A reaction is first order in A and second order in B. How is the rate affected on increasing the concentration of B three times?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Increases nine times.
The rate law shows how the rate depends on the concentration of each reactant, each raised to the power of its order. For a reaction that is first order in A and second order in B:
\( \displaystyle \text{Rate}=k[\text{A}]^{1}[\text{B}]^{2} \)
When the concentration of B is made three times, with [A] unchanged:
\( \displaystyle \text{Rate}^{\prime}=k[\text{A}](3[\text{B}])^{2} \)
\( \displaystyle \text{Rate}^{\prime}=9\,k[\text{A}][\text{B}]^{2} \)
\( \displaystyle \text{Rate}^{\prime}=9\times\text{Rate} \)
So the rate increases nine times.
Shortcut: The rate changes by (factor)order. For B: 32 = 9.
Note: 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):
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?”