MHT CET · Chemistry · Chemical Kinetics
Half life of first order reaction is 20 minutes. What is the time taken to reduce the initial concentration of the reactant to \(\frac{1}{10}\) th ?
- A \(6.6 \mathrm{~min}\)
- B \(66.56 \mathrm{~min}\)
- C \(150 \mathrm{~min}\)
- D \(79 \cdot 68 \mathrm{~min}\)
Answer & Solution
Correct Answer
(B) \(66.56 \mathrm{~min}\)
Step-by-step Solution
Detailed explanation
For first order reaction,
\(k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{20}=0.0346 \mathrm{~min}\)
Here, if \([A]_{0}=1,[A]_{1}-\frac{1}{10}, t_{1}/10=?\)
\(\begin{array}{l}
k=\frac{2.303}{t} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{1}} \\
\therefore t_{1 / 10}=\frac{2.303}{0.0346} \log 10=66.56 \mathrm{~min}
\end{array}\)
\(k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{20}=0.0346 \mathrm{~min}\)
Here, if \([A]_{0}=1,[A]_{1}-\frac{1}{10}, t_{1}/10=?\)
\(\begin{array}{l}
k=\frac{2.303}{t} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]_{1}} \\
\therefore t_{1 / 10}=\frac{2.303}{0.0346} \log 10=66.56 \mathrm{~min}
\end{array}\)
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