### Learning Objectives

- To be able to determine free energy at nonstandard conditions using the standard change in Gibbs free energy, Δ
*G*⁰. - To understand and use the relationship between Δ
*G*⁰ 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, Δ*G*⁰

The change in Gibbs free energy under nonstandard conditions, Δ*G*, can be determined from the standard change in Gibbs free energy, Δ*G*⁰:

Δ*G* = Δ*G*⁰ + *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:

Δ*G* = Δ*G⁰ *(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 *a*A + *b*B ⇌ *c*C + *d*D:

*Q*_{gases} = [latex]\frac{{(P_C)}^c{{(P}_D)}^d}{{(P_A)}^a{{(P}_B)}^b}[/latex] or *Q*_{solutes} = [latex]\frac{{[C]}^c[{D]}^d}{{[A]}^a{[B]}^b}[/latex]

**Example 7**

Consider the following reaction:

4 NH_{3}(g) + 5 O_{2}(g) ⇌ 6 H_{2}O(g) + 4 NO(g)

- Use the thermodynamic data in the appendix to calculate Δ
*G*⁰ at 298 K. - Calculate Δ
*G*at 298 K for a mixture of 2.0 bar NH_{3}(g), 1.0 bar O_{2}(g), 1.5 bar H_{2}O(g), and 1.2 bar NO(g).

Solution

- Δ
*G⁰*= ∑*nΔG*⁰(products) – ∑_{f}*m*Δ*G*⁰(reactants)_{f}

Δ*G*⁰ = [(6 x −228.6 kJ/mol) + (4 x 87.6 kJ/mol)] – [(4 x −16.4 kJ/mol) + (5 x 0.0 kJ/mol)]

Δ*G*⁰ = (-1021.2 kJ/mol) – (-65.6 kJ/mol)

Δ*G*⁰= -955.6 kJ/mol = -9.56 x 10^{5} J/mol

* *2.Δ*G* = Δ*G*⁰ + *RT* ln *Q*

*Q =** [latex]\frac{{(1.5\ bar)}^6{(1.2\ bar)}^4}{{(2.0\ bar)}^4{(1.0\ bar)}^5}[/latex]= 1.5*

Δ*G* = (-9.56 x 10^{5} J/mol) + (8.314 J/mol K)(298 K)ln (1.5)

Δ*G* = (-9.56 x 10^{5} J/mol) + (1.0 x 10^{3} J/mol)

Δ*G* = -9.6 x 10^{5} J/mol

## The Relationship between Δ*G*⁰ and *K*

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

Δ*G* = Δ*G*⁰ + *RT* ln *Q*

0 = Δ*G*⁰ + *RT* ln *K*

Δ*G*⁰ = –* 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 *K*_{sp} for CuI(s) at 25⁰C is 1.27 × 10^{-12}. Determine Δ*G*⁰ for the following:

Cu^{+}(aq) + I^{–}(aq) → CuI(s)

Solution

Δ*G*⁰ = –* RT* ln *K*

The equation given is in the opposite direction to the definition of* K*_{sp}:

* K = *1/*K*_{sp}* = *(1/1.27 × 10^{-12}) = 7.87 x 10^{11}

Δ*G*⁰ = -(8.314 J/mol K)(298 K)ln (7.87 x 10^{11})

Δ*G*⁰= -(8.314 J/mol K)(298 K)(27.4)

Δ*G*⁰ = – 6.79 x 10^{4} J/mol = – 67.9 kJ/mol

### Key Takeaways

- The free energy at nonstandard conditions can be determined using
*G*= Δ*G*⁰ +*RT*ln*Q.* - There is a direct relationship between Δ
*G*⁰ and the equilibrium constant*K:**G*⁰ = –*RT*ln*K.*