JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
Consider the following reaction, the rate expression of which is given below \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}\) \(\text { rate }=\mathrm{k}[\mathrm{A}]^{1 / 2}[\mathrm{~B}]^{1 / 2}\) The reaction is initiated by taking \(1 \mathrm{M}\) concentration \(A\) and \(B\) each. If the rate constant \((k)\) is \(4.6 \times 10^{-2} \mathrm{~s}^{-1}\), then the time taken for \(\mathrm{A}\) to become \(0.1 \mathrm{M}\) is _______ sec. (nearest integer)
- A \(50\)
- B \(40\)
- C \(51\)
- D \(55\)
Answer & Solution
Correct Answer
(A) \(50\)
Step-by-step Solution
Detailed explanation
\(\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{1}{0.1}\) \(4.6 \times 10^{-2}=\frac{2.303}{\mathrm{t}}\) \(\mathrm{t}=50 \mathrm{sec}.\)
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