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 \(19.1 \text{ kcal mol}^{-1}\).

Answer 2

Correct Answer:

Option 3: 19.1

Explanation:

1. Self-ionization of Water:

H₂O(l) ⇌ H⁺(aq) + OH^-(aq)

2. Ionic Product of Water (Kw):

Kw = [H⁺][OH^-] = 10^-14 (at 25°C)

3. Relationship between ΔG° and Kw:

ΔG° = -RT ln(Kw)

where:

• ΔG° = standard Gibbs free energy change
• R = gas constant (8.314 J mol^-1 K^-1)
• T = temperature in Kelvin
• Kw = equilibrium constant

4. Convert Temperature to Kelvin:

T = 25°C + 273.15 = 298.15 K

5. Calculate ΔG° in Joules:

ΔG° = -(8.314 J mol^-1 K^-1) * (298.15 K) * ln(10^-14)

ΔG° ≈ 79899.9 J mol^-1

6. Convert ΔG° to Kilocalories:

ΔG° (kcal mol^-1) = 79899.9 J mol^-1 / 4184 J kcal^-1

ΔG° ≈ 19.1 kcal mol^-1