Q. The rate law for a reaction between the substances A and B is given by rate$\ \ = k\lbrack A\rbrack^{n}\lbrack B\rbrack^{m}$ . On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as
A) $\frac{1}{2^{m + n}}$
B) $(m + n)$
C) $(n - m)$
D) $2^{(n - m)}$
Explanation
(d)
Rate becomes $x^{y}$ times if concentration is made x time of a reactant giving $y^{th}$ order reaction.
Rate =k$\lbrack A\rbrack^{n}\lbrack B\rbrack^{m}$
Concentration of A is doubled hence x=2 ,y=n and rate becomes$= 2^{n}$ times
Concentration of B is halved ,hence x=$\frac{1}{2}$ and y=m and rate becomes=$\left( \frac{1}{2} \right)^{m}$ times
Net rate becomes=$(2)^{n\ \ \ }\left( \frac{1}{2} \right)^{m}$times
=$(2)^{n - m}$times
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.