Q72 / 180 · AIIMS 2018 (26 May) · Chemistry

If energy of activation of the reaction is $53.6 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and the temperature changes from $27^{\circ} \mathrm{C}$ to $37^{\circ} \mathrm{C}$ then the value of $\left(\frac{k_{37^{\circ} \mathrm{C}}}{k_{27^{\circ} \mathrm{C}}}\right)$ is

  • A
    $2.5$
  • B
    $1.0$
  • Correct answer
    C
    $2.0$
  • D
    $1.5$

Solution

$\begin{aligned} & \text { (c) : Given : } T_1=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K} \\ & T_2=37^{\circ} \mathrm{C}=37+273=310 \mathrm{~K} \\ & \text { and } E_a=53.6 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \begin{aligned} \log \left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303 R}\left(\frac{T_2-T_1}{T_1 T_2}\right) \\ \log \left(\frac{k_{310 \mathrm{~K}}}{k_{300 \mathrm{~K}}}\right)=\frac{53.6 \times 10^3}{2.303 \times 8.314}\left(\frac{310-300}{300 \times 310}\right) \\ =0.3010\end{aligned} \\ & \therefore \frac{k_{310 \mathrm{~K}}}{k_{300 \mathrm{~K}}}=2\end{aligned}$

Hence, the correct answer is option: 'C'.

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