Question:
For a first-order reaction \( A \rightarrow B \), the rate constant is \( 0.1 \text{ s}^{-1} \). What is the time required for \( 50\% \) completion?
For a first-order reaction \( A \rightarrow B \), the rate constant is \( 0.1 \text{ s}^{-1} \). What is the time required for \( 50\% \) completion?
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For first-order reactions:
\[
t_{1/2}=\frac{0.693}{k}
\]
The half-life is independent of the initial concentration.
Updated On: May 29, 2026
- \( 6.93 \text{ s} \)
- \( 5 \text{ s} \)
- \( 10 \text{ s} \)
- \( 0.693 \text{ s} \)
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The Correct Option is A
Solution and Explanation
Concept:
For a first-order reaction, the time required for \(50\%\) completion is called the half-life:
\[
t_{1/2}=\frac{0.693}{k}
\]
where \(k\) is the rate constant.
Step 1: Write the half-life formula for first-order reactions.
\[ t_{1/2}=\frac{0.693}{k} \]
Step 2: Substitute the given rate constant.
Given: \[ k=0.1\text{ s}^{-1} \] Thus: \[ t_{1/2}=\frac{0.693}{0.1} \] \[ t_{1/2}=6.93\text{ s} \] Hence, the time required for \(50\%\) completion is: \[ \boxed{6.93\text{ s}} \]
Step 1: Write the half-life formula for first-order reactions.
\[ t_{1/2}=\frac{0.693}{k} \]
Step 2: Substitute the given rate constant.
Given: \[ k=0.1\text{ s}^{-1} \] Thus: \[ t_{1/2}=\frac{0.693}{0.1} \] \[ t_{1/2}=6.93\text{ s} \] Hence, the time required for \(50\%\) completion is: \[ \boxed{6.93\text{ s}} \]
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