MHT CET · Chemistry · Chemical Kinetics
The half life of a first order reaction is \(6.0\) hour. How long will it take for the concentration of reactant to decrease from \(0.4 \mathrm{M}\) to \(0 \cdot 12 \mathrm{M} ?\)
- A \(30 \cdot 36 \mathrm{~h}\)
- B \(10.42 \mathrm{~h}\)
- C \(4.25 \mathrm{~h}\)
- D \(9.51 \mathrm{~h}\)
Answer & Solution
Correct Answer
(B) \(10.42 \mathrm{~h}\)
Step-by-step Solution
Detailed explanation
For first order reaction,
\(k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{6.0 h}=0.1155 h^{-1} \)
\( \text {Here, }[A]_{0}=0.4 M,[A]_{t}=0.12 M \)
\( \therefore t=\frac{2.303}{k} \log _{10} \frac{[A]_{0}}{[A]_{1}}=\frac{2.303}{0.1155 h^{-1}} \times \log _{10} \frac{0.4}{0.12}\)\(=\frac{2.303}{0.1155} \times 0.5224 \)
\( \therefore t=10.42 h\)
\(k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{6.0 h}=0.1155 h^{-1} \)
\( \text {Here, }[A]_{0}=0.4 M,[A]_{t}=0.12 M \)
\( \therefore t=\frac{2.303}{k} \log _{10} \frac{[A]_{0}}{[A]_{1}}=\frac{2.303}{0.1155 h^{-1}} \times \log _{10} \frac{0.4}{0.12}\)\(=\frac{2.303}{0.1155} \times 0.5224 \)
\( \therefore t=10.42 h\)
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