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

Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20 kJ mol⁻¹. If \( k_{1} \) and \( k_{2} \) are the rate constants of first and second reaction respectively at 300 K, then \( ln \frac{k_{2}}{k_{1}} \) will be ______.(nearest integer) [\( R=8.3 \, J \, K^{-1} \, mol^{-1} \)]

  1. A 4
  2. B 6
  3. C 8
  4. D 10
Verified Solution

Answer & Solution

Correct Answer

(C) 8

Step-by-step Solution

Detailed explanation

\(A \xrightarrow{ RX ^{ n }(1)}\) product \(E _1\) \(B \xrightarrow{ RX ^{ n }(2)}\) product \(E _2\) Assuming 'A' same for both reaction. \(\ln k _1=\ln A -\frac{ E _1}{300 R }\) \( ln(\frac{k_{2}}{k_{1}}) = \frac{E_{1}-E_{2}}{RT} = \frac{20 \times 1000}{300 \times 8.3}\)…
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