Q25 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Chemistry

The standard Gibb's energy $\left(\Delta G^{\circ}\right)$ for the following reaction is $\begin{aligned} & A(s)+B^{2+}(a q) \rightleftharpoons A^{2+}(a q)+B(s), \\ & K_C=10^{12} \text { at } 25^{\circ} \mathrm{C}\left(K_C=\text { equilibrium constant }\right)\end{aligned}$

  • A
    $-150 \mathrm{~kJ}$
  • B
    $-96.80 \mathrm{~kJ}$
  • Correct answer
    C
    $-68.47 \mathrm{~kJ}$
  • D
    $-100 \mathrm{~kJ}$

Solution

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 $$ \begin{aligned} \Delta G^{\circ} & =-8.314 \times 298 \times 2.303 \log 10^{12} \\ & =-68,47018 \mathrm{~J}=-68.47 \mathrm{~kJ} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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