MHT CET · Chemistry · Chemical Kinetics
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 \(18 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}\)
- B \(90 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}\)
- C \(72 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}\)
- D \(36 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}\)
Answer & Solution
Correct Answer
(B) \(90 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}\)
Step-by-step Solution
Detailed explanation
Rate law is.
\(
\begin{aligned}
& \mathrm{r}=\mathrm{K}[\mathrm{A}]^2[\mathrm{~B}]^2 \\
& 3.6 \times 10^{-2}=\mathrm{k}(0.2)^2(0.1)^2 \\
& \mathrm{~K}=\frac{3.6 \times 10^{-2}}{0.04 \times 0.01} \\
& \mathrm{~K}=90 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{r}=\mathrm{K}[\mathrm{A}]^2[\mathrm{~B}]^2 \\
& 3.6 \times 10^{-2}=\mathrm{k}(0.2)^2(0.1)^2 \\
& \mathrm{~K}=\frac{3.6 \times 10^{-2}}{0.04 \times 0.01} \\
& \mathrm{~K}=90 \mathrm{~mol}^{-3} \mathrm{dm}^9 \mathrm{sec}^{-1}
\end{aligned}
\)
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