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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$10^{- 2}$

$2 \times 10^{- 3}$

$10^{- 3}$

$5 \times 10^{- 4}$

answer is B.

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Detailed Solution

For every $10^{°}$ rise or fall in temperature rate gets doubled or halved.

$rate = 10^{- 3} {[\frac{1}{2}]}^{\frac{t_{2} - t_{1}}{10}} as temperature is decreased from 27^{0} C to 17^{0} C = \frac{10^{- 3}}{2} = 5 \times 10^{- 4} \min^{- 1}$

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