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The standard Gibbs free energy change $\left(\Delta G^{o}\right)$ at $25^{\circ} \mathrm{C}$ for the dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})$ to $\mathrm{NO}_{2}(g)$ is (given, equilibrium constant $=0.15, R=8.314 \mathrm{JK} / \mathrm{mol})$
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The correct answer is:
$4.7 \mathrm{KJ}$
$$
\begin{array}{l}
=-2.303 \times 8.314 \times 298 \times \log 0.15 \\
=-2.303 \times 8.314 \times 298 \times(-0.82) \\
=4678.7 \mathrm{J}=4.67 \mathrm{kJ}
\end{array}
$$
\begin{array}{l}
=-2.303 \times 8.314 \times 298 \times \log 0.15 \\
=-2.303 \times 8.314 \times 298 \times(-0.82) \\
=4678.7 \mathrm{J}=4.67 \mathrm{kJ}
\end{array}
$$
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