Equilibrium constant for a reaction is $20 .$ What is the value of $\Delta \mathrm{G}^{\circ}$ at $300…

Equilibrium constant for a reaction is $20 .$ What is the value of $\Delta \mathrm{G}^{\circ}$ at $300 \mathrm{~K} ?\left(\mathrm{R}=8 \times 10^{-3} \mathrm{~kJ}\right)$
  1. $-5 \cdot 527 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  2. $-1 \cdot 663 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  3. $16 \cdot 63 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  4. $-2 \cdot 763 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

$\mathrm{K}=20, \mathrm{~T}=300 \mathrm{~K}, \quad \mathrm{R}=8 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ $\Delta G^{0}=-2.303 \mathrm{RT} \log _{10} \mathrm{~K}$ $\begin{aligned} &=-2.303 \times 8 \times 10^{-3} \times 300 \times \log _{10}(20) \end{aligned}$

Asked in: MHT CET 2020 (16 Oct Shift 2)

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