MHT CET · Chemistry · Chemical Kinetics
A first order reaction is \(75 \%\) completed in 60 minutes, the time required for it's \(50 \%\) completion is.
- A \(120 \mathrm{~min}\)
- B \(60 \mathrm{~min}\)
- C \(40 \mathrm{~min}\)
- D \(30 \mathrm{~min}\)
Answer & Solution
Correct Answer
(D) \(30 \mathrm{~min}\)
Step-by-step Solution
Detailed explanation
i. \([\mathrm{A}]_{0}=100, \quad[\mathrm{~A}]_{\mathrm{t}}=100-75=25, \mathrm{t}=60 \mathrm{~min}\)
For first order reaction,
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{t}}\)
\(\therefore \mathrm{k}=\frac{2.303}{60 \mathrm{~min}} \log _{10} \frac{100}{25}\)
\(\mathrm{k}=\frac{2.303 \times 0.6020}{60 \mathrm{~min}}=0.0231 \mathrm{~min}^{-1}\)
ii. The time required for \(50 \%\) completion of reaction is,
\(\mathrm{t}_{1 / 2}=\frac{0.693}{0.0231 \mathrm{~min}^{-1}}=30 \mathrm{~min}\)
For first order reaction,
\(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log _{10} \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{t}}\)
\(\therefore \mathrm{k}=\frac{2.303}{60 \mathrm{~min}} \log _{10} \frac{100}{25}\)
\(\mathrm{k}=\frac{2.303 \times 0.6020}{60 \mathrm{~min}}=0.0231 \mathrm{~min}^{-1}\)
ii. The time required for \(50 \%\) completion of reaction is,
\(\mathrm{t}_{1 / 2}=\frac{0.693}{0.0231 \mathrm{~min}^{-1}}=30 \mathrm{~min}\)
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