JEE Main · 2019 · Shift-IImediumTHERMO-081

For the equilibrium 2H2O H3O+ + OH-; the value of G at 298 K is approximately:

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

For the equilibrium 2H2O⇌H3O++OH−2\text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-; the value of ΔG∘\Delta G^\circ at 298 K is approximately:

Options
  1. a

    100 kJ mol−1100\,\text{kJ mol}^{-1}

  2. b

    −80 kJ mol−1-80\,\text{kJ mol}^{-1}

  3. c

    80 kJ mol−180\,\text{kJ mol}^{-1}

    ✓
  4. d

    −100 kJ mol−1-100\,\text{kJ mol}^{-1}

Correct Answerc

80 kJ mol−180\,\text{kJ mol}^{-1}

Detailed Solution

🧠 Autoionization of water: Kw=10−14K_w = 10^{-14} at 298 K, giving a large positive ΔG∘\Delta G^\circ \ceH2O⇌H++OH−\ce{H2O \rightleftharpoons H+ + OH-}: equilibrium constant = Kw=10−14K_w = 10^{-14} (a very unfavourable equilibrium).

🗺️ Calculation ΔG∘=−RTln⁡Kw=−2.303RTlog⁡Kw\Delta G^\circ = -RT\ln K_w = -2.303RT\log K_w =−2.303×8.314×298×log⁡(10−14)= -2.303 \times 8.314 \times 298 \times \log(10^{-14}) =−2.303×8.314×298×(−14)= -2.303 \times 8.314 \times 298 \times (-14) =5706×14=79,884 J≈80 kJ mol−1= 5706 \times 14 = 79{,}884\text{ J} \approx 80\text{ kJ mol}^{-1}

Large positive ΔG∘\Delta G^\circ makes sense: water barely ionises, so the equilibrium lies far to the left.

⚠️ Trap: Kw=10−14K_w = 10^{-14} is very small, so log⁡Kw=−14\log K_w = -14 (negative). The double negative gives a positive ΔG∘\Delta G^\circ. Don't accidentally compute a negative answer.

Answer: (c)\boxed{\text{Answer: (c)}}

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For the equilibrium 2H2O H3O+ + OH-; the value of G at 298 K is approximately: (JEE Main 2019) | Canvas Classes