MHT CET · Chemistry · Chemical Kinetics
What is the time needed to reduce the initial concentration of reactant to \(10 \%\) in a first order reaction if its half life time is 10 minutes?
- A 15 minute
- B 20 minute
- C 25 minute
- D 33 minute
Answer & Solution
Correct Answer
(D) 33 minute
Step-by-step Solution
Detailed explanation
\(\mathrm{k} =\frac{0.693}{\mathrm{t}_{1 / 2}} ; \mathrm{k}=\frac{0.693}{10 \mathrm{~min}}=0.0693 \mathrm{~min}^{-1} \)
\( \mathrm{t} =\frac{2.303}{\mathrm{k}} \log _{10} \frac{[\mathrm{~A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \)
\(\mathrm{t} =\) \(\frac{2.303}{0.0693 \mathrm{~min}^{-1}} \log _{10} \frac{100}{10}=\frac{2.303}{0.0693 \mathrm{~min}^{-1}} \times 1 \)
\( \therefore \mathrm{t} =33 \text { minute }\)
\( \mathrm{t} =\frac{2.303}{\mathrm{k}} \log _{10} \frac{[\mathrm{~A}]_0}{[\mathrm{~A}]_{\mathrm{t}}} \)
\(\mathrm{t} =\) \(\frac{2.303}{0.0693 \mathrm{~min}^{-1}} \log _{10} \frac{100}{10}=\frac{2.303}{0.0693 \mathrm{~min}^{-1}} \times 1 \)
\( \therefore \mathrm{t} =33 \text { minute }\)
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