KEY2CHEM

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. $$K < 1$$

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