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Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, T (K).
$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$
The energy of activation in $\text{kcal mol}^{-1}$ for the reaction is :
(Given : $k \text{ in s}^{-1}$, $R = 1.987 \text{ cal mol}^{-1} \text{ K}^{-1}$)

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
24.84

Calculate First Order Reaction Activation Energy

The problem provides the expression for the rate constant ($k$) of a first-order reaction dependent on temperature ($T$):

$\qquad \ln k = 14.34 - \frac{1.25 \times 10^4}{T}$

This form is comparable to the Arrhenius equation, which relates the rate constant to temperature:

$\qquad k = A' e^{-E_a/RT}$

Taking the natural logarithm of the Arrhenius equation yields:

$\qquad \ln k = \ln A' - \frac{E_a}{RT}$

Comparing Arrhenius Equation Terms

By comparing the given expression to the logarithmic form of the Arrhenius equation, we identify:

  • $\ln A' = 14.34$
  • $\frac{E_a}{RT} = \frac{1.25 \times 10^4}{T}$

From the second comparison, we can determine the ratio of activation energy ($E_a$) to the gas constant ($R$):

$\qquad \frac{E_a}{R} = 1.25 \times 10^4$

Activation Energy Calculation

To find the energy of activation ($E_a$), we use the given value for the gas constant $R$:

  • $R = 1.987 \text{ cal mol}^{-1} \text{ K}^{-1}$

Rearrange the equation to solve for $E_a$:

$\qquad E_a = (1.25 \times 10^4) \times R$

Substitute the value of $R$:

$\qquad E_a = (1.25 \times 10^4) \times (1.987 \text{ cal mol}^{-1} \text{ K}^{-1})$

$\qquad E_a = 24837.5 \text{ cal mol}^{-1}$

Convert to Kilocalories

The question requires the activation energy in kilocalories per mole ($\text{kcal mol}^{-1}$). Convert the result from calories to kilocalories:

$\qquad E_a = \frac{24837.5 \text{ cal mol}^{-1}}{1000 \text{ cal kcal}^{-1}}$

$\qquad E_a = 24.8375 \text{ kcal mol}^{-1}$

Rounding to two decimal places gives $24.84 \text{ kcal mol}^{-1}$.

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