For the reaction $\mathrm{A}+\mathrm{B} \longrightarrow$ product, rate law equation is, rate…

For the reaction $\mathrm{A}+\mathrm{B} \longrightarrow$ product, rate law equation is, rate $=\mathrm{k}[\mathrm{A}]^2[\mathrm{~B}]$. If rate of reaction is $0.22 \mathrm{~mol} \mathrm{~L}^{-1} \cdot \mathrm{~s}^{-1}$, calculate rate constant. $\left([\mathrm{A}]=1 \mathrm{~mol} \mathrm{~L}^{-1},[\mathrm{~B}]=0.25 \mathrm{~mol} \mathrm{~L}^{-1}\right)$
  1. $0.44 \mathrm{~mol}^{-2} \mathrm{~L}^2 \mathrm{~s}^{-1}$
  2. $0.88 \mathrm{~mol}^{-2} \mathrm{~L}^2 \mathrm{~s}^{-1}$
  3. $1.136 \mathrm{~mol}^{-2} \mathrm{~L}^2 \mathrm{~s}^{-1}$
  4. $\quad 3.52 \mathrm{~mol}^{-2} \mathrm{~L}^2 \mathrm{~s}^{-1}$

Solution

$\begin{aligned} & \text { Rate }=\mathrm{k}[\mathrm{A}]^2[\mathrm{~B}] \\ & 0.22 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}=\mathrm{k} \times\left(1 \mathrm{~mol} \mathrm{~L}^{-1}\right)^2 \times 0.25 \mathrm{~mol} \mathrm{~L}^{-1} \\ & \therefore \quad \mathrm{k}=0.88 \mathrm{~mol}^{-2} \mathrm{~L}^2 \mathrm{~s}^{-1}\end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 2)

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