Question

Using the Gibbs energy change, $\Delta G^\circ = +63.3kJ$ , for the following reaction, $Ag_2CO_3 \rightleftharpoons 2Ag^+_{(aq)}+CO^{2^−}_{3(aq)}$ the $K_{sp}$ of $Ag_2CO_3(s)$ in water at $25^∘$C is ($R=8.314JK^{−1}mol^{−1}$)

• A. $ 3.2 \times 10^{-26} $
• B. $ 8.0 \times 10^{-12} $
• C. $ 2.9 \times 10^{-3} $
• D. $ 7.9 \times 10^{-2} $

Answer 1

The Correct Option is B

To solve this problem, we need to relate the Gibbs energy change, \(\Delta G^\circ\), to the solubility product constant, \(K_{sp}\), for the given reaction:

The dissociation reaction for silver carbonate (\(Ag_2CO_3\)) is:

Ag_2CO_3(s) \rightleftharpoons 2Ag^+_{(aq)} + CO^{2-}_{3(aq)}

The relation between Gibbs energy change and the equilibrium constant is given by the equation:

\Delta G^\circ = -RT\ln K_{sp}

Where:

• \(\Delta G^\circ = +63.3 \text{ kJ mol}^{-1} = 63300 \text{ J mol}^{-1}\)
• \(R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}\)
• \(T = 298 \text{ K}\) (since \(25^\circ C = 273 + 25\))

Rearrange the equation to solve for \(K_{sp}\):

\ln K_{sp} = -\frac{\Delta G^\circ}{RT}

Substituting the given values:

\ln K_{sp} = -\frac{63300}{8.314 \times 298}

Calculate:

\ln K_{sp} \approx -25.585

To find \(K_{sp}\), take the exponent:

K_{sp} = e^{-25.585}

Calculate \(K_{sp}\):

K_{sp} \approx 8.0 \times 10^{-12}

Thus, the solubility product constant \(K_{sp}\) for \(Ag_2CO_3\) at \(25^\circ C\) is 8.0 \times 10^{-12}.