If energy of activation of the reaction is 53.6 ~kJ ~mol^ -1 and the temperature changes from 27^ C to 37^ C then the value of ( k_ 37^ C k_ 27^ C ) is

Correct answer: C. 2.0 aligned & (c) : Given : T_1=27^ C=27+273=300 K & T_2=37^ C=37+273=310 K & and E_a=53.6 kJ mol^ -1 & aligned ((k_2)/(k_1))=(E_a)/(2.303 R)((T_2-T_1)/(T_1 T_2)) ( k_ 310 K k_ 300 K )=(53.6 × 10^3)/(2.303 × 8.314)((310-300)/(300 × 310)) =0.3010 aligned & k_ 310 K k_ 300 K =2 aligned

AIIMS 201826 May

Exam: AIIMS
Year: 2018
Date: 26 May
Subject: Chemistry
Chapter: Chemical Kinetics

AIIMS 2018 · 26 May · Chemistry · Chemical Kinetics

If energy of activation of the reaction is 53.6 ~kJ ~mol^ -1 and...

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)

Options

A $2.5$
B $1.0$
C $2.0$
D $1.5$

Correct answer

C. $2.0$

Step-by-step 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{\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}$

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