JEE Main 2026 — Jan 21, Shift 1 — Chemistry solution

JEE Main2026Jan 21, Shift 1Chemistry

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 $ln (k_{2} / k_{1})$ will be ________. (nearest integer) [ $R = 8.3$ J K $^{- 1}$ mol $^{- 1}$ ]

Your answer

Solution

$ln k_{1} = ln A - E_{1} / (R T)$ and $ln k_{2} = ln A - E_{2} / (R T)$ .

$ln (k_{2} / k_{1}) = (E_{1} - E_{2}) / (R T) = 20000/ (8.3 \times 300) = 8.032 \approx 8$ .

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