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Abreaction has an activation energy of $100 \mathrm{~kJ} / \mathrm{mol}$ and a rate constant of $2.5 \times 10^{-3} \mathrm{5}^{-1}$ at $27^{\circ} \mathrm{C}$ . Calculate the rate constant at $37^{\circ} \mathrm{C}$ (Given $\mathrm{R=8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}$ ).
Option: 1
$3 \times 10^{-3} s^{-1}$

Option: 2
$5 \times 10^{-3} s^{-1}$

Option: 3
$7 \times 10^{-3} s^{-1}$

Option: 4
$9 \times 10^{-3} s^{-1}$

Answers (1)

best_answer

For every reaction, for every $10^{\circ} \mathrm{C}$ rise in temperature the rate of reaction becomes double of its initial rate. Therefore $\mathrm{k_{2}=2 k_{1}}$ , when $\mathrm{T_{2}-T_{1}=10}$ . Hence $\mathrm{k_{37\degree c}=2 \times 2.5 \times 10^{-3}}$

$\mathrm{k_{37^{\circ} \mathrm{C}}=2 \times 2.5 \times 10^{-3}=5 \times 10^{-3} s^{-1}}$
$\mathrm{k_{37{ }^{\circ} \mathrm{C}}=5 \times 10^{-3} s^{-1}}$

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