Question

At \(25^\circ \text{C}\), the ionic product of water is \(10^{-14}\). The free energy change (\(\Delta G^\circ\)) for the self-ionization of water in kcal mol\(^{-1}\) is:

• A. 20.5
• B. 14
• C. 19.1
• D. 25.3

Hint:

Alwaysuse∆G◦=−RTlnKforequilibriumconstant-relatedproblems.Rememberto convertunitsasneeded.

Answer 1

The Correct Option is C

The Gibbs Free Energy change (\(\Delta G^\circ\)) relates to the equilibrium constant (K) by:

\( \Delta G^\circ = -RT\ln{K} \)

Where:

• \(K = 10^{-14}\) (ionic product of water),
• \(R = 1.987\) cal mol^-1 K^-1 (universal gas constant),
• \(T = 298\) K (temperature in Kelvin).

Using these values:

\( \Delta G^\circ = -1.987 \times 298 \times \ln(10^{-14}) \)

Since \(\ln(10^{-14}) = -14 \ln(10)\) and \(\ln(10) \approx 2.303\):

\( \Delta G^\circ = -1.987 \times 298 \times (-14 \times 2.303) = 19100 \text{ cal mol}^{-1} \)

Converting to kcal mol^-1:

\( \Delta G^\circ = 19.1 \text{ kcal mol}^{-1} \)

Therefore, the free energy change is approximately \(19.1 \text{ kcal mol}^{-1}\).