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The standard Gibb's energy $\left(\Delta G^{\circ}\right)$ for the following reaction is
$\begin{aligned} & A(s)+B^{2+}(a q) \rightleftharpoons A^{2+}(a q)+B(s), \\ & K_C=10^{12} \text { at } 25^{\circ} \mathrm{C}\left(K_C=\text { equilibrium constant }\right)\end{aligned}$
Options:
$\begin{aligned} & A(s)+B^{2+}(a q) \rightleftharpoons A^{2+}(a q)+B(s), \\ & K_C=10^{12} \text { at } 25^{\circ} \mathrm{C}\left(K_C=\text { equilibrium constant }\right)\end{aligned}$
Solution:
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Verified Answer
The correct answer is:
$-68.47 \mathrm{~kJ}$
The standard Gibb's free energy change $\left(\Delta G^{\circ}\right)$ at standard conditions is given by the formula $\Delta G^{\circ}=-R T \ln K_{\text {eq }}$
where, $R$ is gas constant $=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, $T=(25+273) \mathrm{K}=298 \mathrm{~K}$ and $K_{\mathrm{eq}}$ is the equilibrium constant
$$
\begin{aligned}
\Delta G^{\circ} & =-8.314 \times 298 \times 2.303 \log 10^{12} \\
& =-68,47018 \mathrm{~J}=-68.47 \mathrm{~kJ}
\end{aligned}
$$
where, $R$ is gas constant $=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, $T=(25+273) \mathrm{K}=298 \mathrm{~K}$ and $K_{\mathrm{eq}}$ is the equilibrium constant
$$
\begin{aligned}
\Delta G^{\circ} & =-8.314 \times 298 \times 2.303 \log 10^{12} \\
& =-68,47018 \mathrm{~J}=-68.47 \mathrm{~kJ}
\end{aligned}
$$
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