Question 19.9: Calculating K from the Standard Free-Energy Change (Net Ioni...
Calculating K from the Standard Free-Energy Change (Net lonic Equation)
Calculate the equilibrium constant K_{s p} at 25^{\circ} \mathrm{C} for the reaction
\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)
using standard free energies of formation.
PROBLEM STRATEGY
This is similar to the preceding example, except that you must first calculate \Delta G_{f}^{\circ} for the ionic equation. Note that K_{s p} equals K, the thermodynamic equilibrium constant.
You first calculate \Delta G^{\circ}. Writing \Delta G_{f}^{\circ} values below the formulas in the equation gives
\begin{array}{rcc} \mathrm{AgCl}(s) & \rightleftharpoons \mathrm{Ag}^{+}(a q)&+\mathrm{Cl}^{-}(a q) \\ \Delta G_{f}^{\circ}:-109.8 & 77.1 & -131.3 \mathrm{~kJ} \end{array}
Hence,
\begin{aligned} \Delta G^{\circ} & =[(77.1-131.3)-(-109.8)] \mathrm{kJ} \\ & =55.6 \mathrm{~kJ}\left(\text { or } 55.6 \times 10^{3} \mathrm{~J}\right) \end{aligned}
You now substitute numerical values into the equation relating \ln K and \Delta G^{\circ}.
\ln K=\frac{\Delta G^{\circ}}{-R T}=\frac{55.6 \times 10^{3}}{-8.31 \times 298}=-22.45
Therefore,
K=e^{-22.45}=\mathbf{1 . 8} \times 10^{-10}