Question:
The value of the rate constant for the reaction \(A \rightarrow \text{products}\) is \(5 \times 10^{-5}\ \mathrm{s^{-1}}\) at \(300\ \mathrm{K}\). Its activation energy is \(50\ \mathrm{kJ\ mol^{-1}}\). At temperature \(T\), the rate constant becomes \(1.0 \times 10^{-4}\ \mathrm{s^{-1}}\). What is the value of \(T\) (in K)?
\[
\text{Given: } R = 8.3\ \mathrm{J\ mol^{-1}\ K^{-1}}, \qquad \log 2 = 0.3
\]
The value of the rate constant for the reaction \(A \rightarrow \text{products}\) is \(5 \times 10^{-5}\ \mathrm{s^{-1}}\) at \(300\ \mathrm{K}\). Its activation energy is \(50\ \mathrm{kJ\ mol^{-1}}\). At temperature \(T\), the rate constant becomes \(1.0 \times 10^{-4}\ \mathrm{s^{-1}}\). What is the value of \(T\) (in K)?
\[
\text{Given: } R = 8.3\ \mathrm{J\ mol^{-1}\ K^{-1}}, \qquad \log 2 = 0.3
\]
Show Hint
Whenever the rate constant doubles, first calculate
\(
k_2/k_1
\)
before applying the Arrhenius equation. Most competitive exam questions simplify to \(\log 2 = 0.3\).
Updated On: Jun 10, 2026
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The Correct Option is B
Solution and Explanation
Concept:
When the rate constant of a reaction changes with temperature, the relationship between the two rate constants is given by the Arrhenius equation:
\[
\log\left(\frac{k_2}{k_1}\right)
=
\frac{E_a}{2.303R}
\left(
\frac{1}{T_1}
-
\frac{1}{T_2}
\right)
\]
where
• \(k_1\) and \(k_2\) are the rate constants,
• \(T_1\) and \(T_2\) are the corresponding temperatures,
• \(E_a\) is the activation energy,
• \(R\) is the gas constant.
This equation allows us to calculate the new temperature when the activation energy and change in rate constant are known.
Step 1: Write the given data \[ k_1=5\times10^{-5}\ \mathrm{s^{-1}} \] \[ k_2=1\times10^{-4}\ \mathrm{s^{-1}} \] \[ T_1=300\ \mathrm{K} \] \[ E_a=50\times10^3\ \mathrm{J\ mol^{-1}} \] \[ R=8.3\ \mathrm{J\ mol^{-1}\ K^{-1}} \]
Step 2: Calculate the ratio of rate constants \[ \frac{k_2}{k_1} = \frac{1\times10^{-4}} {5\times10^{-5}} = 2 \] Therefore, \[ \log\left(\frac{k_2}{k_1}\right) = \log 2 = 0.3 \]
Step 3: Substitute into Arrhenius equation \[ 0.3 = \frac{50000} {2.303\times8.3} \left( \frac{1}{300} - \frac{1}{T} \right) \] \[ 0.3 = 2614 \left( \frac{1}{300} - \frac{1}{T} \right) \] \[ \frac{1}{300} - \frac{1}{T} = \frac{0.3}{2614} \] \[ = 1.15\times10^{-4} \]
Step 4: Solve for \(T\) \[ \frac{1}{T} = \frac{1}{300} - 1.15\times10^{-4} \] \[ = 0.003333 - 0.000115 \] \[ = 0.003218 \] \[ T = \frac{1}{0.003218} \] \[ T \approx311\ \mathrm{K} \] \[ \boxed{T=311\ \mathrm{K}} \] Hence, the correct answer is \[ \boxed{\text{Option (B)}} \]
• \(k_1\) and \(k_2\) are the rate constants,
• \(T_1\) and \(T_2\) are the corresponding temperatures,
• \(E_a\) is the activation energy,
• \(R\) is the gas constant.
This equation allows us to calculate the new temperature when the activation energy and change in rate constant are known.
Step 1: Write the given data \[ k_1=5\times10^{-5}\ \mathrm{s^{-1}} \] \[ k_2=1\times10^{-4}\ \mathrm{s^{-1}} \] \[ T_1=300\ \mathrm{K} \] \[ E_a=50\times10^3\ \mathrm{J\ mol^{-1}} \] \[ R=8.3\ \mathrm{J\ mol^{-1}\ K^{-1}} \]
Step 2: Calculate the ratio of rate constants \[ \frac{k_2}{k_1} = \frac{1\times10^{-4}} {5\times10^{-5}} = 2 \] Therefore, \[ \log\left(\frac{k_2}{k_1}\right) = \log 2 = 0.3 \]
Step 3: Substitute into Arrhenius equation \[ 0.3 = \frac{50000} {2.303\times8.3} \left( \frac{1}{300} - \frac{1}{T} \right) \] \[ 0.3 = 2614 \left( \frac{1}{300} - \frac{1}{T} \right) \] \[ \frac{1}{300} - \frac{1}{T} = \frac{0.3}{2614} \] \[ = 1.15\times10^{-4} \]
Step 4: Solve for \(T\) \[ \frac{1}{T} = \frac{1}{300} - 1.15\times10^{-4} \] \[ = 0.003333 - 0.000115 \] \[ = 0.003218 \] \[ T = \frac{1}{0.003218} \] \[ T \approx311\ \mathrm{K} \] \[ \boxed{T=311\ \mathrm{K}} \] Hence, the correct answer is \[ \boxed{\text{Option (B)}} \]
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