Question:
The half-life period of a first order reaction having rate constant $k = 0.231 \times 10^{-10}$ s$^{-1}$ will be}
The half-life period of a first order reaction having rate constant $k = 0.231 \times 10^{-10}$ s$^{-1}$ will be}
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For first-order reactions, half-life is independent of initial concentration.
Updated On: May 8, 2026
- $32 \times 10^{10}$ s
- $2 \times 10^{10}$ s
- $3 \times 10^{10}$ s
- $2 \times 10^{-10}$ s
- $3 \times 10^{-12}$ s
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The Correct Option is C
Solution and Explanation
Concept:
For a first-order reaction:
\[
t_{1/2} = \frac{0.693}{k}
\]
Step 1: Substitute value of $k$. \[ k = 0.231 \times 10^{-10} \text{ s}^{-1} \]
Step 2: Calculate half-life. \[ t_{1/2} = \frac{0.693}{0.231 \times 10^{-10}} \] \[ = \frac{0.693}{0.231} \times 10^{10} \] \[ = 3 \times 10^{10} \text{ s} \]
Step 3: Interpretation.
• Larger half-life → slower reaction
• First-order reactions have constant half-life Final Answer: \[ \boxed{3 \times 10^{10} \text{ s}} \]
Step 1: Substitute value of $k$. \[ k = 0.231 \times 10^{-10} \text{ s}^{-1} \]
Step 2: Calculate half-life. \[ t_{1/2} = \frac{0.693}{0.231 \times 10^{-10}} \] \[ = \frac{0.693}{0.231} \times 10^{10} \] \[ = 3 \times 10^{10} \text{ s} \]
Step 3: Interpretation.
• Larger half-life → slower reaction
• First-order reactions have constant half-life Final Answer: \[ \boxed{3 \times 10^{10} \text{ s}} \]
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