MHT CET · Chemistry · Chemical Kinetics
The rate for reaction \(2 \mathrm{~A}+\mathrm{B} \rightarrow\) product is \(6 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\) Calculate the rate constant if the reaction is first order in \(\mathrm{A}\) and zeroth order in B. \([\operatorname{Given}[\mathrm{A}]=[\mathrm{B}]=0.3 \mathrm{M}]\)
- A \(1 \times 10^{-3} \mathrm{~s}^{-1}\)
- B \(2 \times 10^{-3} \mathrm{~s}^{-1}\)
- C \(3 \times 10^{-3} \mathrm{~s}^{-1}\)
- D \(4 \times 10^{-3} \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(B) \(2 \times 10^{-3} \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\text {Rate}=\mathrm{k}[\mathrm{A}] \)
\( \therefore \mathrm{k}=\frac{\text { Rate }}{[\mathrm{A}]}=\frac{6 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}}{0.3 \mathrm{~mol} \mathrm{dm}^{-3}}\)\(=2 \times 10^{-3} \mathrm{~s}^{-1}\)
\( \therefore \mathrm{k}=\frac{\text { Rate }}{[\mathrm{A}]}=\frac{6 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}}{0.3 \mathrm{~mol} \mathrm{dm}^{-3}}\)\(=2 \times 10^{-3} \mathrm{~s}^{-1}\)
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