A reaction is 2nd order w.r.t A. and 1st order w.r.t B. When the concentrations of both A and B are doubled the rate of reaction will become
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 8 times
The rate law of a reaction that is second order in A and first order in B is
\( \text{Rate} = k[A]^2[B] \)
When both concentrations are doubled:
\( \text{Rate}' = k(2[A])^2(2[B]) = 4 \times 2 \times k[A]^2[B] = 8 \times \text{Rate} \)
The rate becomes 8 times.
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Exercises – “3.9 A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times? (iii) How is the rate affected when the concentrations of both A and B are doubled?”