The rate constant of a first-order reaction at 27 °C is 10−3 min−1. The rate constant of this reaction doubles for every 10 °C rise in temperature. What is its rate constant at 17 °C?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5 × 10−4 min−1
The rate constant of a reaction rises with temperature. Here it doubles for every 10 °C rise, so it is halved for every 10 °C fall.
Going from 27 °C to 17 °C is a fall of 10 °C:
\( \displaystyle \frac{k_{27}}{k_{17}}=2 \)
\( \displaystyle k_{17}=\frac{k_{27}}{2}=\frac{1\times10^{-3}\,\mathrm{min^{-1}}}{2} \)
\( \displaystyle k_{17}=5\times10^{-4}\,\mathrm{min^{-1}} \)