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
Solution
<p>$\mathrm{r}_1=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}$</p>
<p>Now $A$ is doubled \& B is halved in concentration</p>
<p>$$\Rightarrow \mathrm{r}_2=\mathrm{k} 2^{\mathrm{n}}[\mathrm{~A}]^{\mathrm{n}} \cdot \frac{[\mathrm{~B}]^{\mathrm{m}}}{2^{\mathrm{m}}}$$</p>
<p>Now $\frac{r_2}{r_1}=2^{(n-m)}$</p>
About this question
Subject: Chemistry · Chapter: Chemical Kinetics · Topic: Rate of Reaction
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