WBJEE · Chemistry · Chemical Kinetics
For a reaction \(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{P}\), when concentration of \(\mathrm{B}\) alone is doubled, rate does not change and when concentrations of both \(\mathrm{A}\) and \(\mathrm{B}\) is doubled, rate increases by a factor of 4 . The unit of rate constant is,
- A \(\mathrm{s}^{-1}\)
- B \(\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}\)
- C \(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\)
- D \(\mathrm{L}^{2} \mathrm{~mol}^{-2} \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
Hint : Rate law: \(\mathrm{r}=\mathrm{k}[\mathrm{A}]^{\alpha}[\mathrm{B}]^{\beta} \quad \mathrm{B}\) follows first order kinetics. \(\therefore \beta=1\) \(\frac{r_{2}}{r_{1}}=\frac{[2 A]^{\alpha}[2 B]^{p}}{[A]^{\alpha}[B]^{\beta}}\) or, \(4=2^{a+\beta}\) or, \(\alpha+\beta=2\)…
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