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 relationship between Gibbs Free Energy change ($\Delta G^\circ$) and the equilibrium constant (K) is given by:

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

Here:

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

Substituting 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:

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

Thus, the free energy change is approximately $19.1 \text{ kcal mol}^{-1}$.