JEE MAIN - Chemistry (2025 - 4th April Morning Shift - No. 4)
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
$(\mathrm{n}-\mathrm{m})$
$2^{(\mathrm{n}-m)}$
$\frac{1}{2^{m+n}}$
$(\mathrm{m}+\mathrm{n})$
Explanation
$$\mathrm{r}_1=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}$$
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 $\frac{r_2}{r_1}=2^{(n-m)}$
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