MHT CET · Chemistry · Chemical Kinetics
The rate constant for first order reaction is \(0.02232 \mathrm{~min}^{-1}\). Calculate the time required for \(75 \%\) completion of the reaction.
- A \(62 \cdot 12 \mathrm{~min}\)
- B \(\ 28.31 \mathrm{~min}\)
- C \(12.77 \mathrm{~min}\)
- D \(\ 48 \cdot 12 \mathrm{~min}\)
Answer & Solution
Correct Answer
(A) \(62 \cdot 12 \mathrm{~min}\)
Step-by-step Solution
Detailed explanation
If \([\mathrm{A}]_{0}=100, \quad[\mathrm{~A}]_{\mathrm{t}}=100-75=25\)
\(\mathrm{k}=0.02232 \min ^{-1}, \quad t=?\)
\(t=\frac{2.303}{k} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{t}}=\frac{2.303}{0.02232} \log \frac{100}{25}\)
\(\therefore t=62.12 \min\)
\(\mathrm{k}=0.02232 \min ^{-1}, \quad t=?\)
\(t=\frac{2.303}{k} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{t}}=\frac{2.303}{0.02232} \log \frac{100}{25}\)
\(\therefore t=62.12 \min\)
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