The rate law for a reaction between the substances A and B is given by Rate = k[A]n[B]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:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2^{(n - m)}\)
In a rate law each concentration is raised to its order, so a change in concentration changes the rate by that factor raised to the order.
\( \displaystyle r_1=k[A]^n[B]^m \)
With [A] doubled and [B] halved:
\( \displaystyle r_2=k(2[A])^n\left(\frac{[B]}{2}\right)^m=2^n\,2^{-m}\,k[A]^n[B]^m \)
\( \displaystyle \frac{r_2}{r_1}=2^{(n-m)} \)
The new rate is \( 2^{(n-m)} \) times the earlier rate.