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Thermodynamics question

2019 · 11 Jan · Shift 2 · Q9
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Thermodynamics question

2019 · 11 Jan · Shift 2 · Q9

JEE MainChemistryThermodynamicsMCQ+4 / −1
For the equilibrium, 2H2OH_2OH2​O ⇌\rightleftharpoons⇌ H3O+H_3O^+H3​O+ + OHOHOH −-−, the value of Δ\DeltaΔ Go at 298 K is approximately :
  1. A
    −-− 80 kJ mol–1
  2. B
    100 kJ mol −-− 1
  3. C
    −-− 100 kJ mol −-− 1
  4. D
    80 kJ mol–1
View written solutionFree

Correct answer: D

  1. Identify the equilibrium constant

For 2H2O⇌H3O++OH−2H_2O \rightleftharpoons H_3O^+ + OH^-2H2​O⇌H3​O++OH− this is the auto-ionization of water.

At 298 K298\,\text{K}298K, Kw=[H3O+][OH−]=10−14K_w = [H_3O^+][OH^-] = 10^{-14}Kw​=[H3​O+][OH−]=10−14 So for this equilibrium, K=10−14K = 10^{-14}K=10−14

  1. Use the relation between ΔG∘\Delta G^\circΔG∘ and KKK

The standard Gibbs free energy change is ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln KΔG∘=−RTlnK

Substitute:

  • R=8.314 J mol−1K−1R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}R=8.314J mol−1K−1
  • T=298 KT = 298\,\text{K}T=298K
  • K=10−14K = 10^{-14}K=10−14

Thus, ΔG∘=−(8.314)(298)ln⁡(10−14)\Delta G^\circ = - (8.314)(298)\ln(10^{-14})ΔG∘=−(8.314)(298)ln(10−14)

  1. Simplify the logarithm

Since ln⁡(10−14)=−14ln⁡10\ln(10^{-14}) = -14\ln 10ln(10−14)=−14ln10 and ln⁡10≈2.303\ln 10 \approx 2.303ln10≈2.303,

ΔG∘=(8.314)(298)(14)(2.303)\Delta G^\circ = (8.314)(298)(14)(2.303)ΔG∘=(8.314)(298)(14)(2.303)

  1. Calculate numerically

First, 8.314×298≈2477.68.314 \times 298 \approx 2477.68.314×298≈2477.6

Then, 14×2.303≈32.24214 \times 2.303 \approx 32.24214×2.303≈32.242

So, ΔG∘≈2477.6×32.242≈7.99×104 J mol−1\Delta G^\circ \approx 2477.6 \times 32.242 \approx 7.99 \times 10^4\,\text{J mol}^{-1}ΔG∘≈2477.6×32.242≈7.99×104J mol−1

Therefore, ΔG∘≈79.9 kJ mol−1≈80 kJ mol−1\Delta G^\circ \approx 79.9\,\text{kJ mol}^{-1} \approx 80\,\text{kJ mol}^{-1}ΔG∘≈79.9kJ mol−1≈80kJ mol−1

  1. Choose the correct option

The value is positive and approximately 80 kJ mol−180\,\text{kJ mol}^{-1}80kJ mol−1.

So the correct option is: D: 80 kJ mol−180\,\text{kJ mol}^{-1}80kJ mol−1

  1. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.

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