In a first-order reaction, the concentration of the reactant decreases from 800 mol dm−3 to 50 mol dm−3 in 2 × 104 s. What is the rate constant of the reaction (in s−1)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.386 × 10−4
For a first-order reaction, the rate constant follows from the integrated rate equation. The concentration falls from 800 to 50 mol dm−3, so [R]0/[R] = 800/50 = 16.
\( \displaystyle k=\frac{2.303}{t}\log\frac{[R]_0}{[R]} \)
\( \displaystyle k=\frac{2.303}{2\times10^{4}\ \mathrm{s}}\log\frac{800}{50}=\frac{2.303\times1.2041}{2\times10^{4}\ \mathrm{s}} \)
\( \displaystyle k=1.386\times10^{-4}\ \mathrm{s^{-1}} \)