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At $300 \mathrm{~K}$, the equilibrium constant for a reaction is 10 . The standard free energy change (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) for the reaction is
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1625 Upvotes
Verified Answer
The correct answer is:
-57.4
From Gibbs free energy equation,
$$
\Delta G=-2.303 R T \log K
$$
Given, $K=10, T=300 \mathrm{~K}$
$$
\begin{aligned}
& R=8.314 \mathrm{~J} / \mathrm{K} / \mathrm{mol} \\
& \therefore \Delta G=-2.303 \times 8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K} \times 300 \mathrm{~K} \times \log 10 \\
& \Delta G=-2.303 \times 8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K} \times 300 \mathrm{~K} \\
& {[\because \log 10=1]} \\
& \Delta G=-5744.14 \mathrm{~J} / \mathrm{mol}, \Delta G=-5.74 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$
Thus, option (d) is correct.
$$
\Delta G=-2.303 R T \log K
$$
Given, $K=10, T=300 \mathrm{~K}$
$$
\begin{aligned}
& R=8.314 \mathrm{~J} / \mathrm{K} / \mathrm{mol} \\
& \therefore \Delta G=-2.303 \times 8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K} \times 300 \mathrm{~K} \times \log 10 \\
& \Delta G=-2.303 \times 8.314 \mathrm{~J} / \mathrm{mol} / \mathrm{K} \times 300 \mathrm{~K} \\
& {[\because \log 10=1]} \\
& \Delta G=-5744.14 \mathrm{~J} / \mathrm{mol}, \Delta G=-5.74 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$
Thus, option (d) is correct.
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