1. H.W. # 20 Study pp. 810-819 Ans. Ques. p. 826 # 65,67,71,79
Aim # 20: What is the relationship between free energy and chemical equilibrium?
H.W. # Study pp
Ans. Ques. p. 826 # 65,67,71,79

2. I The dependence of Free Energy on pressure
For an isothermal process ∆H = ∆E + ∆(PV) For an ideal gas, since ∆E depends upon temp., and ∆T = 0 ∆H = 0 + ∆(PV) ∆H = 0 + ∆(nRT) but ∆T =0 and ∆H = 0 H is not pressure dependent. But S does depend on on P because it depends on V. As V ↑, S ↑ As P ↓, S ↑ ••• G is pressure dependent because G = H – TS.

3. II The relationship of ΔG0 to equilibrium
For an ideal gas, it can be shown that G = G0 + RT(ln P)
at P atm at 1 atm
∆G = ∑np(Gproducts) -∑nr(Greactants)
leads to
ΔG = ΔG 0 + RT lnQ where R is the ideal gas constant J/K•mol
Note: 1 mol refers to 1 mole or “round” of the reaction T is the Kelvin temp Q is the reaction quotient for the reaction
Problem: Calculate ΔG at 250C for the following reaction if the reaction mixture consists of 1.0 atm N2, atm H2, and 1.0 atm NH3:
N2(g) + 3H2(g) →2NH3(g)

4. Ans: For the balanced equation and the pressures given,
Q = (PNH3)2 = (1.0) = 3.7 x (PN2)(PH2) (1.0)(3.0)3
ΔG0 = 2ΔGf0(NH3(g)) – [3ΔGf0(H2(g)) + ΔGf0(N2(g))]
Using ΔGf0 values from the reference tables we obtain
ΔG0 = -34 kJ
Note: This value is for one mole or “round” of the reaction.
and ΔG = ΔG0 + RTlnQ ΔG = -34 kJ + (8.314 J/K)(298 K)(ln(3.7 x 10-2) note the units! ΔG = -34 kJ + (-8168J) = -34 kJ – 8.2 kJ
ΔG = -42 kJ

5. Note: At pressures above 1
Note: At pressures above 1.0 atm, ΔG < ΔG which is consistent with the fact that fewer moles of gas are produced (Le Chatelier’s principle).
III Free Energy and Equilibrium
Equilibrium occurs at the lowest value of free energy available to the reaction system.
For A(g) ↔ B(g)
A(g)
B(g) G equilibrium
Fraction A reacted

6. At equilibrium: GB = GA or ∆G = GB – GA = 0
∆G = ∆G0 + RT ln Q
∆G = 0 and Q = K
0 = ∆G0 + RT ln K
∆G0 = -RT ln K
If ∆G0 = 0, ln K = 0 and K = Remember, the standard state of a gas is 1 atm.
If ∆G0 < 0, G0products < G0reactants If all reactants & products are at 1 atm, the system is not at equilibrium ln Q = reaction proceeds → products ln K > 0 and K > 1
If ∆G0 > 0, G0products > G0reactants If all reactants & products are at 1 atm, the system is not at equilibrium ln Q = reactants ← reaction proceeds ln K < 0 and K < 1

7. In other words, the more negative ∆G0 is, the larger K is, and the forward reaction becomes spontaneous.
ln K = - ∆G RT
-∆G0/RT and K = e We can calculate equilibrium constants from ∆G0 and T
Problem: Using the following data, calculate the standard free energy change ΔG0, and the equilibrium constant at 250 C for the following reaction
2H2(g) + O2(g) → 2H2O(g)

8. Ans: ΔGf0 (kJ/mol) H2O(g) -229 H2(g) 0 O2(g) 0
ΔG0 = 2mol(-229 kJ/mol) – 0 kJ/mol = -458 kJ
ln K = -ΔG0 = (-458 kJ) ͘ RT (8.314 x 10-3 kJ/K)(298 K) ln K =
K = e = 1.90 x 1080
If K is so large (as is ΔG0), why don’t H2(g) and O2(g) react spontaneously upon mixing to produce water vapor?

9. Problem: The reaction 2NO2(g) ↔ N2O4(g) has ∆G0298 = kJ/mol N2O In a reaction mixture , the partial pressure of NO2 is 0.25 atm and the partial pressure of N2O4 is 0.60 atm. In which direction must this reaction proceed, at C, to reach equilibrium?
Ans: Q = PN2O4 and ∆G = ∆G0 + RT ln (PN2O4 /PNO2) PNO2 ∆G = x 103 J/mol + (8.314 J/mol•K)(298K)ln[0.60/(0.25)2] ∆G = +2.0 x 102 J/mol Since ∆G > 0 , the forward reaction is nonspontaneous. The reverse reaction will occur. Some N2O4 will have to decompose to reach equilibrium.

10. III The temperature dependence of K
∆G0 = -RTln K = ∆H0 - T∆S0
ln K = -∆H0 + ∆S0 = -∆H ∆S0
RT R R T R
y = m (x) + b

11. V Free Energy and Work
The maximum possible useful obtainable work from a process at constant T and P is equal to the change in free energy. For a process that is not spontaneous, the value of ∆G tells us the minimum amount of work that must be expended to make the process occur. Practice Problems Zumdahl (8th ed.) p. 810 # 60,64,69,72