Question:
\(A \rightarrow\) products, is a first order reaction. The time required to decompose \(A\) to half its initial amount is \(60\) minutes. The rate constant of the reaction \((in\ s^{-1})\) is
\(A \rightarrow\) products, is a first order reaction. The time required to decompose \(A\) to half its initial amount is \(60\) minutes. The rate constant of the reaction \((in\ s^{-1})\) is
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For first order reactions:
\[
t_{1/2}=\frac{0.693}{k}
\]
The half-life is independent of initial concentration.
Updated On: Jun 25, 2026
- \(1.05\times10^{-2}\)
- \(1.15\times10^{-2}\)
- \(1.25\times10^{-4}\)
- \(1.92\times10^{-4}\)
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The Correct Option is D
Solution and Explanation
Step 1: Use half-life formula for first order reaction.
For a first order reaction: \[ t_{1/2}=\frac{0.693}{k} \] where: \[ t_{1/2}=\text{half-life} \] and \[ k=\text{rate constant} \]
Step 2: Convert time into seconds.
Given: \[ t_{1/2}=60\ \text{minutes} \] \[ =60\times60 \] \[ =3600\ s \]
Step 3: Calculate rate constant.
Using: \[ k=\frac{0.693}{t_{1/2}} \] \[ k=\frac{0.693}{3600} \] \[ k=1.925\times10^{-4}\ s^{-1} \] \[ \approx1.92\times10^{-4}\ s^{-1} \]
Step 4: Final conclusion.
Hence, the rate constant is \[ \boxed{1.92\times10^{-4}\ s^{-1}} \]
For a first order reaction: \[ t_{1/2}=\frac{0.693}{k} \] where: \[ t_{1/2}=\text{half-life} \] and \[ k=\text{rate constant} \]
Step 2: Convert time into seconds.
Given: \[ t_{1/2}=60\ \text{minutes} \] \[ =60\times60 \] \[ =3600\ s \]
Step 3: Calculate rate constant.
Using: \[ k=\frac{0.693}{t_{1/2}} \] \[ k=\frac{0.693}{3600} \] \[ k=1.925\times10^{-4}\ s^{-1} \] \[ \approx1.92\times10^{-4}\ s^{-1} \]
Step 4: Final conclusion.
Hence, the rate constant is \[ \boxed{1.92\times10^{-4}\ s^{-1}} \]
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