Calculating K from the Standard Free-Energy Change (Net Ionic Equation) Calculate the equilibrium constant Ksp at 25°C for the reaction AgCl(s) ⇌ Ag^+(aq) + Cl^-(aq) using standard free energies of formation. PROBLEM STRATEGY This is similar to the preceding example, except that you must first

Question

Calculating [latex]K[/latex] from the Standard Free-Energy Change (Net lonic Equation)

Calculate the equilibrium constant [latex]K_{s p}[/latex] at [latex]25^{\circ} \mathrm{C}[/latex] for the reaction

[latex]\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q)[/latex]

using standard free energies of formation.

PROBLEM STRATEGY

This is similar to the preceding example, except that you must first calculate [latex]\Delta G_{f}^{\circ}[/latex] for the ionic equation. Note that [latex]K_{s p}[/latex] equals [latex]K[/latex], the thermodynamic equilibrium constant.

Accepted Answer

You first calculate [latex]\Delta G^{\circ}[/latex]. Writing [latex]\Delta G_{f}^{\circ}[/latex] values below the formulas in the equation gives

[latex] \begin{array}{rcc} \mathrm{AgCl}(s) & ightleftharpoons \mathrm{Ag}^{+}(a q)&+\mathrm{Cl}^{-}(a q) \ \Delta G_{f}^{\circ}:-109.8 & 77.1 & -131.3 \mathrm{~kJ} \end{array} [/latex]

Hence,

[latex] \begin{aligned} \Delta G^{\circ} & =[(77.1-131.3)-(-109.8)] \mathrm{kJ} \ & =55.6 \mathrm{~kJ}\left( ext { or } 55.6 imes 10^{3} \mathrm{~J} ight) \end{aligned} [/latex]

You now substitute numerical values into the equation relating [latex]\ln K[/latex] and [latex]\Delta G^{\circ}[/latex].

[latex]\ln K=\frac{\Delta G^{\circ}}{-R T}=\frac{55.6 imes 10^{3}}{-8.31 imes 298}=-22.45[/latex]

Therefore,

[latex]K=e^{-22.45}=\mathbf{1 . 8} imes 10^{-10}[/latex]

Question 19.9: Calculating K from the Standard Free-Energy Change (Net Ioni...

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