**Relating Equilibrium Constant and Free Energy**

A thermodynamically-favorable process is one in which the free energy of the system decreases (\(\Delta G^\circ < 0\)). This is referred to as “exergonic”. Similarly, a process that occurs with a large value of equilibrium constant (\(K > 1\)) is also one that favors products and is thermodynamically-favorable. As such, as quantitative relationship between \(K\) and \(\Delta G^\circ\) is \(\Delta G^\circ = - RT ln K\), where\( R = 8.314 \frac{J}{mol \cdot K}\) (universal gas constant in energy units) and \( T = \text{ temperature in Kelvin}\). Rearranging the equation gives the alternate form \(K = e^\frac{-\Delta G^\circ}{RT}\). A large, positive value of \(K\) corresponds to a large, negative value of \(\Delta G^\circ\).

**Example 1.**

Calculate \(K\) given \(\Delta G^\circ = -50. \text{ kJ/mol}\) at \(25^\circ C\).

A. \(1.02\)

B. \(5.81 \times 10^8\)

C. \(1.51 \times 10^{20}\)

*Solution*

B. \(5.81 \times 10^8\)

A large, negative value of \(\Delta G^\circ\) is thermodyamically favorable. As such, expect \(K >> 1\).

Convert units of \(\Delta G^\circ\) and \(T\) first.

\(K = e^\frac{-\Delta G^\circ}{RT} = e^-\frac{-50,000 \text{ J/mol}}{8.314 \text{ J/mol K} \times 298 K} = e^{20.18} = 5.81 \times 10^8\)

**Example 2.**

A chemical reaction is not thermodynamically favorable under standard conditions. Which statement about this reaction is true?

A. \(\Delta G^\circ < 0, K > 1\)

B. \(\Delta G^\circ > 0, K > 1\)

C. \(\Delta G^\circ > 0, K < 1\)

*Solution*

C. \(\Delta G^\circ > 0, K < 1\)

A process that is not thermodynamically favorable has \(\Delta G^\circ > 0\) (endergonic) and \(K < 1\) (favors reactants).

**Example 3.**

A reaction has \(\Delta G^\circ = +2.0 \text{ kJ/mol}\) at \(25^\circ C\). What is true about \(K\) for this reaction?

A.\( K > 1\)

B. \(K < 1\)

C. \(K = 1\)

*Solution*

B. \(K < 1\)

\(K = e^\frac{-\Delta G^\circ}{RT} = e^-\frac{2500 \text{J/mol}}{8.314 \text{J/mol K} \times 298 K} = e^{-0.807} = 0.45 < 1\)