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The rate for reaction $\mathrm{A}+\mathrm{B} \rightarrow$ product, is $1.8 \times 10^{-2} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$. Calculate the rate constant if the reaction is second order in $\mathrm{A}$ and first order in $\mathrm{B} .([\mathrm{A}]=0.2 \mathrm{M} ;[\mathrm{B}]=0.1 \mathrm{M})$
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Verified Answer
The correct answer is:
$4.5 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}$
Rate $=\mathrm{k}[\mathrm{A}]^2[\mathrm{~B}]$
$\begin{aligned} \therefore \quad \mathrm{k} & =\frac{\text { rate }}{[\mathrm{A}]^2[\mathrm{~B}]} \\ \mathrm{k} & =\frac{1.8 \times 10^{-2} \mathrm{moldm}^{-5} \mathrm{~s}^{-1}}{\left(0.2 \mathrm{moldm}^{-3}\right)^2 \times 0.1 \mathrm{moldm}^{-3}} \\ & =4.5 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\end{aligned}$
$\begin{aligned} \therefore \quad \mathrm{k} & =\frac{\text { rate }}{[\mathrm{A}]^2[\mathrm{~B}]} \\ \mathrm{k} & =\frac{1.8 \times 10^{-2} \mathrm{moldm}^{-5} \mathrm{~s}^{-1}}{\left(0.2 \mathrm{moldm}^{-3}\right)^2 \times 0.1 \mathrm{moldm}^{-3}} \\ & =4.5 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\end{aligned}$
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