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) [\( R=8.3 \, J \, K^{-1} \, mol^{-1} \)]
- A 4
- B 6
- C 8
- D 10
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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