## Free Energy under Nonstandard Conditions

### Learning Objectives

• To be able to determine free energy at nonstandard conditions using the standard change in Gibbs free energy, ΔG⁰.
• To understand and use the relationship between  ΔG⁰ and the equilibrium constant K.

Many reactions do not occur under standard conditions, and therefore we need some ways of determining the free energy under nonstandard conditions.

## Using Standard Change in Gibbs Free Energy, ΔG⁰

The change in Gibbs free energy under nonstandard conditions, ΔG, can be determined from the standard change in Gibbs free energy, ΔG⁰:

ΔG = ΔG⁰ + RT ln Q

where R is the ideal gas constant 8.314 J/mol K, Q is the reaction quotient, and T is the temperature in Kelvin.

Under standard conditions, the reactant and product solution concentrations are 1 M, or the pressure of gases is 1 bar, and Q is equal to 1. Taking the natural logarithm simplifies the equation to:

ΔG = ΔG⁰ (under standard conditions)

Under nonstandard conditions, Q must be calculated (in a manner similar to the calculation for an equilibrium constant). For gases, the concentrations are expressed as partial pressures in the units of either atmospheres or bars, and solutes in the units of molarity.

For the reaction aA + bB ⇌ cC + dD:

Qgases = $\frac{{(P_C)}^c{{(P}_D)}^d}{{(P_A)}^a{{(P}_B)}^b}$  or Qsolutes = $\frac{{[C]}^c[{D]}^d}{{[A]}^a{[B]}^b}$

Example 7

Consider the following reaction:

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

1. Use the thermodynamic data in the appendix to calculate ΔG⁰ at 298 K.
2. Calculate ΔG at 298 K for a mixture of 2.0 bar NH3(g), 1.0 bar O2(g), 1.5 bar H2O(g), and 1.2 bar NO(g).

Solution

1. ΔG⁰ = ∑nΔGf(products) – ∑mΔGf(reactants)

ΔG⁰ = [(6 x −228.6 kJ/mol) + (4 x 87.6 kJ/mol)] – [(4 x −16.4 kJ/mol) + (5 x 0.0 kJ/mol)]

ΔG⁰ = (-1021.2 kJ/mol) – (-65.6 kJ/mol)

ΔG⁰= -955.6 kJ/mol = -9.56 x 105 J/mol

2.ΔG = ΔG⁰ + RT ln Q

Q = $\frac{{(1.5\ bar)}^6{(1.2\ bar)}^4}{{(2.0\ bar)}^4{(1.0\ bar)}^5}$= 1.5

ΔG = (-9.56 x 105 J/mol) + (8.314 J/mol K)(298 K)ln (1.5)

ΔG = (-9.56 x 105 J/mol) + (1.0 x 103 J/mol)

ΔG = -9.6 x 105 J/mol

## The Relationship between ΔG⁰ and K

There is a direct relationship between ΔG⁰ and the equilibrium constant K. We can establish this relationship by substituting the equilibrium values (ΔG = 0, and K = Q) into the equation for determining free energy change under nonstandard conditions:

ΔG = ΔG⁰ + RT ln Q

0 = ΔG⁰ + RT ln K

ΔG⁰ = – RT ln K

We now have a way of  relating the equilibrium constant directly to changes in enthalpy and entropy. As well, we can now determine the equilibrium constant from thermochemical data tables or determine the standard free energy change from equilibrium constants.

Example 8

The Ksp for CuI(s) at 25⁰C is 1.27 × 10-12. Determine ΔG⁰ for the following:

Cu+(aq) + I(aq) → CuI(s)

Solution

ΔG⁰ = – RT ln K

The equation given is in the opposite direction to the definition of Ksp:

K = 1/Ksp = (1/1.27 × 10-12) = 7.87 x 1011

ΔG⁰ = -(8.314 J/mol K)(298 K)ln (7.87 x 1011)

ΔG⁰= -(8.314 J/mol K)(298 K)(27.4)

ΔG⁰ = – 6.79 x 104 J/mol = – 67.9 kJ/mol

### Key Takeaways

• The free energy at nonstandard conditions can be determined using ΔG = ΔG⁰ + RT ln Q.
• There is a direct relationship between ΔG⁰ and the equilibrium constant K: ΔG⁰ = – RT ln K.