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JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics

For the reaction \(\mathrm{A} \rightarrow \mathrm{B}\), the rate constant \(\mathrm{k}\left(\right.\) in \(\left.\mathrm{s}^{-1}\right)\) is given by  \(\log _{10} \mathrm{k}=20.35-\frac{\left(2.47 \times 10^{3}\right)}{\mathrm{T}}\)   . The energy of activation in \(\mathrm{kJ} \,\mathrm{mol}^{-1}\) is ..... . (Nearest integer)  \(\left[\right.\) Given \(\left.: \mathrm{R}=8.314\, \mathrm{~J}\, \mathrm{~K}^{-1}\, \mathrm{~mol}^{-1}\right]\)

  1. A \(85\)
  2. B \(47\)
  3. C \(12\)
  4. D \(4.7\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(47\)

Step-by-step Solution

Detailed explanation

Given \(\log \mathrm{K}=20.35-\frac{2.47 \times 10^{3}}{\mathrm{~T}}\) We know \(\log K=\log A-\frac{E_{a}}{2.303 R T}\) \(\Rightarrow \quad \frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}=2.47 \times 10^{3}\)…
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