MHT CET · Chemistry · Chemical Kinetics
What is the half-life of a first order reaction if time required to decrease concentration of reactant from \(1.0 \mathrm{M}\) to \(0.25 \mathrm{M}\) is 10 hours?
- A 12 hour
- B 4 hour
- C 5 hour
- D 10 hour
Answer & Solution
Correct Answer
(C) 5 hour
Step-by-step Solution
Detailed explanation
For first order reaction
\(
\mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{A}_0}{\mathrm{At}}=\frac{1}{10} \cdot \ln \frac{1}{0.25}=\frac{\ln 4}{10} \mathrm{~h}^{-1}
\)
Half-life, \(\mathrm{t}_{1 / 2}=\frac{\ln 2}{\mathrm{k}}=\frac{\ln 2}{\ln 4 / 10}=\frac{10 \cdot \ln 2}{2 \cdot \ln 2}\)
\(
\mathrm{t}_{1 / 2}=5 \text { hour }
\)
\(
\mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{A}_0}{\mathrm{At}}=\frac{1}{10} \cdot \ln \frac{1}{0.25}=\frac{\ln 4}{10} \mathrm{~h}^{-1}
\)
Half-life, \(\mathrm{t}_{1 / 2}=\frac{\ln 2}{\mathrm{k}}=\frac{\ln 2}{\ln 4 / 10}=\frac{10 \cdot \ln 2}{2 \cdot \ln 2}\)
\(
\mathrm{t}_{1 / 2}=5 \text { hour }
\)
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