16.3 Free Energy & Equilibrium

Earlier in this chapter, we established that when ∆G is < 0, the reaction is favored in the forward direction, and when  ∆G is > 0, the reaction is favored in the backward direction. When ∆G = 0, the reaction is at equilibrium.  Recall your knowledge from equilibrium chapter, we established when Q > K, the reaction is favored in the backward direction, and when Q < K, the reaction is favored in the forward direction. When Q = K, the reaction is at equilibrium.  If we combine these two concepts, Gibbs free energy and equilibrium constant, and their relation to the spontaneity of reaction, we can also relate equilibrium constant K and free energy G by this mathematical equation.

∆G = ∆G⁰ + RT ln Q

Here ∆G⁰ is free energy change for a reaction when all the reactants and products are in the standard state. (Thermodynamic standard state means the pressure is 1 atm and concentration is 1 M and temperature is 25 ⁰C )

∆G is free energy change for a reaction when all the reactants and products are in a nonstandard state.

R is the gas constant 8.3145 J/K.mol, T is the temperature in Kelvin and Q is the reaction quotient.

As we said earlier when the system is at equilibrium,

∆G = 0,

∆G⁰ + RT ln K = 0

∆G⁰ = – RT ln K

We can calculate the equilibrium constant for a reaction using the standard free energy value of reactants and products from the thermodynamic data table.

Question 10: Find the Ksp value for dissolution of CaSO₄ in water at 25 ⁰C.

Ca SO₄ (s)               Ca⁺²(aq) + SO₄ ^-2 (aq)

∆G⁰_reaction = ∑∆ G⁰ (products) – ∑∆ G⁰ (reactants) =

[(-553.04)+ (-744.5) ] –[-1322.0 ] =24.46 Kj/mol

= 24460 J/mol

At equilibrium            ∆G⁰ = – RT ln K

R = 8.3145 J / K.mol, T = 25 ⁰C = (25 + 273.15) K = 298.15 K

24460 = – 8.3145 x 298.15 x ln K

ln K = 24460 / (-8.3145 x 298.15) = – 9.87

K = e^-9.87 = 5.17 x 10^-5

This result for the value of Ksp is consistent with the theoretical value of Ksp for CaSO4, which is 6.1 x 10^-5.

Watch this YouTube video to understand the relation between Delta G and equilibrium constant K.

Question  11: Calculate ∆G for the following reaction under these specified conditions at 25 ⁰C.

N₂(g) + 3H₂(g)                       2NH₃(g)          ΔG°= – 33.0 kJ

P_N2 = 0.170 atm, P_H2 = 0.250 atm, P_NH3 = 41.2 atm

∆G = ∆G⁰ + RT ln Q

Q = = 639000

ΔG°= – 33.0 kJ = – 33000 J

R = 8.314 J/K.mol

∆G = ∆G⁰ + RT ln Q = -33000 + 8.314 x 298.15 x ln 639000

∆G = 136 J

A positive value of ∆G indicates reaction will be non-spontaneous under these conditions.

Question 12: Calculate the free energy change for this same reaction at 455 °C in a 2.50 L mixture containing 1.00 mol of each gas. Is the reaction spontaneous under these conditions?

PV = nRT,

Given: V = 2.50 L,

n = 1.00 mol,

T = 375 °C = 455 + 273 = 728 K

P =  = 23.9 atm

Q =  =  = 0.00175

∆G = ∆G⁰ + RT ln Q = -33000 + 8.314 x 728 x ln 0.00175

∆G = -71424 J

A negative value of ∆G indicates reaction will be spontaneous under these conditions.