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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Chemistry
Chapter
Thermodynamics (C)

JIPMER 2018 · Chemistry · Thermodynamics (C)

For the following reaction, Then, K_ e 0 for this equilibrium is

For the following reaction, For the following reaction, Then, K_ e 0 for this equilibrium is Then, log⁡Ke0\log \mathrm{K}_{\mathrm{e} 0} for this equilibrium is

Options

  1. A−3.41-3.41
  2. B2.382.38
  3. C120120
  4. D15001500

Correct answer

B. 2.382.38

Step-by-step solution

Given, ΔG∘=−13.6 kJ mol−1 T=298 K\begin{aligned} \Delta \mathrm{G}^{\circ} & =-13.6 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{~T} & =298 \mathrm{~K}\end{aligned} For equilibrium, the Gibb's free energy is calculated by using the formula. ΔG∘=−2.303RTlog⁡Keq\Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{eq}} or, log⁡Keq=−ΔG∘2.303RT=−−13.6×103 J mol−12.303×8.314 J mol−1 K−1×298 K=2.38\log \mathrm{K}_{\mathrm{eq}}=-\frac{\Delta \mathrm{G}^{\circ}}{2.303 \mathrm{RT}} =-\frac{-13.6 \times 10^3 \mathrm{~J} \mathrm{~mol}^{-1}}{2.303 \times 8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 298 \mathrm{~K}}=2.38
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