MHT CET · Chemistry · Chemical Kinetics
Half-life and rate constant for first order reaction are related by equation,
- A \(t_{1 / 2}=\frac{k}{0.693}\)
- B \(\mathrm{k}=\frac{[\mathrm{A}]_0-[\mathrm{A}]_{\mathrm{t}}}{\mathrm{t}_{1 / 2}}\)
- C \(\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{k}}\)
- D \(\mathrm{t}_{1 / 2}=\frac{[\mathrm{A}]_{\mathrm{t}}-[\mathrm{A}]_0}{\mathrm{k}}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{k}}\)
Step-by-step Solution
Detailed explanation
For first order reaction,
\(\mathrm{t}=\frac{2.303}{\mathrm{k}} \cdot \log \frac{\mathrm{a}_0}{\mathrm{a}_0-\mathrm{x}}\)
At half-life, \(x=a_0 / 2\)
\(\begin{aligned}
& \mathrm{t}=\frac{2.303}{\mathrm{k}} \cdot \log \frac{\mathrm{a}_0}{\mathrm{a}_0-\frac{\mathrm{a}_0}{2}} \\
& \mathrm{t}_{1 / 2}=\frac{2.303}{\mathrm{k}} \cdot \log 2=\frac{2.303 \times 0.3010}{\mathrm{k}} \\
& \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{k}}
\end{aligned}\)
\(\mathrm{t}=\frac{2.303}{\mathrm{k}} \cdot \log \frac{\mathrm{a}_0}{\mathrm{a}_0-\mathrm{x}}\)
At half-life, \(x=a_0 / 2\)
\(\begin{aligned}
& \mathrm{t}=\frac{2.303}{\mathrm{k}} \cdot \log \frac{\mathrm{a}_0}{\mathrm{a}_0-\frac{\mathrm{a}_0}{2}} \\
& \mathrm{t}_{1 / 2}=\frac{2.303}{\mathrm{k}} \cdot \log 2=\frac{2.303 \times 0.3010}{\mathrm{k}} \\
& \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{k}}
\end{aligned}\)
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