Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, T (K). ln k = 14.34 − T 1.25 × 1 0 4 The energy of activation in kcal mol − 1 for the

Question

Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, $T$ (K).
$ln k = 14.34 - \frac{1.25 \times 1 0 ^{4}}{T}$
The energy of activation in kcal mol $^{- 1}$ for the reaction is :
(Given : $k$ in s $^{- 1}$ , $R = 1.987$ cal mol $^{- 1}$ K $^{- 1}$ )NEET 2026 (Cancelled)

TestHub · Accepted Answer

The Arrhenius equation is given by: ln k = ln A − RT E a ​ ​ The given equation is: ln k = 14.34 − T 1.25 × 1 0 4 ​ Comparing the two equations: R E a ​ ​ = 1.25 × 1 0 4 E a ​ = 1.25 × 1 0 4 × R Substituting R = 1.987 cal mol − 1 K − 1 : E a ​ = 1.25 × 1 0 4 × 1.987 E a ​ = 24837.5 cal mol − 1 E a ​ = 24.8375 kcal mol − 1 ≈ 24.84 kcal mol − 1 Answer: 24.84

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