For the equilibrium, \(2\mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{OH^-}\), the value of \(\Delta G^\circ\) at 298 K is approximately :
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- \(-80\ \mathrm{kJ\ mol^{-1}}\)
- \(100\ \mathrm{kJ\ mol^{-1}}\)
- \(-100\ \mathrm{kJ\ mol^{-1}}\)
- \(80\ \mathrm{kJ\ mol^{-1}}\)
The Correct Option is D
Solution and Explanation
\(\Delta G^\circ = -RT\ln K = -2.303\,RT\log K_w\)
\(= 2.303 \times 8.314 \times 298 \times 14\ \mathrm{J\ mol^{-1}} \approx 79.9\times 10^{3}\ \mathrm{J\ mol^{-1}}\)
\(\Delta G^\circ \approx +80\ \mathrm{kJ\ mol^{-1}}\) (positive, since ionisation of water is non-spontaneous).
Answer: (4) 80 kJ mol\(^{-1}\)
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