Conversion of molecule X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → the rate will increase 9 times
The rate law shows how the rate depends on concentration. For a reaction of order n with respect to X:
\( \displaystyle \text{Rate}=k[X]^{n} \)
If [X] is multiplied by a factor a, the rate is multiplied by an. With a = 3:
Zero order (n = 0): 30 = 1, the rate does not change.
First order (n = 1): 31 = 3, the rate becomes 3 times.
Second order (n = 2): 32 = 9, the rate becomes 9 times.
The conversion of X to Y is second order:
\( \displaystyle \text{Rate}_1=k[X]^2 \)
\( \displaystyle \text{Rate}_2=k(3[X])^2=9\,k[X]^2 \)
\( \displaystyle \frac{\text{Rate}_2}{\text{Rate}_1}=9 \)
The rate of formation of Y equals the rate of the reaction, so it increases 9 times.
Note 1: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Molecularity of a Reaction – “… The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?”
Note 2: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| 25-05 S2 | 14-05 S2 | – | – | – |