Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature T(K): lnk = 14.34− 1.25× 10⁴/T .The energy of activation in kcal mol⁻¹ for the reaction is: R = 1.987 cal mol⁻¹K⁻¹

We are given the Arrhenius‑type expression for the rate constant of a first‑order reaction:

$ln k\; =\; 14.34\; \backslash ;-\backslash ;\; \backslash frac\{1.25\backslash times10^\{4\}\}\{T\}$

Recall the general Arrhenius equation written in logarithmic form:

$ln k\; =\; ln A\; \backslash ;-\backslash ;\; \backslash frac\{E\_\{\backslash text\; a\}\}\{R\backslash ,T\}$

Comparing the two equations, the term that multiplies $1/T$ is $\backslash frac\{E\_\{\backslash text\; a\}\}\{R\}$. Therefore:

$\backslash frac\{E\_\{\backslash text\; a\}\}\{R\}=1.25\backslash times10^\{4\}$

We are given the gas constant

$R\; =\; 1.987\backslash ;\backslash text\{cal mol\}^\{-1\}\backslash ,\backslash text\{K\}^\{-1\}$

Multiply both sides by $R$ to obtain the activation energy in calories per mole:

$E\_\{\backslash text\; a\}=\; (1.25\backslash times10^\{4\})\backslash times\; 1.987\backslash ;\backslash text\{cal mol\}^\{-1\}$

Calculate the product:

$1.25\backslash times10^\{4\}=12\{,\}500$
$12\{,\}500\; \backslash times\; 1.987\; =\; 24\{,\}837.5\backslash ;\backslash text\{cal mol\}^\{-1\}$

Convert calories to kilocalories by dividing by 1000:

$E\_\{\backslash text\; a\}=\; \backslash frac\{24\{,\}837.5\}\{1000\}=\; 24.8375\backslash ;\backslash text\{kcal mol\}^\{-1\}$

Rounding to two decimal places gives:

$E\_\{\backslash text\; a\}\backslash approx\; 24.84\backslash ;\backslash text\{kcal mol\}^\{-1\}$

Thus the activation energy for the reaction is **24.84 kcal mol⁻¹**, which matches the provided correct option.