Thermodynamics of Reactions
I.
Entropy and Chemical Reactions
A.
Example: N2(g) + 3H2(g)
1)
2NH3(g)
System: positional probability
a) 4 reactant particles
2 product particles
b) Fewer possible configurations, DS = 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, DS = +
3) Example: Predict sign of DS
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; DS = +
1) We can calculate S for any substance if we know how S depends on T
[Standard So values, Appendix IIB]
2) DSoreaction = SnSoprod - SnSoreactants
4) Example: Find DSo at 25 oC for
2NiS(s) + 3O2(g)
So(J/K mol) 53
205
2SO2(g) + 2NiO(s)
248
38
DSo = 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 Reactions
A.
Standard Free Energy Change = DGo = 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) DGo = -33.3 kJ
3) DGo can’t be measured directly, it must be calculated from DHo and DS
4) Usefulness of DGo: The more negative DGo is, the more likely reaction is
i. If DGo is negative, the reaction is spontaneous as written
ii. If DGo is positive, the reaction is not spontaneous
5) Why use standard states? Because DG changes with P, T, concentration
B.
Calculating DGo: DGo = DHo - TDSo
1) Example: C(s) + O2(g)
CO2(g)
i.
DHo = -393.5 kJ DSo = 3.05 J/K
ii. DGo = (-3.935 x 105 J) – (298 K)(3.05 J/K) = -394.4 kJ
2) Example: Find DHo, DSo, DGo for
2SO2 + O2
DHo (kJ/mol) -297
0
DSo (J/Kmol) 248 205
T = 298 K
2SO3
-396
257
3) Use the DGo of known reactions to find DGo of unknown reactions
a) 2CO + O2
2CO2 DGo = ?
b) Known Reactions
i) 2CH4 + 3O2
2CO + 4H2O DGo = -1088 kJ/mol
ii) CH4 + 2O2
CO2 + 2H2O DGo = -801 kJ/mol
c) If we reverse the first reaction and double the second…
i) 2CO + 4H2O
2CH4 + 3O2
DGo = +1088 kJ/mol
ii) 2(CH4 + 2O2
CO2 + 2H2O) DGo = -1602 kJ/mol
iii) 2CO + O2
2CO2
DGo = -514 kJ/mol
d) Example: Find DGo for Cdiamond
Cgraphite
Given that Cd + O2
CO2 DGo = -397 kJ/mol
Cg + O2
CO2 DGo = -394 kJ/mol
4) The DGfo Method
a) Free energy of formation DGfo is tabulated for many compounds
b) We can sum these for reactants and products to find DGo
c)
Example: 6C(s) + 6H2(g) + 3O2(g)
d)
DGfo = 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)
DGfo (kJ/mol)
-163
0
-394
-229
DGo = SnGfoprod - SnGforeactants
DGo = 2(-394) + 4(-229) - 2(-163) - 3(0) = -1378 kJ/mol
III. DG and Pressure
A.
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
C6H12O6(s) (glucose)
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) DG = 2G(NH3) – G(N2) – 3G(H2)
ΔG  2[G o (NH 3 )  RTlnP(NH 3 )]
- [G o (N 2 )  RTlnP(N 2 )] - 3[G o (H 2 )  RTlnP(H 2 )]
DG  2G o (NH 3 ) - G o (N 2 ) - 3G o (H 2 )  RT[2lnP(NH 3 )
- lnP(N 2 ) - 3lnP(H 2 )]
2


(P(NH
))
o
3

DG  DG reaction  RTln 
3 
 P(N 2 )(P(H 2 )) 
DG  DG o  RTlnQ
At Equilibriu m, DG  DG o  RTlnK
K = PH2O(g)
= 0.0313
1
1
2) Example: CO(g) + 2H2(g)
CH3OH(l) K 

2
2
(
5
)(
3
)

)

)
P
P
CO
H
Find DG at 25 oC, P(CO) = 5 atm, P(H2) = 3 atm
a) From Appendix IIB, find DGo = -166 – (-137) – 0 = -29 kJ/mol
b) Find RTlnQ = (8.3145)(298)ln(1/(5)(3)2) = -9.4 kJ/mol
c) DG = DGo + RTlnQ = -38 kJ/mol
d) More spontaneous at these conditions than at standard conditions
2
IV. Meaning of DG for Reaction at Equilibrium
A.
Even if DG 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
4) The equilibrium mixture of reactants and products might be more stable
(lower DG) 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 PBE
a) G is no longer decreasing, it is at its minimum point
b)No further driving force for the reaction to proceed (DG = 0)
Reaction Starts
Reaction Proceeds
Equilibrium
5) Example: A(g)
1 mol A, 2 atm
B(g)
1 mol B, 2 atm
1 mol A/B, 2 atm
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 (DG = Gprod – Greact = 0)
8) At Equilibrium: DG = 0 = DGo + RTlnK
DGo = -RTlnK
a) If DGo = 0, then K = 1 and we are at equilibrium
b) If DGo < 0, then K > 1 and the reaction proceeds forward
c) If DGo > 0, then K < 1 and the reaction proceeds in reverse
9) Example: N2(g) + 3H2(g)
2NH3(g)
a) Given DGo = -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)
DHfo
0
0
So
27
205
Find K
11) Dependence of K on T
o
ΔG  RTlnK  ΔH o  TDSo
lnK
ΔH o  1  ΔSo
lnK   
R T R
2Fe2O3(s) at 25 oC
-826 kJ/mol
90 J/Kmol
Slope = -DH/R
Intercept = DS/R
1/T
1) DG and Work
a) Wmax = DG All the free energy produced from a spontaneous reaction could be used
to do work (Reversible Process only)
b) For a non-spontaneous reaction, DG 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
DG = -50.5 kJ
DS = -80.8 J/K
Irreversible (real)
Process
Increases Ssurr to make DSuniv = +