99X10-5 1

K)

1kJ

1000J

æ

èç

ö

ø÷ = 301kJ

Since difference largely due to rounding off, should quote average: ∆Hr = 310(± 9) kJ

6.9 A) Using thermodynamic values for ∆G° for ions given below and tabled values for H2O(l):

∆G

r

o = 1.0mol∆Gf ,NH

4

+

o

(aq)+1.0mol∆G

f ,OH-(aq)

oé

ëêù

ûú- 1.0mol∆G

f ,NH3(aq)

o +1.0mol∆Gf ,H

2O(l)

oéëê

ùûú

= -79.31 + (-157.24)éë

ùû

- -26.5 + (-237.5)éë

ùûkJ = -27.0kJ

B) Kc = Kp since ∆n = 0, then:

lnK

C= -

∆G°

RT= -

-27.0X103 J

molO2

8.314 J

mol-K(298K)

= -10.90 K

C= e-10.90 = 1.85X10-5

So the thermodynamic values appear accurate, since the literature value is 1.8 X 10-5 for the dissociation.

C) • Will need the values of ∆H° and ∆S° for the reaction to determine their signs.

∆H

r

o = 1.0mol∆Hf ,NH

4

+

o

(aq)+1.0mol∆H

f ,OH-(aq)

oé

ëêù

ûú- 1.0mol∆H

f ,NH3(aq)

o +1.0mol∆Hf ,H

2O(l)

oéëê

ùûú

= -132.5 + (-229.99)éë

ùû

- -80.29 + (-285.5)éë

ùûkJ = 3.59kJ

∆Sr

o = 1.0mol Sf ,NH

4

+

o

(aq)+1.0mol S

f ,OH-(aq)

oé

ëêù

ûú- 1.0mol S

f ,NH3(aq)

o +1.0mol Sf ,H

2O(l)

oéëê

ùûú

= 113.4 + (-10.79)éë

ùû

- 111.3 + (69.91)éë

ùû

J

K= -78.6

J

K

No, given ∆H° is positive and ∆S° is negative, ∆G° will be positive at all temperatures and cannot change

sign.


88

6.10 A) Convert data so that a plot of ln K versus 1/T can be made:

B)

From equation for the line:

∆Hr

= -slope(R) = -(-7080K) 8.314 J

mol-K( ) = 58.9kJ / mol

C) Have to calculate ln K for 20°C (293K) using equation for the line

on graph:

lnK =-7080K

293K+19.005 = -24.16 +19.005 = -5.16

Then:∆G° = -RT lnK = -8.314J

mol - K(293K)(-5.16) = +12,567J = 12.57kJ / mol

Assuming ∆H is constant: ∆S293K

o =∆H° - ∆G°

T=

(58.9 -12.6)

293K

kJ

mol

1000 J

1kJ

æ

èç

ö

ø÷ = 158

J

mol - K

6.11 A) (a) logKP

= 7.55 -4844K

673K= 7.75 - 7.198 = 0.352 K

P= 100.352 = 2.25

∆G° = -2.303RT logK = -8.314J

mol - K(2.303)(673K)(0.352) = -4536 = -4.54kJ / mol

(b) Given the form of the Van’t Hoff equation is:

lnK = -∆H

r

o

RT+∆S

r

o

RÞ lnK =

∆Sr

o

R-∆H

r

o / R( )T

Since lnK = 2.303logK =∆S

r

o

R-∆H

r

o

RTÞ logK =

∆Sr

o

2.303R

æ

èç

ö

ø÷ -

∆Hr

o

2.303R

é

ë

êê

ù

û

úú´

1

T then can plot log K versus 1/T(K)

to get ∆H° from slope and ∆S° from y-intercept of plot.

So that: y - intercept = 7.55 =∆S

r

o

2.303R

æ

èç

ö

ø÷ Þ ∆S

r

o = 7.55(2.303)(8.314 J

mol-K) = 144.6 J

mol-K and

slope = -4844K = -∆H

r

o

2.303R

é

ë

êê

ù

û

úúÞ ∆H

r

o = 4844K(2.303)(8.314 J

mol-K) = 86,823 J

mol= 86.8 kJ

mol

B) KP

= KC(RT)∆n Þ K

C=

KP

(RT)∆n=

2.25

(0.08206 L-atm

mol-K(673K))(3-2)

=2.25

55.23= 0.0407

C) Given KC

=[HI(g)]2[C

5H

8(g)]

[I2(g)][C

5H

10(g)]

= 0.0407 =(2x)2(x)

(0.10 - x)2

Since K is fairly large, the method of successive approximations must be used to estimate the value of

“x”. A minimum of 4 successive approximation yields the likely estimated value of x= 0.0345, so that

[HI] ≈ 0.069 M at equilibrium

6.12 Q versus K question. Given current combination calculate Q and compare to K from ∆G°:

Q =[glu - 6 - phosphate]

[glu][HPO4

-2]=

1.6X10-4

(0.0045)(2.70X10-3)= 1.32

lnKC

= -∆G°

RT= -

-13.4X103 J

molO2

8.314 J

mol-K(310K)

= -5.20 KC

= e-5.20 = 5.52X10-3

Q > K so there are too many products. The reverse reaction will dominate and glu-6-phosphate will be

lowered in concentration, not increased.

6.13 (A) Coupled reaction: PEP + ADP Py + ATP, yields:


89

∆G° = ∆G°(reaction1)+ ∆G°(reaction2) = (-53.6 + 29.3)kJ = -24.3kJ

lnK = -∆G°

RT=

-(-24.3X103 J)

8.314 J

mol-K(298K)

= 9.808 KC

= e9.808 = 1.82X104

B) Consider iCe table, then substitute and solve for x:

K =[Py][ATP]

[PEP][ADP]= 1.82X104 =

x2

(0.010 - x)2Þ 1.82X104 =

x

(0.010 - x)= 134.9

x =1.35

135.9= 0.00993 = [ATP]

eq %conversion =

0.00993

0.010´100 = 99.3%

C) Without coupling reaction 2: lnKp

= -∆G°

RT= -

29.3X103 J

mol

8.314 J

mol-K(298K)

= -11.8 Kp

= e-11.8 = 7.31X10-6

Without coupling, the percent conversion of ADP to ATP would have been much lower, probably only a

few % based on the low value of K (≈ 2.7%). The 99% conversion ADP would be impossible without

coupling.

6.14 A) • Must reverse the second reaction to produce correct overall

glutamate + pyruvate ® ketoglutarate + alanine

+(ketoglutarate + aspartate ® glutamate + oxoxacetate)

pyruvate + aspartate ® alanine + oxoxacetate

K =[alanine][oxolacetate]

[pyruvate][asparate]

Then: ∆G°overall

= ∆G°(reaction1)éë

ùû

+ -∆G°(reaction2)éë

ùû

= (-1004 + (4812))J = 3808 J

lnK = -∆G°

RT=

-(3808 J)

8.314 J

mol-K(303K)

= -1.512 KC

= e-1.512 = 0.221

B) • Must calculate value of Q and then ∆G for the reaction.

Q =10-4éë

ùû

10-5éë

ùû

10-2éë

ùû

10-2éë

ùû

= 1.0X10-5

∆G = ∆G° + RT lnQ = 3808J( ) + 8.314J

mol - K(303K)ln(1.0X10-5)

= 3808J( ) + (-2.90X10-4) = -25.2kJ

So the reaction will be spontaneous under the cell conditions given.

6.15 A) Both FAD and NADH are the reactants, so the second reaction must be reversed before adding.

2e- + FAD + 2H+ ® FADH2

+ (NADH ® NAD+ + H+ + 2e- )

H+ + FAD + NADH ® FADH2

+ NAD+

∆G°overall

= (42.3 + (-65.5)) = -23.2kJ

B) lnKc

= -∆G°

RT=

+(23.2X103 J)

8.314 J

mol-K(298K)

= 9.364 Kc

= e9.364 = 1.17X104

C) The Kc term involves [H+], so the value of all equilibrium concentrations will depend on the pH. This

includes the thermodynamic values as well. Therefore to create a consistent reference value, the pH must

be set for biochemical reactions, like those involved in this example, and that has been determined to be

pH = 7.0 as part of the “biochemical standard state” conditions.

6.16 A)

∆GSTEP I

o = å∆Gf

o

,products- å∆G

f

o

,reactants= 1mol∆G

f

o

,cis-aconitate(aq)+1mol∆G

f

o

,H2O(l)

éëê

ùûú

- 1mol∆Gf

o

,citrate(aq)éë

ùû

= -921.7 + (-237.0)éë

ùû

- (-1167)éë

ùû

= 8.3kJ = ∆GSTEP I

o

B) lnKc(STEP I) = -

∆G°

RT=

-(8.30X103 J)

8.314 J

mol-K(298K)

= -3.35 Kc(STEP I) = e-3.35 = 0.0351

KSTEP I

=[cis - aconitate]

[citrate]= 0.0351 Þ 0.0351 =

12.0mM

[citrate]Þ [citrate] = 342mM = 0.342M


90

C) ∆G°overall

= 13.3kJ = ∆G°STEP I

+ ∆G°STEP II

= (8.3 + ∆G°STEP II

) Þ ∆G°STEP II

= 5.0kJ

D) Since Koverall

= KSTEP I

´ KSTEP II

Þ KSTEP II

=K

overall

KSTEP I

• Need to calculate Koverall

lnKc,overall

= -∆G°

RT=

-(13.30X103 J)

8.314 J

mol-K(298K)

= -5.37 Kc,overall

= e-5.37 = 4.65X10-3

KSTEP II

=K

overall

KSTEP I

=4.65X10-3

0.0351= 0.133

6.17 A) Convert data and plot ln K versus (1/T(K)):

Then calculate ∆H° and ∆S° from trendline equation:

∆H° = -R(slope) = -8.314 J

mol-K(-56.838K) = 473 J

mol

∆S° = R(y - intercept) = 8.314 J

mol-K(-1.6645) = -13.8 J

mol-K

B) Can use equation to get ln K and then ∆G° OR use ∆H° and ∆S° to calculate ∆G°:

∆G° = (473 J

mol) - (298.15K)(-13.38 J

K-mol) = 473 + 4123 = 4597 J

mol= 4.60 kJ

mol

C) Will need K for 298.15K, can use equation or ∆G° to calculate:

lnK =-58.838K

298.15K-1.6645 = -1.855

K298

= e-1.855 = 0.1565

K =2 - Pglyéë

ùû

3 - Pglyéë

ùû

= 0.1565 Þxéë

ùû

0.150 - xéë

ùû

= 0.1565 Þ 2 - Pglyéë

ùû

= x = 0.02029 = 0.020M

3 - Pglyéë

ùû

= 0.150 - x = 0.1297 = 0.130M

6.18

A) Dilute Solution or pure water: Higher ionic strength solutions:

Ka,HA =

[H3O+][A-]

[HA] K

a,HA =

aH

3O+

aA-

[HA]=

gH

3O+

[H3O+]´ g

A-[A-]

[HA]= g

H3O+

gA-

æè

öø´

[H3O+][A-]

[HA]

Kb,B =

[BH+][OH-]

[B] K

b,B =

aBH+

aOH-

[B]=

gBH+

[BH+]´ gOH-

[OH-]

[B]= g

BH+g

OH-( ) ´[BH+][OH-]

[B]

Ksp

,M2X = [M+]2[X -] = 4s3 → K

sp,M

2X = (a

M+)2(a

X-) = (g

M+)2(g

X-) ´ 4s3

6.19 A) (a) CaCO3(s) Ca+2(aq) + CO3-2(aq) K

sp= a

Ca+2(a

CO3

-2) = g

Ca+2(g

CO3

-2)[Ca+2][CO

3

-2]

The values for the activity coefficients will be less than 1.0, so the product will be a decimal value. In

order to equal the same Ksp value (set by the thermodynamics), the solubility or dissolved concentrations

of ions must increase.

(b) B(aq) + H2O(l) BH+(aq) + OH-(aq)

Kb

=a

BH+(a

OH-)

aB

= gBH+

(gOH-

)[BH+][OH-]

[B]

The values for the activity coefficients for the two ions, BH+ and OH- will be less than 1.0 but that for the

neutral weak base, B, will be 1.0. Consequently, in order to keep K the same, the ratio of ions to neutral

weak base at equilibrium must increase.


91

B) Fe+3 + SCN- [Fe(SCN)]+2 Kf

=a

[Fe(SCN)]+2

aFe+3

(aSCN-

)=

g[Fe(SCN)]+2

gFe+3

(gSCN-

)

[Fe(SCN)+2]

[Fe+3][SCN-]=

g[Fe(SCN)]+2

gFe+3

(gSCN-

)Q

As the ionic strength increases, the ratio of activity coefficients will get smaller since there are two ions

terms in the denominator, but only one in the numerator. So assuming the activities are close in value,

the effective ratio is 1/γ and gets larger as I increases, since γ will get smaller. Consequently, Q which

equals the ratio of complex to the ions should get smaller as I increases, so that the K stays the same.

6.20 A) Solubility in pure water

Ksp,BaCrO

4

= [Ba+2][CrO4

-2] = 2.11´10-10 = s2 s = 2.11´10-10 = 1.45 ´10-5 M

B) Solubility in 0.10 M NaCl

Ksp,BaCrO

4

= 2.11´10-10 = aBa+2

aCrO

4

-2= g

Ba+2g

CrO4

-2[Ba+2][CrO

4

-2] = gBa+2

gCrO

4

-2

æè

öøs2

CNa+

= CCl-

= 0.10M and CBa+2

= CCrO

4

-2» 10-5 M

I =1

2(0.1)(+1)2 + (0.1)(-1)2éë

ùû

= 0.10 and I = 0.10 = 0.316

log gBa+2

=-0.51(2)2(0.316)

1 +500

305(0.316)

= -0.645

1.518= -0.424 g

Ba+2= 10(logg ) = 10-0.424 = 0.376

log gCrO

4

-2=

-0.51(-2)2(0.316)

1 +400

305(0.316)

= -0.645

1.415= -0.456 g

CrO4

-2= 10(logg ) = 10-0.456 = 0.350

Ksp,BaCrO

4

gBa+2

gCrO

4

-2

æè

öø

=2.11´10-10

(0.376)(0.350)= s

obs

2 = 1.60 ´10-9 Þ sobs

= 4.0 ´10-5

So the solubility of the compound is increased to four times its value in water alone. More solid will

dissolve in the NaCl solution than in pure water.

6.21 For a weak acid dissociation HA + H2O(l) H3O+ + A-,

A) Ka

=[H+][A-]

[HA]=

aH+

´ aA-

[HA]=

[gH+

x][gA-

x]

[HA]o

- x becomes

Ka

= gH+

gA-

x2

[HA]o

- x» g

H+g

A-

x2

[HA]o

x2 =K

a[HA]

o

gH+

gA-

Þ x = [H+] =K

a[HA]

o

gH+

gA-

(a) pH in pure water, 0.10 M CH3CO2H

Ka

=[H+][A-]

[HA]=

[x][x]

[HA]o

- x»

x2

[HA]o

Þ x = 1.8X10-5 (0.10) = 1.34X10-3 pH = - log(1.34X10-3) = 2.87

(b) With activities in 1.0M KCl:

CK+

= CCl-

= 1.00M and CH+

= CC

2H

3O-

» 10-3 M I = 1.0 and I = 1.0

log gH+

=-0.51(1)2(1.0)

1 +900

305(1.0)

= -0.51

3.95= -0.129 Þ g

H+= 10-0.129 = 0.743

log gC

2H

3O

2

-=

-0.51(-1)2(1.0)

1 +450

305(1.0)

= -0.51

2.48= -0.206 Þ g

C2H

3O

2

-= 10-0.206 = 0.622

x = [H+] =K

a[HA]

o

gH+

gA-

=1.8X10-5(0.10)

(.743)(0.622)= 6.24X10-3 pH

obs= - log(6.24X10-3) = 2.20


92

6.22 A) Prove that pH = pKa,acid form

+ log[base form]

[acid form] becomes:

pHobs

= pKa,acid form

+ logmol base

mol acid

é

ëê

ù

ûú + log

gbase

gH+

gacid

é

ë

êê

ù

û

úú when activities included.

B) Suppose you combine 2.00 g NaH2PO4(s) [MW = 104] with 9.00 g Na2HPO4(s) [MW = 126] in 500 mL

of water. What will be the pH in the solution?

(a) Neglecting activities:

Equilibrium : H2PO4-1 + H2O H3O+ + HPO4

-2

no. mol NaH2PO

4= 2.0g

1mol NaH2PO

4

104g

é

ë

êê

ù

û

úú

= 0.0192mol NaH2PO

4

no. mol Na2HPO

4= 8.77g

1mol Na2HPO

4

126g

é

ë

êê

ù

û

úú

= 0.0696mol Na2HPO

4

pH = pKa,H

2PO

4-

+ log[mol HPO

4

-2]

[mol H2PO

4

-]= 6.78 + log

0.0696

0.0192

é

ëê

ù

ûú = 6.78 + 0.56 = 7.34

(b) With activities must first calculate ionic strength:

(• Can neglect [H3O+] ions, concentration too small)

M H2PO

4

- =0.0192mol H

2PO

4

-

0.500L

é

ë

êê

ù

û

úú

= 0.0384M M HPO4

-2 =0.0696mol HPO

4

-2

0.500L

é

ë

êê

ù

û

úú

= 0.1392M

M Na+ =0.1584mol Na+

0.500L

é

ëêê

ù

ûúú

= 0.3168M

I =1

2(0.317)(+1)2 + (0.0384)(-1)2 + (0.1392)(-2)2éë

ùû

=1

2(0.317) + (0.0384) + (1.112)éë

ùû

= 0.734

I = 0.734 = 0.856

Then apply DHHL for coeffcients:

log gH

2PO

4

-=

-0.51(-1)2(0.856)

1 +450

305(0.856)

= -0.437

2.26= -0.193 Þ g

H2PO

4

-(acid)= 10-0.193 = 0.641

log gHPO

4

-2=

-0.51(-2)2(0.856)

1 +400

305(0.856)

= -1.75

2.12= -0.824 Þ g

HPO4

-2(base)= 10-0.824 = 0.149

log gH+

=-0.51(1)2(0.856)

1 +900

305(0.856)

= -0.437

3.52= -0.124 Þ g

H+= 10-0.124 = 0.751

Final result: pHobs

= 6.78 + 0.56 + log(0.149)(0.751)

(0.641)

é

ëê

ù

ûú = 7.34 + (-.758) = 6.58 =pH

obs

6.23 A) ∆E° must be positive to produce a spontaneous reaction so that:

∆E° = E°ClO

4

- /ClO3

-- E°

Cd+2 /Cd(OH)2

= 0.36 - (-0.81)V = 1.17V so ClO4- reduced.

• ClO4- half reaction must be the cathode reaction

B) Overall Reaction: ClO4- + Cd(s) +H2O(l) → Cd(OH)2(s) + ClO3

- and K =[ClO

3

-]

[ClO4

-]

C) Cell notation: Cd| Cd(OH)2(s)| OH-||OH-, ClO3-, ClO4

-|Pt

D) Since n = 2 mol e’s in balanced reaction, then:

∆G° = -nF ∆E° = -2.0mol(9.65X104 coul

mol)(1.17V) = -2.558X105 J = -226kJ

lnK = -∆G°

RT=

2.258X105 J

8.314 J

mol-K(298K)

= 91.14 and K = e91.14 = 3.82X1039


93

E) Since the K value is so high, it is very likely that the ratio of ClO3- ion to ClO4

- could not be determined

by any direct measurements since ClO4- would always be extremely small at equilibrium.

6.24 • Work = -nF∆E°= ∆G° for overall reaction, then must make per mole of metal oxidized

Half reactions: ∆E° n ∆G°/mol metal oxidized

(a) Red. Cu+2+ 2 e- → Cu(s)

Oxid. Zn(s) → Zn+2+ 2 e- 0.34-(-0.76)=1.10 V 2

(b) Red. 2H+ + 2e- → H2(g)

Oxid. Ca(s) → Ca+2+ 2 e- 0.0-(-2.87) = 2.87 V 2

(c) Red. 2H2O + 2e- → H2(g)+ 2OH-

Oxid. Li(s) → Li++ e- -0.83 -(-3.05) = 2.22 V 1

(d) Red. O2(g)+4H++ 4e- → 2H2O

Oxid. Ni(s) → Ni+2+ 2 e- 1.23-(-0.23)=1.46V 4

6.25 A) Need to have the overall reaction be: Hg2SO4(s) Hg2+2 + SO4

-2

Cathode: Hg2SO4(s) + 2e- → 2 Hg(l) + SO4-2 E° = 0.6125 V

Anode: Hg2+2 + 2e- → 2 Hg(l) E° = 0.7973 V

∆E° = 0.6125 - (-0.7973) V = - 0.1848 V

B) lnKsp

=nF

RT∆E '° =

2(9.65X104 coul

mol)

8.314 J

mol-K(298K)

(-0.1848V) = -14.395 Ksp

Hg2SO

4(s) = e-14.395 = 5.60X10-7

C) The literature value of 6.5 X10-7 is quite close to the calculated value. A difference of a few hundredths

of a volt in the ∆E° could account for the difference.

6.26 A) Need the formation reaction: Hg+2 + 4 CN- Hg(CN)4-2

• Reverse the first half reaction and add it to the second half reaction, so that:

Cathode: Hg+2 + 2e- → Hg(l) E° = 0.86 V

Anode: Hg(l) + 4 CN- → Hg(CN)4-2 + 2e- E° = -0.37 V

∆E° = 0.86 - (-0.37) V = 1.23 V

B) ∆G° = -2.0mol(9.65X104 coul

mol)(1.23V) = -2.374X105 J = -23.7kJ

lnK =-∆G°

RT=

2.374X105 J

8.314 J

mol-K(298K)

= 95.8 Kformation

= e95.8 = 4.09X1041

C) (a) Cell notation: Hg(l)|Hg(CN)4-2, CN-||Hg+2|Hg(l)

(b) Need an electrical connection to the Hg(l) even though it can act as the metal electrode.

6.27 A) The Cl- ion is a spectator ion since both CuCl and CuCl2 are aqueous species, so that the half

reactions do not involve Cl- ion.

Reduction: Cu+2+ 2 e- → Cu(s) E° = 0.34 V Oxidation: Cu+1→ Cu+2 + e- E° = 0.16 V

∆E° = 0.34 - (0.16)V = 0.18V

B) lnK =nF ∆E°

RT=

(2mol)(9.65X104 coul

mol)(0.18V)

8.314 J

mol-K(298K)

=34740

2477.5= 14.02 K = e14.02 = 1.23X10

6

C) Calculate Q and apply the Nernst Equation

(a)Q =[Cu+2]

[Cu+]2=

(x)

(0.20 - x)2=

(0.10)

(0.10)2= 10 → ∆E = ∆E° -

0.05916

nlogQ = 0.18V - 0.0296log(10) = 0.15V

(b) The ∆E would be the same because the ratio Q would have the same value, if 50% conversion occurs,

independent of the starting concentration.

D) As long as the same concentration of HCl is added to both cells, the voltage should not change.

∆G° = -2nF(1.10) =-212.3kJ

1mol Zn= w

∆G° = -2nF(2.87) =-554kJ

1molCa= w

∆G° = -nF(2.22) =-214kJ

1mol Li= w

∆G° = -nF(1.46) =-564kJ

1molNi= w


94

6.28 A) • Use Nernst equation for half reaction: EMnO

4

- /Mn+2= E°

MnO4

- /Mn+2-

0.05916

5log

[Mn+2]

[MnO4

-2][H+]8

If assume standard conditions for other ions, [Mn+2]=[MnO4-2] = 1.0 M then at:

(a) pH = 6.0 EMnO

4

- /Mn+2= 1.51V - 0.01183V log

1

[10-6]8

æ

èç

ö

ø÷ = 1.51 - 0.01183V(48) = 0.942V

(b) pH = 2.0 EMnO

4

- /Mn+2= 1.51V - 0.01183V log

1

[10-2]8

æ

èç

ö

ø÷ = 1.51 - 0.01183V(16) = 1.32V

B) The lower pH produces a more positive value of E, so when coupled in a redox reaction, you should

favor getting more product.

6.29 Combustion/Fuel cell reaction: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

• CH4 → CO2 loss of 8 electrons (-4 → +4 change in oxidation number)

• 4 O atoms gaining 2 electrons each (0 → -2 change in oxidation number)

A) • Can get ∆E° from ∆G° for reaction.

∆G° = [1mol∆G

f

o

,CO2(g)

+ 2mol∆Gf

o

,H2O(l)

]- [1mol∆Gf

o

,CH4(g)

+ 2mol∆Gf

o

,O2(g)

]

= [-394.4 + 2(-237.1)]kJ - [-50.8 + 2(0)]kJ = -818kJ

Given n= 8 mole electrons: ∆E° = -∆G°

nF=

(8.18X105 J)

8mol(9.65X104 coul

mol)

= 1.059V

Q = 0.03806 =[Cu+2][Fe+2]2

[Fe+3]2=

[Cu+2][0.060]2

[0.20 - 0.060]2Þ [Cu+2] =

0.03806(0.0196)

(0.0036)= 0.207M

Yes, this fuel cell produces just a little over 1.0 volt at 25°C.

B) Half reaction for CH4: CH4(g) + 2 H2O(l) → CO2(g) + 8 H+ + 8 e-

∆E° = E°O

2/H

2O

- E°CH

4/CO

2

Þ E°CH

4/CO

2

= E°O

2/H

2O

- ∆E° = 1.23V -1.06V = 0.17V

C) Need moles CH4 reacted:

n =PV

RT=

20.0atm(30.0L)

0.08206 L-atm

mol-K(298K)

= 2.45mol-818kJ

1molCH4

é

ë

êê

ù

û

úú

= -2008kJ = ∆G = wmax

6.30 Cathode: Fe+3+ e- → Fe+2 E° =0.77V Anode: Cu+2+ 2e- → Cu(s) E° =0.34V

Overall reaction: 2 Fe+3+ Cu(s) → 2Fe+2 +Cu+2 ∆E° =0.43V

∆E = ∆E° -0.0257

nlnQ Þ 0.472V = 0.43V - 0.01285V lnQ Þ lnQ =

(0.472 - 0.43)V

-0.1285= -3.268

Q = 0.03806 =[Cu+2][Fe+2]2

[Fe+3]2=

[Cu+2][0.060]2

[0.14]2Þ [Cu+2] = 0.03806

0.14

0.060

é

ëê

ù

ûú

2

= 0.207M

6.31 A) Given half reaction is: M+n + ne- ®M(s)

Ecathode

= E°M+ n/M

-0.0592

nlog

1

[M+n]high

é

ë

êê

ù

û

úú and E

anode= E°

M+ n/M-

0.0592

nlog

1

[M+n]low

é

ë

êê

ù

û

úú then:

∆E = Ecathode

- Eanode

= E°M+ n /M

-0.0592

nlog

1

[M+n]high

é

ë

êê

ù

û

úú

æ

è

çç

ö

ø

÷÷

- E°M+ n /M

-0.0592

nlog

1

[M+n]low

é

ë

êê

ù

û

úú

æ

èçç

ö

ø÷÷

=0.0592

nlog

1

[M+n]low

é

ë

êê

ù

û

úú

- log1

[M+n]high

é

ë

êê

ù

û

úú

æ

è

çç

ö

ø

÷÷

=0.0592

nlog

[M+n]high

[M+n]low

é

ë

êê

ù

û

úú

B) Cell I: ∆E =0.0592

3log

0.300

0.020

é

ëê

ù

ûú = 0.01973(1.176) = 0.0232V • Highest voltage


95

Cell II: ∆E =0.0592

2log

1.00

0.20

é

ëê

ù

ûú = 0.0296(0.699) = 0.0207V • Same as Cell IV

Cell III: ∆E =0.0592

2log

2.00

1.00

é

ëê

ù

ûú = 0.0296(0.301) = 0.00891V • Lowest voltage

Cell III: ∆E =0.0592

3log

2.60

0.232

é

ëê

ù

ûú = 0.0197(1.0495) = 0.0207V

6.32 A) Since ∆E negative means cell concentration inside is less than the outside concentration.

∆E = 0.0592log[K +]

inside

[K+]outside

é

ë

êê

ù

û

úú → -0.067V = 0.0592V log

[K+]inside

[K+]outside

é

ë

êê

ù

û

úúÞ

[K+]inside

[K+]outside

é

ë

êê

ù

û

úú

= 10-1.133 = 0.0736

6.33 A) Ecathode

= E°H+ /H

2(g)

-0.0592

1log

1

[H+]high

é

ë

êê

ù

û

úú

= 0.0V - 0.0592 - log[H+]high( ) = -0.0592pH

high

Eanode

= E°H+ /H

2(g)

-0.0592

1log

1

[H+]low

é

ë

êê

ù

û

úú

= -0.0592pHlow

so ∆E = 0.0592 pHanode

- pHcathode( )

Ecathode

- Eanode

= (300 - 216)mV = 59.2mV pHanode

- 3.00( ) Þ1.419 + 3.00 = 4.42 = pHunknown

• Higher pH means lower [H+] in unknown

6.34 Left side: Cu++ e- → Cu(s) Right side: CuCl(s)+ e- → Cu(s)+ Cl-

Overall reaction: CuCl(s) Cu+)+ Cl- which is the Ksp of CuCl(s)

• But given ∆E, not ∆E°, so can’t calculate K from ∆E°. Must find [Cu+] on right side since that equals “s”,

the molar solubility of CuCl(s) and then apply Ksp = s2.

∆E = 0.175V =0.0592

1log

[Cu+]Left

[Cu+]Right

é

ë

êê

ù

û

úúÞ

0.175V

0.0592V= log

1.00M

[Cu+]Right

é

ë

êê

ù

û

úúÞ

1.00M

[Cu+]Right

é

ë

êê

ù

û

úú

= 102.956

[Cu+]Right

= molarsolubility, s =1

903.8= 1.106X10-3M K

sp= 1.106X10-3( )

2

= 1.22X10-6

6.35 A) ∆E° must be positive to produce a spontaneous reaction so that:

∆E° = E°NAD+ /NADH

- E°CO

2/HCO

2

-= -0.320 - (-0.42)V = 0.10V so NAD+ reduced.

• NAD+ reduction must be the cathode reaction

B) Since n = 2 mol e’s in balanced reaction:

∆G '° = -nF ∆E '° = -2.0mol(9.65X104 coul

mol)(0.10V) = -19,300J = -19.3kJ

lnK ' = -∆G '°

RT= -

(-1.93X104 J)

8.314 J

mol-K(298K)

= 7.78 K ' = e7.79 = 2416

6.36 A) Cathode: NAD++ 2e- + H+ → NADH Anode: O2(g)+ 2H++ 2e- → H2O2

B) Pt|O2(g) (P = 1.0 atm)|H+, H2O2||NADH, H+, NAD+|Pt

C) (a) ∆E’° = 0.295-(-0.320)V = 0.615 V

(b) ∆G '° = -nF ∆E '° = -2.0mol(9.65X104 coul

mol)(0.615V) = -3.14X105 J = -314kJ

(c) lnK ' = -∆G '°

RT=

-3.14X105 J

8.314 J

mol-K(298K)

= 126.7

6.37 A) Cathode: FAD + 2H++ 2e- → FADH2 E’° = -0.219V

Anode: NAD++ 2e- + H+ → NADH E’° = -0.320V

∆E°’ = -0.219-(-0.320) V = 0.101V

∆G '° = -2.0mol(9.65X104 coul

mol)(0.101V) = -1.95X105 J = -19.5kJ

B) No, the ADP to ATP conversion requires 30.5-35.0 kJ per mole of ADP depending on pH. This reaction

would only provide about 2/3 of the energy to convert one mole of ADP.


96

6.38 A) cytochrome c(Fe+3) + e- → cytochrome c(Fe+2) E’° = 0.254 V

pyruvate + 2H+ + 2e- → lactate E’°= -0.185 V

Balanced reaction: 2 cytochrome c(Fe+3) + lactate 2 cytochrome c(Fe+2) + pyruvate

Then: ∆E’° = 0.254 - (-0.185) V = 0.438 V

B) (a) lnK ' =-∆G '°

RT=

nF

RT∆E '° =

n(9.65X104 coul

mol)

8.314 J

mol-K(298.15K)

(∆E '°) = 38.93V -1(n∆E '°)

(b) lnK ' = 38.93V -1(2 ´ 0.438)V = 34.1 K ' = e34.1 = 6.47X1014

C) Since K ' =[cytochrome - c(Fe+2)]2[pyruvate]

[cytochrome - c(Fe+3)]2[lactate]= 1.76X1014 then:

K ' =[1000]2

[1]2´

[pyruvate]

[lactate]= 6.47X1014 Þ

[pyruvate]

[lactate]=

6.47X1014

1.0X106= 6.47X108

D) Q =[cytochrome - c(Fe+2)]2[pyruvate]

[cytochrome - c(Fe+3)]2[lactate]=

[0.100M]2[0.20M]

[0.100M]2[1.0X10-5M]= 2.0X107

So Q < K, [pyruvate] increases

6.39 A)

B) lnK

pH=7.0

KpH=5.0

æ

èçç

ö

ø÷÷

=-∆G

pH=7.0+ ∆G

pH=5.0

RT=∆G

pH=5.0- ∆G

pH=7.0

RT=

3.425X104 J / mol

8.314J

mol-K(298K)

= 13.82 → K

pH=7.0

KpH=5.0

= 1.00X106

• So K is a million times larger at pH = 7.0 than at pH =5.0

C) (a) % inc =87.5 - 70.35

70.35´100 = 24.4% (b) 19.5% (c) 48.7%

6.40 A) Applying the standard thermodynamics calculation:

∆G '° = 2mol(∆Gf ,cytochrome-c(Fe+2)

'° , kJ

mol) +1mol(∆G

f ,pyruvate

'° , kJ

mol)é

ëêùûú

- 2mol(∆Gf ,cytochrome-c(Fe+3)

'° , kJ

mol) +1mol(∆G

f ,lactate

'° , kJ

mol)é

ëêùûú

At I = 0: ∆G '° = 2(-24.54) +1(-352.4)éë

ùû

- 2(0) +1(-316.94)éë

ùûkJ = -84.5kJ

At I = 0.10: ∆G '° = 2(-26.96) +1(-351.2)éë

ùû

- 2(-5.51) +1(-314.5)éë

ùûkJ = -79.6kJ

At I = 0.25: ∆G '° = 2(-27.75) +1(-350.8)éë

ùû

- 2(-7.29) +1(-313.7)éë

ùûkJ = -78.0kJ

B) (a) Use lnK ' = -∆G '°

RT → K = 6.6 X 1014(I=0), K = 9.0 X 1013(I=0.10), K = 4.8 X 1013(I=0.25)

(b) They agree very well 6.6 X 1014 versus 6.74 X 1014 from ∆E’°.

C) K’ is 7.33 times larger at I = 0 than at I =0.10, while K’ is 13.8 times larger at I = 0 than at I = 0.25,

so K is decreasing as ionic strength increases in this reaction.

D) Cellular environments have higher values of ionic strength so that knowing how K’ depends on ionic

strength gives a better picture of the reaction in its real environment.

6.41 A) Balanced reaction: NAD+ + CH3CH2OH NADH + CH3CHO + H+

B) Applying the standard thermodynamics calculation:

∆G '° = 1mol(∆Gf ,NADH

'° , kJ

mol) +1mol(∆G

f ,CH3CHO

'° , kJ

mol)é

ëêùûú

- 1mol(∆Gf ,NAD+

'° , kJ

mol) +1mol(∆G

f ,ethanol

'° , kJ

mol)é

ëùû

At I = 0: ∆G '° = +25.3kJ Use lnK ' = -∆G '°

RTthen K’ = 3.72 X 10-5

At I = 0.25: ∆G '° = +22.1kJ leads to: K’ = 1.33 X 10-4


97

C) In this reaction, K is increasing with increasing ionic strength, opposite to the situation in Problem

6.40.

D) (a) K ' = Kc

´107x In this reaction x = +1.0 since H+ is a product, so that:

Kc

=K '

107=

3.72X10-5

107= 3.72X10-12

(b) If [H+] = 1.0 M for Kc and 1.0 X 10-7 for K’ then a shift towards product should occur since the

numerator in Q has decreased dramatically, so it is consistent.

6.42 A) Balanced reaction: O2(g) + 2 cysteine 2 H2O(l) + 4 cystine, ∆E’° (0.814 – (-0.34)V = 1.16 V,

∆G '° = -nF ∆E '° = -4.0mol(9.65X104 coul

mol)(1.156V) = -4.46X105 J = -446kJ

B) The concentration should decrease as the O2(g) converts cysteine spontaneously to cystine in the very

product favored reaction.

6.43 A) (a) 2 AgCl(s) + Zn(s) → 2 Ag(s) + 2 Cl- + Zn+2 (b) ∆E° = (0.22 – 0.76) V = 0.98 V

(c) (1) AgCl(s) needed for cathode reaction is coating the electrode at the bottom

(2) Ag(s) needed for cathode reaction and functions as one of the metal electrodes

(3) Zn(s) needed for the anode reaction and functions as the second metal electrode

(4) Solution of ZnCl2, containing Zn+2 ions needed for anode reaction

(5) The function of this part is for the electrical connections needed to measure voltage.

(d) Since the reactants are both solids, they cannot come into contact and exchange electrons except

through the external wires so we don’t need to separate them.

B) • Must define Q in terms of the concentration of [Zn+2] = x then solve for Q from Nernst Equation:

[Zn+2] = x,[Cl-] = 2x leadsto:Q = [Zn+2][Cl-]2 = (x)(2x)2 = 4x3

∆E - ∆E° = -0.0257

nlnQ Þ (1.015 - 0.98)V = -0.01285lnQ Þ lnQ = -

0.035

0.01285= -2.724

Q = e-2.724 = 0.06563 = 4x3 Þ x = [Zn+2] = 0.016413= 0.254M

C) ∆G° = -nF ∆E° = -2.0mol(9.65X104 coul

mol)(0.98V) = -1.89X105 J = -189kJ

∆S° = nFd∆E°

dT

é

ëê

ù

ûúP

= 2.0mol(9.65X104 coul

mol)(-4.02X10-4 V

K) = -77.6

J

K

∆H° = ∆G° + (T ∆S°) = -195kJ + 298K(-77.6J

K)

1 kJ

1000J

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= -195 - 23 = -218kJ

6.44 Overall reaction: Hg2Cl2(s) +H2(g) Hg(l) + 2 HCl n= 2 mol e-

• Need to interpolate value of ∆E° at 298K from data to calculate ∆G° at 298K

∆E298

o = ∆E293

o -5K

10K

æ

èç

ö

ø÷ (∆E

303

o - ∆E293

o ) = 2.699V - 0.5(0.0030)V = 2.6975V

∆G° = -nF ∆E° = -2.0mol(9.65X104 coul

mol)(2.6975V) = -5.206X105 J = -520.6kJ

• Estimate derivative as: (∆E°303 - ∆E°303)/(303-293) K

∆S° = nFd∆E°

dT

é

ëê

ù

ûúP

» nF∆(∆E°)

∆T

é

ëê

ù

ûú = 2.0mol(9.65X104 coul

mol)

-3.0X10-3

10

V

K

é

ëêê

ù

ûúú

= -57.9J

K

∆H° = ∆G° + (T ∆S°) = -520.6kJ + 298K(-57.9J

K)

1 kJ

1000J

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= -520.6 -17.25 = -537.8kJ

6.45 A) Cathode: Hg2Cl2(s) + 2e- → 2 Hg(l) + 2 Cl- E° = 0.268 V

Anode: AgCl(s) + e- → Ag(s) + Cl- E° = 0.222 V

Overall: Hg2Cl2(s) + 2 Ag(s) 2 Hg(l) + 2 AgCl(s) ∆E° = 0.268 -(0.222) V = 0.046 V

B) ∆G° = -2.0mol(9.65X104 coul

mol)(0.046V) = -8878J = -8.88kJ

C) • Must plot data, as ∆E(volts), not millivolts, versus T(K)

• From equation: slope = 3.544 X10-4 V/K


98

∆S° = nFd∆E°

dT

é

ëê

ù

ûúP

= 2.0mol(9.65X104 coul

mol)(3.54X10-4 V

K) = 68.4

J

K

∆H° = ∆G° + (T ∆S°)

= -8.878kJ + 298K(68.4J

K)

1 kJ

1000J

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= -8.878 + 20.38 = 11.5kJ

D) Since in this cell ∆H°(+) and ∆S°(+), then ∆G° will change sign with

T, as will ∆E°. It should spontaneous at higher T’s, non-spontaneous at

lower T’s. Since the ∆E° is positive we are above the changeover

temperature at 298K.

6.46 A) Calculated ∆G°, ln K, and K values:

B) Plot of ∆E° versus T(K) and C) Plot of lnk versus 1/T(K)

B) The plot of ∆E° versus T(K) is linear, so ∆S° is

constant over the T range. The value of ∆S° is:

∆S° = nFd∆E°

dT

é

ëê

ù

ûúP

= 1.0mol(9.65X104 coul

mol)(-6.21X10-4 V

K) = -60.0

J

K

C) The plot of ln K versus 1/T(K) is linear, so ∆H° is

constant. The value of ∆H° is then:

∆H° = -R(slope) == 8.314 J

K-mol(4.722X103K) = -39.25kJ


99

D) There is excellent agreement between the values of ∆H calculated from each ∆E° using the equation

and the plot. They differ only by hundredths of a kilojoule.

6.47 A) Balanced reaction: AgBr(s) + ½ H2(g) → Ag(s) + HBr(aq) and

Kdissociation

=[H+][Br -]

[HBr]

B) (a) Calculated results for ∆G°:

(b) As T increases, ∆G° for the dissociation decreases indicating the dissociation is less spontaneous at

the higher T’s. This indicates the sign of ∆H is negative for the dissociation.

(c) When the % ethanol increases, the ∆G° decreases indicating the change in solvent is impeding the

dissociation. So the dielectric strength of the solvents

is affecting the dissociation of the acid. In addition the

activity effects in the binary mixture discussed earlier in Part

5 (page 45, 47), may also be playing a role in impeding the

dissociation.

C) (a) Calculated results

for Kc for the dissociation:

(b) In pure water, the K values, although greater than 1.0, decrease

dramatically as T increases. The same trend is observed in the other

two solvents. The ratio of the greatest value of Kdiss to the least is about

3 in each solvent. This would indicate that the ∆H are similar for each solvent.

D) Plot of data for ∆S°:

(a) The results are not linear equations, but would be

polynomials in T. So ∆S° is not staying constant over the T range

studied.

(b) Estimating the ∆S° value within each 10 K region using:

∆S° =∆(∆E°

T2

- ∆E°T1

)

(T2

- T1)

and Taverage

=T

2+ T

1

2

• The estimated ∆S° values change over the T regions,

becoming more negative at the higher T’s, with some exceptions

for parts of the 20% and 50% ethanol data. From the graph, one

can observe that the 50% ethanol produces larger shifts in ∆S°

as the T is increased and that is also shown in the estimated

values. Considering the interactions of ethanol and water are

quite strong in this weight % region (50% ethanol by mass =

mol fraction of 0.28 for ethanol), it is not surprising the entropy

for the dissociation of the ions is being affected.

6.48 • Need ionic strength I =1

2c

iz

i

2

all ions

å =1

20.010(+1)2 + 0.005(-2)2éë

ùû

= 0.0150m so I = 0.1225

A) logg±

=-0.509| z

+z

-| I

1 + I=

-0.509(2)(0.1225)

1.1225= -0.111 then g

±= 10-0.111 = 0.774

m±

= (m+

n+m-

n-)1/n = (0.005) ´ (0.010)2( )1/3

= 5.0X10-73= 7.94X10-3m

a±

= g±m

±= 0.774(7.94X10-3) = 6.14X10-3

B) I =1

2c

iz

i

2

all ions

å =1

20.010(+1)2 + 0.005(-2)2 + 0.010(+1)2 + 0.010(+1)2éë

ùû

= 0.0350m so I = 0.187

logg±

=-0.509| z

+z

-| I

1 + I=

-0.509(2)(0.187)

1.187= -0.1604 and g

±= 10-0.1604 = 0.691

For activity, m± stays the same, so that a±

= g±m

±= 0.691(7.94X10-3) = 5.49X10-3m


100

C) Using the Davies equation:

logg±

=0.509| z+z

-|

- I

1 + I+ 0.300I

é

ë

êê

ù

û

úú

= 0.509(2)-(0.187)

1.187+ 0.3(0.035)

é

ëê

ù

ûú = -0.150 g

±= 10-0.150 = 0.708

a±

= g±m

±= 0.674(7.94X10-3) = 5.62X10-3m

No, the activity coefficients are close and the mean activities are very much the same.

6.49 A) I =1

2c

iz

i

2

all ions

å =1

20.40(+1)2 + 0.20(-2)2éë

ùû

= 0.600m so I = 0.775

logg±

=-0.509| z

+z

-| I

1 + I=

-0.509(2)(0.775)

1.775= -0.444 and g

±= 10-0.444 = 0.360

m±

= (m+

n+m-

n-)1/n = (0.40) ´ (0.20)2( )1/3

= 0.0163= 0.252m a

±= g

±m

±= 0.360(0.252) = 0.0907m

B) logg±

=0.509| z+z

-|

- I

1 + I+ 0.300I

é

ë

êê

ù

û

úú

= 0.509(2)-(0.775)

1.775+ 0.3(0.600)

é

ëê

ù

ûú = -0.2614

and g±

= 10-0.2614 = 0.548 a±

= g±m

±= 0.548(0.252) = 0.138m

6.50 A) Results of the calculations:

B) For BaBr2, both the Debye-Huckel and

Davies equation predict well up to 0.010 M.

After that the equations do not predict well.

Although not equal, the mean ionic activity

is close to the measured value at 0.10 m,

for both equations.

For CsF, both equations predict well up to

0.10m and then Davies equation result is

higher than actual for 0.10m and the

Debye-Huckel lower, they are both close to

the actual value. Above 0.50 m the Debye-

Huckel equation predicts higher than actual

values of the coefficient, and the results of the Davies equation are even higher.

6.51 A) m±NaBrO

3= (m

+

n+m-

n-)1/n = (0.50) ´ (0.50)( )1/2

= 0.50 a±,NaBrO

3= g

±m

±= 0.605(0.50) = 0.303m

Activity of the compound, NaBrO3: a = a±

n = (.3025)2 = 0.0915m2

B) m±CuBr

2= (m

+

n+m-

n-)1/n = (0.20) ´ (0.40)2( )1/3

= 0.0323= 0.317m

a±,CuBr

2= g

±m

±= 0.523(0.317) = 0.166m Activity of the compound: a = a

±

n = (.166)3 = 4.56X10-3 m3

6.52 A) mBaCl2

=370.43g

1.0kgH2O

1molBaCl2

208.4g

é

ë

êê

ù

û

úú

= 1.778mBaCl2

B) I =1

2c

iz

i

2

all ions

å =1

20.1.778(+2)2 + 3.558(-1)2éë

ùû

= 5.34m and I = 2.31

C) Ksp

= a±

3 = g±

3m±

3 m±BaCl

2= (m

Bam

Cl

2 )1/3 = (1.778) ´ (3.558)2( )1/3

= 22.513= 2.823m

Ksp

= 176.94 = g±

3(2.823)3 then g±

= 176.94 /22.51 = 2.00

This seeming high value for the coefficient is possible. Many salts such as CaCl2 and BaCl2 can show mean

activity coefficients greater than 1.0 in high ionic strength solutions. Since this a saturated solution, the

ionic strength would be relatively high, and the interactions are very strong at this temperature.

6.53 Cell reaction: ½H2(g) + AgCl(s) HCl(aq) + Ag(s) then n =1, ∆E° = 0.2223 V


101

∆E = ∆E° - 0.0257ln(a±

2) where a±

2 = aH+

aCl-

and ∆E - ∆E° = 0.2411V - 0.2223V = -0.0257(2)ln(g±m

±)

m±HCl = (1.0m) ´ (1.0m)2( )

1/2

= 1.00m Then ln(g±m

±) = -0.365 = ln(g

±) and g

±= e-0.365 = 0.694

6.54 Cell reaction: Hg2Cl2(s) + Zn(s) ZnCl2(aq) + 2 Hg(l) then n =2, ∆E° = 1.0304 V

So: ∆E = ∆E° - 0.0257ln(a±

3) where a±

3 = aZn+2

(aCl-

)2 = g±

3m±

3

Given: m±

ZnCl2

= (0.0050m) ´ (0.010m)2( )1/3

= 7.94X10-3m then

∆E - ∆E° = 1.2272V -1.0304V = -0.0257(3

2)ln(g

±m

±) = 0.03855 lng

±+ ln(7.94X10-3)( )

0.1968 = -0.03855lng±

+ 0.1864 Þ lng±

=0.1968 - 0.1864

-0.03855= -0.2688 Then g

±= e-0.2688 = 0.764

6.55 A) Overall reaction: 2 AgCl(s) + Cd(Hg) 2 Ag(s) + CdCl2(aq)

Cell notation: Cd(Hg)|CdCl2 (aq)|AgCl(s)|Ag

B) ∆E = ∆E° - 0.0257ln(a±

3) Þ ∆E = ∆E° -0.0257(3)

2ln(a

±)where a

±= g

±m

±

∆E° = 0.5732 V , n = 2 and m±CdCl

2= (0.010m) ´ (0.020m)2( )

1/3

= 1.587X10-4m

∆E = 0.5732V - 0.03855V ln(g±m

±) = 0.5732V - 0.03855V ln 0.679(0.01587)é

ëùû

Then ∆Emeasured

= 0.5732V + 0.1746V = 0.7478V

∆H° = ∆G° + (T ∆S°) = -195kJ + 298K(-77.6J

K)

1 kJ

1000J

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= -195 - 23 = -218kJ

6.44 Overall reaction: Hg2Cl2(s) +H2(g) Hg(l) + 2 HCl n= 2 mol e-

• Need to interpolate value of ∆E° at 298K from data to calculate ∆G° at 298K

∆E298

o = ∆E293

o -5K

10K

æ

èç

ö

ø÷ (∆E

303

o - ∆E293

o ) = 2.699V - 0.5(0.0030)V = 2.6975V

∆G° = -nF ∆E° = -2.0mol(9.65X104 coul

mol)(2.6975V) = -5.206X105 J = -520.6kJ

• Estimate derivative as: (∆E°303 - ∆E°303)/(303-293) K

∆S° = nFd∆E°

dT

é

ëê

ù

ûúP

» nF∆(∆E°)

∆T

é

ëê

ù

ûú = 2.0mol(9.65X104 coul

mol)

-3.0X10-3

10

V

K

é

ëêê

ù

ûúú

= -57.9J

K

∆H° = ∆G° + (T ∆S°) = -520.6kJ + 298K(-57.9J

K)

1 kJ

1000J

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= -520.6 -17.25 = -537.8kJ

6.45 A) Cathode: Hg2Cl2(s) + 2e- → 2 Hg(l) + 2 Cl- E° = 0.268 V

Anode: AgCl(s) + e- → Ag(s) + Cl- E° = 0.222 V

Overall: Hg2Cl2(s) + 2 Ag(s) 2 Hg(l) + 2 AgCl(s) ∆E° = 0.268 -(0.222) V = 0.046 V

B) ∆G° = -2.0mol(9.65X104 coul

mol)(0.046V) = -8878J = -8.88kJ

C) • Must plot data, as ∆E(volts), not millivolts, versus T(K)

• From equation: slope = 3.544 X10-4 V/K

∆S° = nFd∆E°

dT

é

ëê

ù

ûúP

= 2.0mol(9.65X104 coul

mol)(3.54X10-4 V

K) = 68.4

J

K

∆H° = ∆G° + (T ∆S°)

= -8.878kJ + 298K(68.4J

K)

1 kJ

1000J

æ

èç

ö

ø÷

é

ëêê

ù

ûúú

= -8.878 + 20.38 = 11.5kJ


102

D) Since in this cell ∆H°(+) and ∆S°(+), then ∆G° will

change sign with T, as will ∆E°. It should spontaneous at

higher T’s, non-spontaneous at lower T’s. Since the ∆E° is

positive we are above the changeover temperature at 298K.

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