MHT CET · Chemistry · Chemical Kinetics
Half-life of first order reaction X \(\longrightarrow\) \(\mathrm{Y}+\mathrm{Z}\) is 3 minutes. What is the time required to reduce the concentration of \({ }^{\prime} \mathrm{X}^{\prime}\) by \(90 \%\) of it's initial concentration?
- A \(4 \cdot 12\) minutes
- B \(9.969\) minutes
- C \(9 \cdot 105\) minutes
- D \(12 \cdot 05\) minutes
Answer & Solution
Correct Answer
(B) \(9.969\) minutes
Step-by-step Solution
Detailed explanation
(A)
\(\mathrm{t}_{1 / 2}=3 \mathrm{~min} \)
\( \therefore \mathrm{k}=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{3}=0.231 \mathrm{~min}^{-1}\)
\({[\mathrm{A}]_{0}=\text { Original amount of reactant }=100} \)
\([\mathrm{A}]_{t}=\text {Reactant remaining unreacted}=\) \(100-90=10\)
For first order reaction,
\(t=\frac{2.303}{k} \log _{10} \frac{[A]_{0}}{[A]_{t}}=\frac{2.303}{0.231 \mathrm{~min}^{-1}} \log _{10} \frac{100}{10}\)\(=9.969 \mathrm{~min}\)
\(\mathrm{t}_{1 / 2}=3 \mathrm{~min} \)
\( \therefore \mathrm{k}=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{3}=0.231 \mathrm{~min}^{-1}\)
\({[\mathrm{A}]_{0}=\text { Original amount of reactant }=100} \)
\([\mathrm{A}]_{t}=\text {Reactant remaining unreacted}=\) \(100-90=10\)
For first order reaction,
\(t=\frac{2.303}{k} \log _{10} \frac{[A]_{0}}{[A]_{t}}=\frac{2.303}{0.231 \mathrm{~min}^{-1}} \log _{10} \frac{100}{10}\)\(=9.969 \mathrm{~min}\)
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