Question

The rate law for a reaction between the substances $A$ and $B$ is given by, $Rate=K\left[A{]}^n\right[B{]}^m$. 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 as:

• A. $\frac{1}{2^{m+n}}$
• B. $m+n$
• C. $n-m$
• D. $2^{(n-m)}$

Answer 1

The correct option is D $2^{(n-m)}$

$Rate1=k\left[A{]}^n\right[B{]}^m$

Now doubling concentration of $A$ and halving the concentration of $B$

$Rate2=k[2A{]}^n.{\left[\frac{B}{2}\right]}^m$

$Rate2=k{2}^{n-m}\left[A{]}^n\right[B{]}^m$

$\frac{Rate2}{Rate1}=2^{n-m}$

Hence, option D is correct.