Question

Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature \(T(K)\): \[ \ln k = 14.34-\frac{1.25\times10^4}{T} \] Given \(R=1.987\text{ cal mol}^{-1}\text{K}^{-1}\). The activation energy of the reaction is:

• A. \(14.34\text{ kcal mol}^{-1}\)
• B. \(18.63\text{ kcal mol}^{-1}\)
• C. \(24.84\text{ kcal mol}^{-1}\)
• D. \(12.42\text{ kcal mol}^{-1}\)

Hint:

Compare the given equation with \(\ln k=\ln A-\frac{E_a}{RT}\). The coefficient of \(\frac{1}{T}\) gives \(\frac{E_a}{R}\).

Answer 1

The Correct Option is C

Step 1: Recall Arrhenius equation.
The Arrhenius equation in logarithmic form is: \[ \ln k=\ln A-\frac{E_a}{RT} \]

Step 2: Compare with the given expression.
Given: \[ \ln k=14.34-\frac{1.25\times10^4}{T} \] Compare with: \[ \ln k=\ln A-\frac{E_a}{RT} \] Therefore: \[ \frac{E_a}{R}=1.25\times10^4 \]

Step 3: Calculate activation energy.
\[ E_a=R\times1.25\times10^4 \] Substitute: \[ R=1.987\text{ cal mol}^{-1}\text{K}^{-1} \] \[ E_a=1.987\times1.25\times10^4 \] \[ E_a=24837.5\text{ cal mol}^{-1} \]

Step 4: Convert cal to kcal.
\[ 1000\text{ cal}=1\text{ kcal} \] \[ E_a=\frac{24837.5}{1000} \] \[ E_a=24.84\text{ kcal mol}^{-1} \] Therefore, the correct answer is: \[ 24.84\text{ kcal mol}^{-1} \]