WBJEE · Chemistry · Thermodynamics (C)
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 \(20301 \mathrm{~cal} / \mathrm{mol}\)
- B \(19091 \mathrm{~cal} / \mathrm{mol}\)
- C \(20096 \mathrm{~cal} / \mathrm{mol}\)
- D \(21301 \mathrm{~cal} / \mathrm{mol}\)
Answer & Solution
Correct Answer
(B) \(19091 \mathrm{~cal} / \mathrm{mol}\)
Step-by-step Solution
Detailed explanation
\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…
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