MHT CET · Chemistry · Thermodynamics (C)
Equilibrium constant for a reaction is \(20 .\) What is the value of \(\Delta \mathrm{G}^{\circ}\) at \(300 \mathrm{~K} ?\left(\mathrm{R}=8 \times 10^{-3} \mathrm{~kJ}\right)\)
- A \(-5 \cdot 527 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- B \(-1 \cdot 663 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- C \(16 \cdot 63 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
- D \(-2 \cdot 763 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Answer & Solution
Correct Answer
(B) \(-1 \cdot 663 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
Step-by-step Solution
Detailed explanation
(A)
\(\mathrm{K}=20, \mathrm{~T}=300 \mathrm{~K}, \mathrm{R}=8 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) \(\Delta G^{0}=-2.303 \mathrm{RT} \log _{10} \mathrm{~K}\)
\(\begin{aligned} &=-2.303 \times 8 \times 10^{-3} \times 300 \times \log _{10}(20) \end{aligned}\)
\(\mathrm{K}=20, \mathrm{~T}=300 \mathrm{~K}, \mathrm{R}=8 \times 10^{-3} \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) \(\Delta G^{0}=-2.303 \mathrm{RT} \log _{10} \mathrm{~K}\)
\(\begin{aligned} &=-2.303 \times 8 \times 10^{-3} \times 300 \times \log _{10}(20) \end{aligned}\)
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