NEET 2026 — Chemistry solution

NEET2026Chemistry

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 is s $^{- 1}$ , R = 1.987 cal mol $^{- 1}$ K $^{- 1}$ )

Solution

Comparing with the Arrhenius form $ln k = ln A - \frac{E _{a}}{R T}$ :

$\frac{E _{a}}{R} = 1.25 \times 1 0^{4}$

$E_{a} = 1.25 \times 1 0^{4} \times 1.987$ cal mol $^{- 1} = 24837.5$ cal mol $^{- 1} \approx 24.84$ kcal mol $^{- 1}$ .

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