The standard free energy change associated with the dissolution of ammonium nitrate in water is -6.73 kJ at 298 K. NH₄NO₃(s) ? NH₄NO₃(aq)
What is the equilibrium constant for the reaction? (R = 8.31 × 10?³ kJ/K)

Question

Quizplus · Accepted Answer

The equilibrium constant, $K$, can be calculated using the formula $\Delta G^\circ = -RT \ln K$, where $\Delta G^\circ$ is the standard free energy change, $R$ is the gas constant, $T$ is the temperature in kelvin, and $K$ is the equilibrium constant. Rearranging this formula gives: $K = \mathrm{e}^{-\Delta G^\circ/RT}$. Substituting the given values gives:
$$K = \mathrm{e}^{-(6.73 \ \mathrm{kJ})/(8.31 	imes 10^{-3} \ \mathrm{kJ/K/mol} 	imes 298 \ \mathrm{K})} \approx 6.6 	imes 10^{-12}$$
Therefore, the correct answer is (B).

The Standard Free Energy Change Associated with the Dissolution of Ammonium