A reaction is second order with respect to a reactant. If the concentration of the reactant is increased by 1.5 times, the rate of the reaction
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Increases by 2.25 times.
For a reaction of order n in a reactant, rate \( = k[\mathrm{A}]^{n} \). Changing the concentration by a factor changes the rate by that factor raised to the power of the order:
\( \displaystyle \frac{r_2}{r_1} = \left(\frac{[\mathrm{A}]_2}{[\mathrm{A}]_1}\right)^{n} \)
Here the reaction is second order (n = 2) and the concentration becomes 1.5 times:
\( \displaystyle r_1 = k[\mathrm{A}]^{2} \)
\( \displaystyle r_2 = k(1.5[\mathrm{A}])^{2} = 2.25\,k[\mathrm{A}]^{2} \)
\( \displaystyle \frac{r_2}{r_1} = 2.25 \)
The rate increases by 2.25 times. Since a higher concentration can only raise the rate of a second order reaction, the two “decreases” options are ruled out at once.