Free Energy and Pressure Example Problem

Finding Free Energy at Nonstandard States

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Helmenstine, Todd. "Free Energy and Pressure Example Problem." ThoughtCo, Apr. 19, 2016, thoughtco.com/free-energy-and-pressure-example-problem-609491. Helmenstine, Todd. (2016, April 19). Free Energy and Pressure Example Problem. Retrieved from https://www.thoughtco.com/free-energy-and-pressure-example-problem-609491 Helmenstine, Todd. "Free Energy and Pressure Example Problem." ThoughtCo. https://www.thoughtco.com/free-energy-and-pressure-example-problem-609491 (accessed October 19, 2017).
It's a big more work, but you can calculate the free energy for a reaction when the reactants are not under standard state conditions.
It's a big more work, but you can calculate the free energy for a reaction when the reactants are not under standard state conditions. Steve McAlister, Getty Images

This example problem demonstrates how to determine free energy of a reaction at conditions that are not standard states.

Free Energy for Reactants not at Standard State

Find ΔG at 700 K for the following reaction

C(s,graphite) + H2O(g) ↔ CO(g) + H2(g)

Given:

Initial pressures:

PH2O = 0.85 atm
PCO = 1.0 x 10-4 atm
PH2 = 2.0 x 10-4 atm

ΔG°f values:

ΔG°f(CO(g)) = -137 kJ/mol
ΔG°f(H2(g)) = 0 kJ/mol
ΔG°f(C(s,graphite)) = 0 kJ/mol
ΔG°f(H2O(g)) = -229 kJ/mol

Solution

Entropy is affected by pressure.

There are more positional possibilities for a gas at low pressure than a gas at high pressure. Since entropy is part of the free energy equation, the change in free energy can be expressed by the equation

ΔG = ΔG + RTln(Q)

where

ΔG° is the standard molar free energy
R is the ideal gas constant = 8.3145 J/K·mol
T is the absolute temperature in Kelvin
Q is the reaction quotient for the initial conditions

Step 1 - Find ΔG° at standard state.

ΔG° = Σ npΔG°products - Σ nrΔG°reactants

ΔG° = (ΔG°f(CO(g)) + ΔG°f(H2(g))) - (ΔG°f(C(s,graphite)) + ΔG°f(H2O(g)))

ΔG° = (-137 kJ/mol + 0 kJ/mol) - (0 kJ/mol + -229 kJ/mol)

ΔG° = -137 kJ/mol - (-229 kJ/mol)

ΔG° = -137 kJ/mol + 229 kJ/mol

ΔG° = +92 kJ/mol

Step 2 - Find the reaction quotient Q

Using the information in the equilibrium constant for gas reactions example problem and the equilibrium constant and reaction quotient example problem

Q = PCO·PH2O/PH2

Q = (1.0 x 10-4 atm)·(2.0 x 10-4 atm)/(0.85 atm)

Q = 2.35 x 10-8

Step 3 - Find ΔG

ΔG = ΔG + RTln(Q)

ΔG = +92 kJ/mol + (8.3145 J/K·mol)(700 K)ln(2.35 x 10-8)
ΔG = (+92 kJ/mol x 1000 J/1 kJ) + (5820.15 J/mol)(-17.57)
ΔG = +9.2 x 104 J/mol + (-1.0 x 105 J/mol)
ΔG = -1.02 x 104 J/mol = -10.2 kJ/mol

Answer:

The reaction has a free energy of -10.2 kJ/mol at 700 K.



Note the reaction at standard pressure was not spontaneous. (ΔG > 0 from Step 1). Raising the temperature to 700 K lowered the free energy to less than zero and made the reaction spontaneous.