Determine the standard free energy change, ΔG_f^° , for the formation of \({{\rm{S}}^{2 - }}(aq)\) given that theΔG_f^° for \({\rm{A}}{{\rm{g}}^ + }(aq)\) and \({\rm{A}}{{\rm{g}}_2}\;{\rm{S}}(s)\) are \(77.1\;{\rm{kJ}}/mole\) and \( - 39.5\;{\rm{kJ}}/{\rm{mole}}\) respectively, and the solubility product for \(Ag{\rm{S}}(s)\)is \(8 \times {10^{ - 51}}.\)

The equilibrium constant for the dissolution of a solid substance into an aqueous solution is the solubility product constant. The symbol \({K_{sp}}\)is used to represent it.

The reaction in question is:

\(A{g_2}S(s) \to 2A{g^ + }(aq) + {S^{2 - }}(aq)\)

Since we know all of the standard formation free energy changes except the one for \({S^{2 - }}\) (aq), which we are supposed to determine, we can figure it out from the formula

\(\Delta G = {G_f}({\rm{ products }}) - {G_f}({\rm{ reactants }})\)

In order to calculate the\(\Delta G\), we can use the formula

\(\Delta G = - RT\ln K\)

where\(K\) is the solubility product and the temperature is room temperature: \(25C\)or\(298K\).

\(\Delta G = - RT\ln K\)

\(\Delta G = - 8.314J{K^{ - 1}}\;{\rm{mo}}{{\rm{l}}^{ - 1}} \cdot 298\;{\rm{K}} \cdot \ln \left( {8 \cdot {{10}^{ - 51}}} \right)\)

\(\Delta G = 285.79\frac{{kJ}}{{mol}}\)

We can convert the free energy change from \(\frac{J}{{mol}}\) to \(\frac{{kJ}}{{mol}}\) by dividing the energy change by \(1000(1kJ = 1000J)\).

\(\Delta G = 2{G_f}(Ag + ,aq) + {G_f}\left( {{S^{2 - }},aq} \right) - {G_f}\left( {A{g_2}S,s} \right)\)

\(\begin{array}{l}{G_f}\left( {{S^{2 - }},aq} \right) = \Delta G - 2{G_f}\left( {A{g^ + },aq} \right) + {G_f}\left( {A{g_2}S,s} \right)\\{G_f}\left( {{S^{2 - }},aq} \right) = (285.79 - 2 \cdot 77 \times 1) + ( - 39.5))\frac{{kJ}}{{mol}}\end{array}\)

\({G_f}\left( {{S^{2 - }},aq} \right) = 92.1\frac{{kJ}}{{mol}}\)

Therefore, the standard free energy formation change for the aquatized sulphide anion is \(92.1\frac{{kJ}}{{mol}}\).