Rate law for a reaction between A and B is given by…

Rate law for a reaction between A and B is given by
$\mathrm{R}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}$
If concentration of A is doubled and concentration of $B$ is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{r_2}{r_1}\right)$ is
  1. $2^{(n-m)}$
  2. $(n-m)$
  3. $(m+n)$
  4. $\frac{1}{2^{m+n}}$

Solution

$\mathrm{r}_1=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}$
Now A is doubled \& B is halved in concentration
$\Rightarrow \mathrm{r}_2=\mathrm{k} 2^{\mathrm{n}}[\mathrm{~A}]^{\mathrm{n}} \cdot \frac{[\mathrm{~B}]^{\mathrm{m}}}{2^{\mathrm{m}}}$
Now $\frac{r_2}{r_1}=2^{(n-m)}$

Asked in: JEE Main 2025 (04 Apr Shift 1)

Practice more Chemical Kinetics questions on Aicharya