Equilibrium - Result Question 72
####76. Using the Gibbs energy change, $\Delta G^{\circ}=+63.3 kJ$, for the following reaction, [2014]
$ Ag_2 CO_3 \rightarrow 2 Ag^{+}(aq)+CO_3{ }^{2-}(aq) $
the $K _{sp}$ of $Ag_2 CO_3(s)$ in water at $25^{\circ} C$ is:( $R=8.314 J K^{-1} mol^{-1}$ )
(a) $3.2 \times 10^{-26}$
(b) $8.0 \times 10^{-12}$
(c) $2.9 \times 10^{-3}$
(d) $7.9 \times 10^{-2}$
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Solution:
- (b) $\Delta G=-2.303 \log K$
here $K=[Ag^{+}]^{2}[CO_3^{2-}]=K _{s p}$
$\therefore \quad 63.3 \times 10^{+3}=-2.303 \times 8.314 \times 298 \log K _{s p}$
$\therefore \quad \log K _{s p}=-\frac{63.3 \times 10^{+3}}{5705.8}=-11.09$
$\therefore \quad K _{s p}=Antilog(-11.09)=8 \times 10^{-12}$
