Kelvi

Kelvi

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Kelvi

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Using the Gibbs free energy change, $\Delta G^{\circ} = + 63.3kJ$ ,for the reaction $Ag_2CO_3 (s) \rightleftharpoons 2Ag^+ (aq) + CO_3^{2-} (aq)$ , the solubility product $(K_{sp})$ of Ag $_2$ CO $_3(s)$ in water at $25^{\circ}C$ is $(R = 8.314 JK^{-1} mol^{-1})$

hard

Equilibrium

2014

chemistry

$3.2 × 10^{-26}$

$8.0 × 10^{-12}$

$2.9 × 10^{-3}$

Explanation

To find the solubility product $(K_{sp})$ of $Ag_2CO_3(s),$

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$7.9 × 10^{-2}$

we will use the relation between

\Delta G^{\circ}

and

K_{sp}. \\

The relationship is given by:

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

Given:

\\ \Delta G^{\circ} = +63.3 \, \mathrm{kJ/mol} = 63300 \, \mathrm{J/mol} \\ R = 8.314 \, \mathrm{J/K \cdot mol} \\ T = 25^{\circ}C = 298 \, \mathrm{K} \\

Substitute the known values into the equation:

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

Calculate

-RT: \\ -RT = -8.314 \times 298 = -2477.572 \\

Substitute back into the equation:

\\ 63300 = -2477.572 \ln K_{sp} \\

Solve for

\ln K_{sp}: \\ \ln K_{sp} = \frac{63300}{-2477.572} \\ \ln K_{sp} = -25.549 \\

To find

K_{sp},

take the exponential of both sides:

\\ K_{sp} = e^{-25.549} \\ K_{sp} \approx 8.0 \times 10^{-12} \\

Therefore, the solubility product

K_{sp}

of

Ag_2CO_3(s)

is

8.0 \times 10^{-12}. \\

This corresponds to Option 2.