KCET · Chemistry · Chemical Kinetics
The time required for \(60 \%\) completion of a first order reaction is \(50 \mathrm{~min}\).
The time required for \(93.6 \%\) completion of the same reaction will be
- A \(100 \mathrm{~min}\)
- B \(83.8 \mathrm{~min}\)
- C \(50 \mathrm{~min}\)
- D \(150 \mathrm{~min}\)
Answer & Solution
Correct Answer
(D) \(150 \mathrm{~min}\)
Step-by-step Solution
Detailed explanation
Time for \(60 \%\) completion \(=50 \mathrm{~min}\) for first order reaction.
Time required for \(93.6 \%\) completion \(=\) ? \(k=\frac{2.303}{50} \log \frac{a}{a-0.6 a} \Rightarrow k=\frac{2.303}{50} \log \frac{a}{0.4 a}\) \(k=\frac{2.303}{50} \log 2.5\)
Also, \(k=\frac{2.303}{x} \log \frac{a}{a-.936 a}\)
\(\therefore \frac{2.303 \log 2.5}{50}=\frac{2.303}{x} \log \frac{a}{0.064 a}\)
\(x=50 \times \frac{\log \frac{1000}{64}}{\log 2.5}\)
\(=50 \times \frac{\log 15.625}{\log 2.5}=\frac{1.19 \times 50}{0.398}=\frac{59.5}{0.398}\)
\(=150 \mathrm{~min}\)
Time required for \(93.6 \%\) completion \(=\) ? \(k=\frac{2.303}{50} \log \frac{a}{a-0.6 a} \Rightarrow k=\frac{2.303}{50} \log \frac{a}{0.4 a}\) \(k=\frac{2.303}{50} \log 2.5\)
Also, \(k=\frac{2.303}{x} \log \frac{a}{a-.936 a}\)
\(\therefore \frac{2.303 \log 2.5}{50}=\frac{2.303}{x} \log \frac{a}{0.064 a}\)
\(x=50 \times \frac{\log \frac{1000}{64}}{\log 2.5}\)
\(=50 \times \frac{\log 15.625}{\log 2.5}=\frac{1.19 \times 50}{0.398}=\frac{59.5}{0.398}\)
\(=150 \mathrm{~min}\)
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