Q14 / 155 · WBJEE 2023 · Chemistry
At STP, the dissociation reaction of water is \(\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}\)(aq.) \(+\mathrm{OH}^{-}\)(aq.), and the \(\mathrm{pH}\) of water is 7.0. The change of standard free energy \(\left(\Delta \mathrm{G}^{\circ}\right)\) for the above dissociation process is given by
- A\(20301 \mathrm{~cal} / \mathrm{mol}\)
- Correct answerB\(19091 \mathrm{~cal} / \mathrm{mol}\)
- C\(20096 \mathrm{~cal} / \mathrm{mol}\)
- D\(21301 \mathrm{~cal} / \mathrm{mol}\)
Solution
\(\begin{aligned}
& \text {Hint : } \Delta G^{\circ}=-2.303 R T \log \mathrm{K}_{\mathrm{w}} \\
& =-2.303 \times 1.987 \times 298 \log 10^{-14} \\
& =+2.303 \times 1.987 \times 298 \times 14 \mathrm{~cal} / \mathrm{mol} \quad \quad\left[\mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right][\mathrm{OH}]=10^{-7} \times 10^{-7}=10^{-14} \mathrm{~as~} \mathrm{pH}=7\right] \\
& =19091.3 \mathrm{cal} / \mathrm{mol}=19091 \mathrm{cal} / \mathrm{mol}
\end{aligned}\)
Hence, the correct answer is option: 'B'.