For the reaction $\mathrm{A}+\mathrm{B} \rightarrow$ product, rate of reaction is $3.6 \times 10^{-2}$…

For the reaction $\mathrm{A}+\mathrm{B} \rightarrow$ product, rate of reaction is $3.6 \times 10^{-2}$ $\mathrm{mol} \mathrm{dm}{ }^{-3} \mathrm{sec}^{-1}$. When $[\mathrm{A}]=0.2 \mathrm{~mol} \mathrm{dm}^{-3}$ and $[\mathrm{B}]=0.1 \mathrm{~mol} \mathrm{dm}^{-3}$, find rate constant of reaction if it is second order with respective to both reactants.
  1. $18 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}$
  2. $90 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}$
  3. $72 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}$
  4. $36 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}$

Solution

Rate law is. $\begin{aligned} & \mathrm{r}=\mathrm{K}[\mathrm{A}]^2[\mathrm{~B}]^2 \\ & 3.6 \times 10^{-2}=\mathrm{k}(0.2)^2(0.1)^2 \\ & \mathrm{~K}=\frac{3.6 \times 10^{-2}}{0.04 \times 0.01} \\ & \mathrm{~K}=90 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1} \end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 1)

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