Question

The rate law for a reaction between the substances $A$ and $B$ is given by rate$=k{\left[A\right]}^n{\left[B\right]}^m$. On doubling the concentration of $A$ and having the concentration of $B$ halved, 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 $2^{(n-m)}$

Given that

$R=K\left[A{]}^n\right[B{]}^m$

after doubling the concentration of $A$ and concentration of $B$ is halfed

$R^1=K\left[2A{]}^n\right[\frac{B}{2}{]}^m$

$R^1=(2{)}^{n-m}R$

$\frac{R^1}{R}=\frac{2^{n-m}}{1}$