Question:
The rate constant of a first-order reaction becomes four times when temperature changes from 300 K to 320 K. Find activation energy.
The rate constant of a first-order reaction becomes four times when temperature changes from 300 K to 320 K. Find activation energy.
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A large increase in rate constant with temperature indicates a high activation energy.
Updated On: Jun 17, 2026
- 55.05
- 550.5
- 27.57
- 225.25
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The Correct Option is A
Solution and Explanation
Concept:
Arrhenius equation:
\[
\log\frac{k_2}{k_1}
=
\frac{E_a}{2.303R}
\left(
\frac{T_2-T_1}{T_1T_2}
\right)
\]
Step 1: Substitute values.
\[ \log 4=0.6 \] \[ T_1=300K,\quad T_2=320K \] \[ 0.6= \frac{E_a}{2.303\times8.3} \left( \frac{20}{300\times320} \right) \]
Step 2: Solve for \(E_a\).
\[ E_a = 55.05\times10^3Jmol^{-1} \] \[ E_a = 55.05kJmol^{-1} \] \[ \boxed{55.05kJmol^{-1}} \]
Step 1: Substitute values.
\[ \log 4=0.6 \] \[ T_1=300K,\quad T_2=320K \] \[ 0.6= \frac{E_a}{2.303\times8.3} \left( \frac{20}{300\times320} \right) \]
Step 2: Solve for \(E_a\).
\[ E_a = 55.05\times10^3Jmol^{-1} \] \[ E_a = 55.05kJmol^{-1} \] \[ \boxed{55.05kJmol^{-1}} \]
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