Question 19.8: Calculating K from the Standard Free-Energy Change (Molecula...

Calculating K from the Standard Free-Energy Change (Molecular Equation)

Find the value of the equilibrium constant $K$ at $25^{\circ} \mathrm{C}(298 \mathrm{~K})$ for the reaction

$2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$

The standard free-energy change, $\Delta G^{\circ}$, at $25^{\circ} \mathrm{C}$ equals $-13.6 \mathrm{~kJ}$. (We calculated this value of $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$ just before Section 19.4.)

PROBLEM STRATEGY

Rearrange the equation $\Delta G^{\circ}=-R T \ln K$ to give

$\ln K=\frac{\Delta G^{\circ}}{-R T}$

$\Delta G^{\circ}$ and $R$ must be in compatible units. You normally express $\Delta G^{\circ}$ in joules and set $R$ equal to $8.31 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})$.