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

Answer

Correct option: A.
$2^{(n-m)}$
(A) $2^{(n-m)}$
$r_{1}=k[A]^{n}[B]^{m}$
Now A is doubled $\& \mathrm{~B}$ 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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