Reaction rate between two substance $A$ and $B$ is expressed as following: rate $=k[A]^n[B]^m$ If the…

Reaction rate between two substance $A$ and $B$ is expressed as following: rate $=k[A]^n[B]^m$ If the concentration of $\mathrm{A}$ is doubled and concentration of $\mathrm{B}$ is made half of initial concentration, the ratio of the new rate to the earlier rate will be:
  1. $m+n$
  2. $n-m$
  3. $\frac{1}{\left.2^{(m+n}\right)}$
  4. $\left.2^{(n-m}\right)$

Solution

$ \text { } \begin{aligned} \text { Rate }_1 & =k[A]^n[B]^m \\ \text { Rate }_2 & =k[2 A]^n\left[\frac{1}{2} B\right]^m \\ \therefore \frac{\text { Rate }_2}{\text { Rate }_1} & =\frac{k[2 \mathrm{~A}]^n\left[\frac{1}{2} \mathrm{~B}\right]^m}{k[\mathrm{~A}]^n[\mathrm{~B}]^m}=(2)^n\left(\frac{1}{2}\right)^m \\ & =2^n \cdot(2)^{-m}=2^{n-m} \end{aligned} $

Asked in: JEE Main 2012 (07 May Online)

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