Question:
The following equation is obtained for a first order reaction: $\log k = 14 - \frac{1.25 \times 10^4 K}{T}$. The $E_a$ (in kJ $mol^{-1}$) and frequency factor A (in $s^{-1}$) of the reaction are respectively:
The following equation is obtained for a first order reaction: $\log k = 14 - \frac{1.25 \times 10^4 K}{T}$. The $E_a$ (in kJ $mol^{-1}$) and frequency factor A (in $s^{-1}$) of the reaction are respectively:
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Compare the given $\log k$ equation directly with the Arrhenius $\log k$ form to find $A$ and $E_a$.
Updated On: Jun 10, 2026
- 238.93; 14
- 238.93; $10^{14}$
- 23.89; $10^{14}$
- 23.89; 14
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Arrhenius equation: $\log k = \log A - \frac{E_a}{2.303 RT}$.
Step 2: Analysis
$\log A = 14 \implies A = 10^{14}$. $E_a / (2.303 \times 8.3) = 1.25 \times 10^4$. $E_a = 1.25 \times 10^4 \times 19.11 \approx 238875$ J/mol $\approx 238.93$ kJ/mol.
Step 3: Conclusion
$E_a \approx 238.93$ kJ/mol, $A = 10^{14}$ $s^{-1}$.
Final Answer: (B)
Arrhenius equation: $\log k = \log A - \frac{E_a}{2.303 RT}$.
Step 2: Analysis
$\log A = 14 \implies A = 10^{14}$. $E_a / (2.303 \times 8.3) = 1.25 \times 10^4$. $E_a = 1.25 \times 10^4 \times 19.11 \approx 238875$ J/mol $\approx 238.93$ kJ/mol.
Step 3: Conclusion
$E_a \approx 238.93$ kJ/mol, $A = 10^{14}$ $s^{-1}$.
Final Answer: (B)
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