Thermodynamics of Reactions

I. Entropy and Chemical ReactionsA. Example: N2(g) + 3H2(g) 2NH3(g)

1) System: positional probability

a) 4 reactant particles 2 product particles

b) Fewer possible configurations, S = -

B. An increase in number of gas particles is entropically favored

1) 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

2) Nine particles 10 particles, S = +

3) Example: Predict sign of S

a) CaCO3(s) CaO(s) + CO2(g)

b) 2SO2(g) + O2(g) 2SO3(g)

C. Third Law of Thermodynamics = the entropy of a perfect crystal at 0K = 0 (there is only one possible configuration)

1) As temperature increases, random vibrations occur; S = +

1) We can calculate S for any substance if we know how S depends on T [Standard So values, Appendix IIB]

2) Soreaction = nSo

prod - nSoreactants

4) Example: Find So at 25 oC for

2NiS(s) + 3O2(g) 2SO2(g) + 2NiO(s)

So(J/K mol) 53 205 248 38

So = 2(248) + 2(38) – 2(53) – 3(205) = -149 J/Kmol

(3 gas particles 2 gas particles)

D. Factors Affecting Standard Entropies

1) Phase of Matter: as discussed earlier, Sgas >> Sliquid > Ssolid

a) So H2O(l) = 70.0 J/Kmol

b) So H2O(g) = 188.8 J/Kmol

2) Molar Mass: heavy elements have more

entropy than light (in same state)

Has to do with translation energy states

3) Allotropes: more rigid structures have less entropy

4) Molecular Complexity: more complex = more entropy

a) So Ar(g) = 154.8 (39.948 g/mol)

b) So NO(g) = 210.8 (30.006 g/mol)

II. Free Energy and Chemical ReactionsA. Standard Free Energy Change = Go = reactants/products at standard states

1) Gases at 1 atm, solution = 1 M, element at 25 oC and 1 atm has Go = 0

2) N2(g) + 3H2(g) 2NH3(g) Go = -33.3 kJ

3) Go can’t be measured directly, it must be calculated from Ho and S

4) Usefulness of Go: The more negative Go is, the more likely reaction is

i. If Go is negative, the reaction is spontaneous as written

ii. If Go is positive, the reaction is not spontaneous

5) Why use standard states? Because G changes with P, T, concentration

B. Calculating Go: Go = Ho - TSo

1) Example: C(s) + O2(g) CO2(g)

i. Ho = -393.5 kJ So = 3.05 J/K

ii. Go = (-3.935 x 105 J) – (298 K)(3.05 J/K) = -394.4 kJ

2) Example: Find Ho, So, Go for 2SO2 + O2 2SO3

Ho (kJ/mol) -297 0 -396

So (J/Kmol) 248 205 257

T = 298 K

3) Use the Go of known reactions to find Go of unknown reactions

a) 2CO + O2 2CO2 Go = ?

b) Known Reactions

i) 2CH4 + 3O2 2CO + 4H2O Go = -1088 kJ/mol

ii) CH4 + 2O2 CO2 + 2H2O Go = -801 kJ/mol

c) If we reverse the first reaction and double the second…

i) 2CO + 4H2O 2CH4 + 3O2 Go = +1088 kJ/mol

ii) 2(CH4 + 2O2 CO2 + 2H2O) Go = -1602 kJ/mol

iii) 2CO + O2 2CO2 Go = -514 kJ/mol

d) Example: Find Go for Cdiamond Cgraphite

Given that Cd + O2 CO2 Go = -397 kJ/mol

Cg + O2 CO2 Go = -394 kJ/mol

4) The Gfo Method

a) Free energy of formation Gfo is tabulated for many compounds

b) We can sum these for reactants and products to find Go

c) Example: 6C(s) + 6H2(g) + 3O2(g) C6H12O6(s) (glucose)

d) Gfo = free energy change of formation of one mole of the product

from its elements in their standard states (-911 kJ/mol for glucose)

e) Example: 2CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g)

Gfo (kJ/mol) -163 0 -394 -229

Go = nGfo

prod - nGfo

reactants

Go = 2(-394) + 4(-229) - 2(-163) - 3(0) = -1378 kJ/mol

III. G and PressureA. Pressure dependencies of the state functions

1) H is not pressure dependent

2) S is pressure dependent

a) S(large volume) > S(small volume)

b) S(low pressure) > S(high pressure)

c) PV = nRT

3) G must depend on pressure since G = H – TS

G = Go + RTlnP Use R = 8.3145 J/Kmol

B. Example: N2(g) + 3H2(g) 2NH3(g)

1) G = 2G(NH3) – G(N2) – 3G(H2)

RTlnKGG m,EquilibriuAt

RTlnQGG

)))(P(HP(N

))(P(NHRTlnGG

)]3lnP(H-)lnP(N-

)RT[2lnP(NH)(H3G-)(NG-)(NH2GG

)]RTlnP(H)(H3[G-)]RTlnP(N)(N[G-

)]RTlnP(NH)(NH2[GΔG

o

o

322

23

reactiono

22

32o

2o

3o

22o

22o

33o

= 0.0313

K = PH2O(g)

2) Example: CO(g) + 2H2(g) CH3OH(l)

Find G at 25 oC, P(CO) = 5 atm, P(H2) = 3 atm

a) From Appendix IIB, find Go = -166 – (-137) – 0 = -29 kJ/mol

b) Find RTlnQ = (8.3145)(298)ln(1/(5)(3)2) = -9.4 kJ/mol

c) G = Go + RTlnQ = -38 kJ/mol

d) More spontaneous at these conditions than at standard conditions

IV. Meaning of G for Reaction at EquilibriumA. Even if G is negative, the spontaneous reaction doesn’t necessarily go to

completion

1) Phase changes always go to completion if spontaneous

2) Reactions often have a minimum G that is somewhere before completion of the reaction

) ) 22 )3)(5(

11

2

HCO PP

K

4) The equilibrium mixture of reactants and products might be more stable (lower G) than the completely formed product alone

CO(g) + 2H2(g) CH3OH(l)

B. Equilibrium

1)Kinetics: forward and reverse reaction rates are the same at equilibrium

2)Thermodynamics: the lowest free energy state is at equilibrium

3)A(g) B(g)

a)GA = GAo + RTlnPA (this is decreasing as the reaction proceeds)

b)GB = GBo + RTlnPB (this is increasing as the reaction proceeds)

c)G = GA + GB (this is decreasing as the reaction proceeds)

4)At equilibrium GA = GB and we have a new PAE and PB

E

a)G is no longer decreasing, it is at its minimum point

b)No further driving force for the reaction to proceed (G = 0)

Reaction Starts Reaction Proceeds Equilibrium

5) Example: A(g) B(g)

a) PAE = 0.25(2 atm) = 0.5 atm

b) PBE = 0.75(2 atm) = 1.5 atm

c) K = PB/PA = 1.5/.5 = 3.0

6) The same Equilibrium Position would be reached from any initial condition where A + B = 2.0 atm

7) At Equilibrium: Gprod = Greact (G = Gprod – Greact = 0)

1 mol A, 2 atm 1 mol B, 2 atm 1 mol A/B, 2 atm

8) At Equilibrium: G = 0 = Go + RTlnK Go = -RTlnK

a) If Go = 0, then K = 1 and we are at equilibrium

b) If Go < 0, then K > 1 and the reaction proceeds forward

c) If Go > 0, then K < 1 and the reaction proceeds in reverse

9) Example: N2(g) + 3H2(g) 2NH3(g)

a) Given Go = -33.3 kJ/mol at 25 oC

b) Predict the direction when P(NH3) = 1 atm, P(N2) = 1.47 atm and P(H2) = 0.01 atm

c) Predict direction when P(NH3) = P(N2) = P(H2) = 1 atm

10) Example: 4Fe(s) + 3O2(g) 2Fe2O3(s) at 25 oC

Hfo 0 0 -826 kJ/mol

So 27 205 90 J/Kmol

Find K

11) Dependence of K on T

R

ΔS

T

1

R

ΔH- lnK

STΔHRTlnKΔGoo

ooo

1/T

lnK

Slope = -H/RIntercept = S/R

1) G and Worka) Wmax = G All the free energy produced from a spontaneous reaction could be used

to do work (Reversible Process only)

b) For a non-spontaneous reaction, G tells us how much work we would have to do on the system to get the reaction to occur

c) Wactual < Wmax We always lose some energy to heat in any process (Irreversible Processes)

d) Theoretically, Reversible Process utilize all energy for work, Unfortunately, all real processes are Irreversible

G = -50.5 kJ S = -80.8 J/K

Irreversible (real)Process

Increases Ssurr to make Suniv = +