Question:
The rate constants of a reaction at \(500\,K\) and \(700\,K\) are \(0.02\,s^{-1}\) and \(0.2\,s^{-1}\) respectively. The activation energy of the reaction in \(kJ\,mol^{-1}\) is
\((R=8.3\,JK^{-1}mol^{-1})\)
The rate constants of a reaction at \(500\,K\) and \(700\,K\) are \(0.02\,s^{-1}\) and \(0.2\,s^{-1}\) respectively. The activation energy of the reaction in \(kJ\,mol^{-1}\) is
\((R=8.3\,JK^{-1}mol^{-1})\)
\((R=8.3\,JK^{-1}mol^{-1})\)
Show Hint
For Arrhenius equation using two temperatures:
\[
\log\frac{k_2}{k_1}
=
\frac{E_a}{2.303R}
\left[
\frac{T_2-T_1}{T_1T_2}
\right]
\]
Always use temperature in Kelvin and convert \(J\) to \(kJ\) at the end.
Updated On: Jun 22, 2026
- \(66.90\)
- \(33.45\)
- \(22.30\)
- \(44.45\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Use Arrhenius equation for two temperatures.
The relation between rate constants and temperature is \[ \log\frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2-T_1}{T_1T_2} \right] \]
Step 2: Write the given values.
Given, \[ k_1=0.02\,s^{-1} \] \[ k_2=0.2\,s^{-1} \] \[ T_1=500\,K \] \[ T_2=700\,K \] \[ R=8.3\,JK^{-1}mol^{-1} \]
Step 3: Substitute the values.
\[ \log\frac{0.2}{0.02} = \frac{E_a}{2.303\times 8.3} \left[ \frac{700-500}{500\times 700} \right] \] \[ \log 10 = \frac{E_a}{2.303\times 8.3} \left[ \frac{200}{350000} \right] \] Since, \[ \log 10=1 \] \[ 1= \frac{E_a}{2.303\times 8.3} \left[ \frac{1}{1750} \right] \]
Step 4: Calculate activation energy.
\[ E_a=2.303\times 8.3\times 1750 \] \[ E_a=33453.575\,J\,mol^{-1} \] Converting into \(kJ\,mol^{-1}\), \[ E_a=33.45\,kJ\,mol^{-1} \]
Step 5: Final conclusion.
Therefore, the activation energy of the reaction is \[ \boxed{33.45\,kJ\,mol^{-1}} \]
The relation between rate constants and temperature is \[ \log\frac{k_2}{k_1} = \frac{E_a}{2.303R} \left[ \frac{T_2-T_1}{T_1T_2} \right] \]
Step 2: Write the given values.
Given, \[ k_1=0.02\,s^{-1} \] \[ k_2=0.2\,s^{-1} \] \[ T_1=500\,K \] \[ T_2=700\,K \] \[ R=8.3\,JK^{-1}mol^{-1} \]
Step 3: Substitute the values.
\[ \log\frac{0.2}{0.02} = \frac{E_a}{2.303\times 8.3} \left[ \frac{700-500}{500\times 700} \right] \] \[ \log 10 = \frac{E_a}{2.303\times 8.3} \left[ \frac{200}{350000} \right] \] Since, \[ \log 10=1 \] \[ 1= \frac{E_a}{2.303\times 8.3} \left[ \frac{1}{1750} \right] \]
Step 4: Calculate activation energy.
\[ E_a=2.303\times 8.3\times 1750 \] \[ E_a=33453.575\,J\,mol^{-1} \] Converting into \(kJ\,mol^{-1}\), \[ E_a=33.45\,kJ\,mol^{-1} \]
Step 5: Final conclusion.
Therefore, the activation energy of the reaction is \[ \boxed{33.45\,kJ\,mol^{-1}} \]
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