MHT CET · Chemistry · Chemical Kinetics
Rate of the reaction \(\mathrm{A}+\mathrm{B} \rightarrow\) product is \(3.6 \times 10^{-2} \mathrm{~mol~} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\) and rate law is \(\mathrm{r}=\mathrm{k}[\mathrm{A}][\mathrm{B}]^2\). What is rate constant of the reaction if \([\mathrm{A}]=0 \cdot 2 \mathrm{M}\) and \([\mathrm{B}]=0 \cdot 1 \mathrm{M}\) ?
- A \(10 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\)
- B \(18 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\)
- C \(24 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\)
- D \(4.8 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(B) \(18 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
\(k = \frac{r}{[\mathrm{A}][\mathrm{B}]^2}\) \(k = \frac{3.6 \times 10^{-2}}{(0.2)(0.1)^2} = 18 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}\)
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