The conversion of molecules A to B follows second order kinetics. If the concentration of A is increased to four times, how will the rate of formation of B be affected?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Increases by 16 times
For a reaction of order n in A, the rate depends on [A]n. A second order reaction has the rate law:
\( \text{Rate} = k[A]^2 \)
When [A] is made four times:
\( \text{Rate}' = k(4[A])^2 = 16\,k[A]^2 = 16 \times \text{Rate} \)
So the rate of formation of B increases 16 times.
Why the other options are incorrect
Option (1) Increases by 9 times — Incorrect. A ninefold increase would come from tripling [A] in a second order reaction (32 = 9), not from making it four times.
Option (2) Increases by 4 times — Incorrect. This is the result for a first order reaction (41 = 4); the square of the concentration term is left out.
Option (4) Increases by 10 times — Incorrect. No whole-number order gives a factor of 10 for a fourfold change in concentration.
NCERT (2026-27 edition):
Rate Expression and Rate Constant – “… In the rate equation (3.4) Rate = k [A]x [B]y x and y indicate how sensitive the rate is to the change in concentration of A and B.”
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | 15-05 S2 | – | 10-06 S3 | – |