For a reaction aA → bB, when [A] = 0.56 M, the rate was found to be 4.8 M s−1. On reducing the concentration of A to 50%, the rate changes to 1.2 M s−1.
The order of reaction with respect to A is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2.0
For a reaction whose rate depends only on [A], the rate law is:
\( \displaystyle \text{Rate}=k[\mathrm{A}]^n \)
Comparing two rates at the same temperature removes k, so the order is found from the ratio of rates and the ratio of concentrations:
\( \displaystyle \frac{\text{Rate}_2}{\text{Rate}_1}=\left(\frac{[\mathrm{A}]_2}{[\mathrm{A}]_1}\right)^n \)
Reducing the concentration to 50% means [A]2 = ½[A]1:
\( \displaystyle \frac{1.2}{4.8}=\left(\frac{1}{2}\right)^n \)
\( \displaystyle \frac{1}{4}=\left(\frac{1}{2}\right)^n \)
\( \displaystyle n=2 \)
So the order with respect to A is 2.0. The actual value [A] = 0.56 M is not needed; only the ratio matters.
Shortcut: Halving the concentration made the rate one-fourth. (1/2)2 = 1/4, so the order is 2.