1

Physical Chemistry I for Biochemists

Chem340

Lecture 17 (2/21/11)

Yoshitaka Ishii

Ch5.2, 5.4, 5.5-5.10 & HW5, HW6

Announcement

• HW6 is uploaded in the web site

• Questions from HW6 to be covered in Quiz 2:

P5.2, 5.5, 5.6, 5.7, 5.8, 5.13, 5.14, 5.20, 5.22, 5.30 & Q1-Q2,

where questions in red will be studiedwhere questions in red will be studied today.

• Review class on Wednesday/Friday

2

Incorrect answers in HW5

• P4.1(a) -1815 kJmol-1; -1818 kJmol-1

• P4.8 -1811 kJ

• P4.18 H-F 566 kJmol-1

• P4.19(b) 331 kJmol-1 328 kJmol-1

• (c) 590 kJmol-1 586 kJmol-1

• P4.29 (a) 5629 kJmol-1

Correction:

• P4.19) Given the data in Tables 4.1 and 4.2 and the data tables, calculate the bond

h l d f h f ll ienthalpy and energy of the following:

a. The CH bond in CH4

b. The CC single bond in C2H6

In b,c, use the data from a for CH bond enthalpy.

c. The CC double bond in C2H4

3

Ch 5.3 Introducing Entropy

T

DqdS reversibleEntropy: S [5.12]

0 T

DqdS reversible

S is a new state function

For a “natural” process S 0 in an isolated systemSo our world’s entropy is always increasing!

Second Law of Thermodynamics:

5.4 Calculating Changes in Entropy for Ideal Gas

T

DqdS reversible V=1L P = 1 bar

T =300 KV=1L P’ = P bar

Case 1. Adiabatic reversible processdS = Dqreversible/T = 0

Case 2. Isothermal reversible process (Vi, Ti) (Vf, Ti)

qreversible = -wreversible =

S = 0

)/ln( if

V

V

V

V

VVnRTdVV

nRTPdV

f

i

f

i

)/ln( ifreversiblereversible VVnR

T

q

T

DqS

Q. What is the sign of S for isothermal reversible compression?

4

Alternative Definition of Entropy in Statical Thermodynamics

S = kB ln(W),

where W is the number of possible configuration (or the degree of freedom)

For a “natural” process S 0 in an isolated system

Nature prefers a state with more randomness (or higher degree of freedom)

p y

Enthalpy Change for Irreversible Process

Case 2’ For isothermal irreversible process (Vi, Ti) (Vf, Ti)S must be calculated for an equivalent reversible process.

Ex. Isothermal adiabatic irreversible expansion (Vi, Ti) (Vf, Ti) Isothermal reversible expansion (Vi, Ti) (Vf, Ti)

)/ln( ifreversiblereversible VVnR

T

q

T

DqS

Q. What is the sign of S for isothermal irreversible expansion?

Q2. Does isothermal irreversible compression occur naturally?

V=1L P = 1 barT =300 K

V=1L P’ = 0 bar

Irreversible adiabatic & isothermal expansion

(1L, 1 bar, 300K) (2L, 0.5 bar, 300K)

2V P = 1/2 barT = 300K

Q. What is the sign of S for isothermal irreversible compression?

5

• dSL = -Dq/TL=CvdTL/TL

• dSR = Dq/TR=CvdTR/TR500K 300K

5.5 Using Enthalpy to Calculate the Natural Direction of a Process in an Isolated System

Dq

• dS = Dq(1/TR-1/TL)

400

300

400

500

R

K

K R

VL

K

K L

V dTT

CdT

T

CS

Dq > 0 dS >0 A

dS>0 or dS<0?

400K

0116

16

100400

400

100400100400

400

300400500400

22

2

2

lnln

ln

)/ln()/ln(

vv

v

VV

CKK

KC

KKKK

KC

KKCKKC

Q.How much is S for BA? Is S positive?

B

Calculated Change in Enthalpy (continued)• Case3 (3’). Reversible change in T for a fixed V

Case4 (4’) Reversible change in T for a fixed P

)/ln(,,

ifmVmvreversible TTnC

T

dTnC

T

DqS

Case4 (4 ). Reversible change in T for a fixed P

Case 5 (reversible) & 5’(irreversible) process: (Ti, Vi) (Tf, Vf)

By calculating S for (T V ) (T V ) (T V )

)/ln(,,

ifmPmPreversible TTnC

T

dTnC

T

DqS

Note: We assume Cv & Cp constant

By calculating S for (Ti, Vi) (Ti, Vf) (Tf, Vf)

Case 6 & 6’ (Ti, Pi) (Tf, Pf)

By calculating S for (Ti, Pi) (Ti, Pf) (Tf, Pf) (Pi/Pf = Vf/Vi)

)/ln()/ln( , ifmVif TTnCVVnRS

)/ln()/ln()/ln()/ln( ,, ifmPififmPif TTnCPPnRTTnCVVnRS

6

HW6• P5.5) Calculate S if the temperature of 1 mol of an

ideal gas with CV = 3/2 R is increased from 150 K to 350 K under conditions of (a) constant pressure and (b) t t l Ch th t(b) constant volume. Choose the correct one.

(1) (2)

(3) (4)

)/ln( KKR

3501502

3

)/ln( KKR

1503502

3

)/ln( KKR 3501502

)/ln( KKR 1503502

(5) (6) )/ln( KKR

3501502

5)/ln( KKR

1503502

5

P5.6) One mole of N2 at 20.5°C and 6.00 bar undergoes a transformation to the state described by 145°C and 2.75 bar. Calculate S if

C T T 2

When Cp or Cv is not constant

•

CP,m

J mol1 K1 30.8111.87 103 T

K 2.3968105 T 2

K2

1.0176108

T 3

K3

Tmp,f

bar6 00

bar 2.75lnndT

T

C n

p

plnnR ΔS

f

R[Q1]

1

Ti

KJ K

Tdn

bar6.00Tpi

K

K T

KT

dKT

cKT

ba15418

65293

32

.

.

7

Q1-2 How can we prove that S is a state function?HW6.

Dqreversible = dU -Dw

PdVdTCdVPP

T V

(U/V)T

PdVdTCdVPT

T VV

dTCT

dVT

P

T

DqdS V

V

ereveresibl 1

1

For ideal gas,

dTCT

dVVnRdS V

1 /

For liquid/solid,

dTT

CdVdTC

TdV

T

V

V

PdS v

VPT

1

5.7 The Change of Entropy in the Surroundings and Stotal = S + Ssurroundings

• The entropy of an isolated system increases in a spontaneous process.

Q Then is it tr e that a process is spontaneo s• Q. Then, is it true that a process is spontaneous

if S >0? True only for an isolated system.

• Consider surroundings for which P or V is constant.

• For a constant V, Dqsurroundings = dU

• For a constant P Dq di = dHState Functions!

For a constant P, Dqsurroundings = dH

dSsurroundings = dqsurroundings/Tsurroundings

dStotal = dS+ dSsurroundings = Dqrev/T + dqsurroundings/Tsurrounding

A spontaneous change occurs when dStotal > 0

8

HW6 P5.7, P5.8P5.7) One mole of an ideal gas with CV = 3R/2

undergoes the transformations described in the following list from an initial state described by Ti = 300. K and Pi = 1.00 bar. Calculate q, w, U, H,i q, , , ,and S for each process.

(c) The gas undergoes a reversible isothermal expansion at 300 K until P is half of its initial value.

P5.8) Calculate Ssurroundings and Stotal for part (c) of Problem P5.7. Is the process spontaneous?Problem P5.7. Is the process spontaneous?

(c) S = -nRln(Pf/Pi)+ nCp,mln(Tf/Ti) = –nRln(0.5/1)

Because T is const Ssurrounding= -q/Tsurrounding = -q/T

So Stotal = 0

)/ln()/ln( ifif PPnRTVVnRTPdVwq [Q1]

[Q2]

5.2 Heat Engines and the Second Law ofThermodyamics

What is Efficiency =|wcycle|/|qab|?

9

5.2 Heat Engines and the Second Law of Thermodyamics

Isotherm: -PdV = -(nRT/V)dVa b: qab = -wab =nRThotln(Vb/Va)>0

d RT l (V /V )<0

Carnot Cycle

c d: qcd = -wcd = nRTcoldln(Vd/Vc)<0Adiabatic: q= 0da & bc: qcd = qbc=0

U = 0 wcycle + qab+qcd =0 wcycle = -(qab+qcd) <0 (|qab|>|qcd|)

Efficiency =|w|/|qab| = |qab+qcd|/|qab| < 1

S=0 for Carnot Cycle?• S = qab/Thot + qcd/Tcold

= nRThotln(Vb/Va)/Thot + nRTcoldln(Vd/Vc)/Tcold

Rl (V /V ) Rl (V /V )= nRln(Vb/Va) + nRln(Vd/Vc)= nRln(VbVd/VaVc)

d a & bc are adiabatic processes:• TcoldVd

-1 = ThotVa-1 & ThotVb

-1 = TcoldVc-1

(VdVb )-1 = (VaVc)-1 VbVd/VaVc=1

S = qab/Thot + qcd/Tcold = 0

wcycle = -(qab+qcd) = nRThotln(Vb/Va) + nRTcoldln(Vd/Vc)= nRThotln(Vb/Va) - nRTcoldln(Vb/Va)

= nR(Thot-Tcold)ln(Vb/Va)

10

5.6 Clausius Inequality (p95)dU=Dq – PextdV (5.29)

• For a reversible change,

dU=Dqreversible – PdV = TdS – PdV (5.30)

When dU and dV for (5 29) and (5 30) are common• When dU and dV for (5.29) and (5.30) are common,

TdS – PdV = Dq-PextdV

TdS – Dq = (P – Pext )dV (5.31)

If P – Pext >0 The system will expand or dV > 0.

If P – Pext <0 The system will shrink or dV < 0. ext y

Namely, TdS – Dq > 0 when P Pext (when 5.29 is irreversible)

T

DqdS leirreversib

0T

DqdS leirreversib For an irreversible process in

an isolated system

T

DqdS leirreversib0

Sample Question

• Which relationship is correct for an i ibl i l t dirreversible process on an isolated system? Choose all.

(a) (b)

(c) (d)

T

DqdS leirreversib T

DqdS leirreversib

0dS 0dS(c) (d)

(e) (f)

0dS 0dS

T

Dq leirreversib0 T

Dq leirreversib0

11

Sample Question2• Which relationship is correct for an

irreversible process? Choose all. p

(a) (b)

(c) (d)

T

DqdS leirreversib

T

DqdS leirreversib

0dS 0dS

(e) (f)

(g)

T

Dq leirreversib0 T

Dq

T

dS leirreversib

T

DqdS leirreversib

S in vaporization and fusion (p90)

• During fusion and vaporization, T is const.Li idG d PLiquidGas under a constant P

Solid Liquid under a constant P

onvaporizati

onvaporizati

onvaporizati

reversiblereversibleonvavorizati T

H

T

q

T

DqS

fusion

fusion

fusion

reversiblereversiblefusion T

H

T

q

T

DqS

12

5.8 Absolute Entropies and The Third Law of Thermodynamics

• The entropy of an element or a compound is experimentally determined from – Dqreversible =CpdT

Cp,m for O2

T gasTb Liquid

f

mfusionTf Solid

pmmm

dTCHdTC

T

H

T

dTCKSTS

'""

'

')()( ,

0

0

Molar Entropy for Gas

T

Tb

gasmp

b

monvaporizatiTb

Tf

Liquidmp

T

dTC

T

H

T

dTC

'"

'"

"

" ,,,

The entropy of a pure, perfectly crystalline

substance (element or compound) is zero at 0K.

Third Law of thermodynamics - What is Sm(0K)?

T Dependence of SmCp,m /T > 0 and Svaporization, Sfusion > 0 Sm increases as T increases

13

The origin of Cp,m and Sm for Solids• Number of degrees of freedom for molecule made

of n atoms.

NT: Translational: 3

NR: Rotational: Atom 0, Linear 2, Non-Linear 3

NV: Vibrational: 3n – NT – NR

Cp,m & Sm higher for a molecule with more atoms.

Calculated Change in Enthalpy (continued)• Case3 (3’). Reversible change in T for a fixed V

Case4 (4’) Reversible change in T for a fixed P

)/ln(,,

ifmVmvreversible TTnC

T

dTnC

T

DqS

Case4 (4 ). Reversible change in T for a fixed P

Case 5 (reversible) & 5’(irreversible) process: (Ti, Vi) (Tf, Vf)

By calculating S for (T V ) (T V ) (T V )

)/ln(,,

ifmPmPreversible TTnC

T

dTnC

T

DqS

By calculating S for (Ti, Vi) (Ti, Vf) (Tf, Vf)

Case 6 & 6’ (Ti, Pi) (Tf, Pf)

By calculating S for (Ti, Pi) (Ti, Pf) (Tf, Pf) (Pi/Pf = Vf/Vi)

)/ln()/ln( , ifmVif TTnCVVnRS

)/ln()/ln()/ln()/ln( ,, ifmPififmPif TTnCPPnRTTnCVVnRS

14

5.9 Standard States in Entropy Calculation (P102

• For enthalpy, we defined Hf,A0 for the most stable

pure elements at 298.15K (T0) and 1 bar (P0) as 0.

• We define standard state of Entropy as S0

m = Sm(P0, T0) • What is the relationship between S0

m(P) and S0

m(1 bar)?

S (1 bar P) = −Rln(P/P ) Ent

ropy

Sm(1 bar P) = −Rln(P/P0)

Mol

ar E

S0m

Sm(P) = S0m(P0) −Rln(P/P0)

5.10 Entropy Change in Chemical Reaction

At 298.15 K and 1 barA + 2B 2C + D • Sreaction

0 = 2SC,m0 + SD,m

0 – SA,m0 – 2SB,m

0

I lIn general

A(T) + 2B(T) 2C(T) + D(T)For a constant pressure

X

mXXR STS 00,)(

T

p

T

dTCKSTS

298

0

298'

')()(

00

XmpXp CC ,

Q. How much is Cp0 for the above reaction?

00000 22BmpAmpDmpCmpp CCCCC ,,,,

15

• P5.20) Consider the formation of glucose from carbon dioxide and water, that is, the reaction of the following photosynthetic process: 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g). The following table of information will be useful in working this problem:p

• Calculate the entropy and enthalpy changes for this chemical system at (a) T = 298 K and (b) T = 330. K. Calculate also the entropy of the surrounding and the universe at both temperatures.

(a) SR0 = XSX

0

(b)

T

p

T

dTCKSTS

298

0

298'

')()(

00 pXXp CC

HW5 P4.32) Using the protein DSC data from Problem P 4.31, calculate the enthalpy change ΔH between the T=288 K and T=318 K. Give your answer in units of kilojoules per mole. Assume the molecular weight of the protein is 14000. grams.

CPtrs

CP

ΔH /g

Area

(1) Obtain the orange area in mm2

ΔH/mol =ΔH /g Mlysozyme

Tm

CPint

(2) Obtain the area for 0.418 Jg-1x 10

in mm2

(3) Convert (1) into Jg-1

16

• P4.2) Calculate and for the oxidation of benzene. Also calculate

Hreaction Ureaction

• The chemical equation for the oxidation of benzene is:

lOH 3gCO 6gO 7lHC 22221

66 (*)

Hreaction

in P4.2

• P4.3) Use the tabulated values of the enthalpy of combustion of benzene and the enthalpies of formation of CO2(g) and H2O(l) to determine for benzene.H f

• Q3(Mod). Now, we would like to derive Hreaction(P) at a pressure slightly higher than 1 bar at a standard temperature. Derive Hreaction(P) using J-T, Cp,A, Hreaction(1 bar). Use dHA = CpAdT – Cp, AJ-T dP + (H/n)T PdnA (A = B X or Y) assuming that J T and(H/n)T, PdnA (A B, X, or Y) assuming that J-T and Cp,A are constant. Specify the path of (P, T, n) you would like to use. Assume a reaction 3B X + 2Y.

Step 1: 3B(P) 3B(1 bar) H1 = – 3Cp, BJ-T,B (1 bar-P)

• Step 2: 3B(1 bar) X(1 bar) + 2Y(1bar)

H H (1 b )H2 = Hreaction(1 bar)

• Step 3: X(1 bar) + 2Y(1bar) X(P) + 2Y(P)

H3 = – (Cp, XJ-T,X + 2Cp, YJ-T,Y) (P- 1 bar)

Hreaction(P) = H1 + H2 +H3

[Q1]

[Q2]

17

P4.19) Given the data in Tables 4.1 and 4.2 and the data tables, calculate the bond enthalpy and energy of the following:

a. The CH bond in CH4

b. The CC single bond in C2H6g 2 6

In b,c, use the data from a for CH bond enthalpy.

c. The CC double bond in C2H4

a) CH4(g) C(g) + 4H(g)

∆Hreaction0= ∆Hf

0(C(g))+ 4∆Hf0(H(g)) - ∆Hf

0(CH4(g)) [Q1]

∆Hbond0(C-H) = ∆Hreaction

0/4

b) C2H6(g) 2C(g) + 6H(g)

∆Hreaction0= 6 ∆Hbond

0(C-H) + ∆Hbond0(C-C)

[Q2]

[Q3]

• P4.8) Calculate at 650 K for the reaction 4NH3(g) +6NO(g) 5N2(g) + 6H2O(g) using the temperature dependence of the heat capacities from the data tables. (-1811 kJ)

T

dTTCHH 00 ')'( K

PKreactionTreaction dTTCHH15298

1529800

.

.,, ')'(

(See Table 2.4 in Appendix B for Hf0)

CP = {-4ANH3(1) - 6ANO(1) + 5 AN2(1) + 6AH2O(1) }+ {XAX(2) }(T/K) + {XAX(3) }(T/K)2 + {XAX(2) }(T/K)3

XXX

T

K

n

n

T

K

P nAnBn

TnBdTTC )}({)(;

')(')'(

.

298

4

115298