For the reaction $2 \mathrm{~A}+2 \mathrm{~B} \rightarrow 2 \mathrm{C}+\mathrm{D}$, the rate law is…

For the reaction $2 \mathrm{~A}+2 \mathrm{~B} \rightarrow 2 \mathrm{C}+\mathrm{D}$, the rate law is expressed as rate $=k[A]^2[B]$. Calculate the rate constant if rate of reaction is $0.24 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$. $([\mathrm{A}]=0.5 \mathrm{M}$ and $[\mathrm{B}]=0.2 \mathrm{M}$ )
  1. $4.8 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}$
  2. $9.6 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}$
  3. $12.1 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}$
  4. $14.4 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}$

Solution

Rate $=k[A]^2[B]$ $\therefore \quad \mathrm{k}=\frac{\text { Rate }}{[\mathrm{A}]^2[\mathrm{~B}]}$ $=\frac{0.24 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}}{\left(0.5 \mathrm{~mol} \mathrm{dm}^{-3}\right)^2\left(0.2 \mathrm{~mol} \mathrm{dm}^{-3}\right)}$ $=4.8 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}$

Asked in: MHT CET 2023 (13 May Shift 2)

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