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

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$2 \times 10^{- 3}$

$10^{- 2}$

$5 \times 10^{- 4}$

$10^{- 3}$

answer is B.

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Detailed Solution

$Temperature coefficient = \frac{Rate constant (at 27^{\circ} C)}{Rate constant (at 17°C)} 2 (given) = \frac{10^{- 3} (given)}{?} So, rate constant (at 17^{\circ} C) = \frac{10^{- 3}}{2} = 5 \times 10^{- 4}$

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