For the reaction A+B arrow product, rate of reaction is 3.6 × 10^-2 mol dm^-3 sec^-1 . When [A]=0.2 ~mol dm^-3 and [B]=0.1 ~mol dm^-3 , find rate constant of reaction if it is second order with respective to both reactants.

Correct answer: (B) 90 ~mol^-3 dm^9 sec^-1. Rate law is. r=K[A]^2[~B]^2 3.6 × 10^-2=k(0.2)^2(0.1)^2 ~K=frac3.6 × 10^-20.04 × 0.01 ~K=90 ~mol^-3 dm^9 sec^-1

Q15 / 150 · MHT CET 2021 (23 Sep Shift 1) · Chemistry

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.

A
$18 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}$
Correct answer
B
$90 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}$
C
$72 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}$
D
$36 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}$

Solution

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

Hence, the correct answer is option: 'B'.

Solution

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