The rate constant of a first order reaction a | vedclass

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Question

(B) The temperature coefficient is defined as the ratio of rate constants at temperatures differing by $10^{\circ} C$: $	ext{Temperature coefficient}

Vedclass · Accepted Answer

$5 	imes 10^{-4}$

Vedclass · Answer

(B) The temperature coefficient is defined as the ratio of rate constants at temperatures differing by $10^{\circ} C$: $	ext{Temperature coefficient} = \frac{k_{(t+10)}}{k_t}$.Given,$	ext{Temperature coefficient} = 2$,$k_{(27^{\circ} C)} = 10^{-3} \ min^{-1}$,and we need to find $k_{(17^{\circ} C)}$.Substituting the values: $2 = \frac{k_{(27^{\circ} C)}}{k_{(17^{\circ} C)}}$.$2 = \frac{10^{-3}}{k_{(17^{\circ} C)}}$.$k_{(17^{\circ} C)} = \frac{10^{-3}}{2} = 0.5 	imes 10^{-3} = 5 	imes 10^{-4} \ min^{-1}$.

The rate constant of a first order reaction at $27^{\circ} C$ is $10^{-3} \ min^{-1}$. The temperature coefficient of this reaction is $2$. What is the rate constant (in $min^{-1}$) at $17^{\circ} C$ for this reaction?

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