Question:
Half-life of a first order reaction is \(10\) minutes. What is the rate of reaction after \(20\) minutes, if the initial concentration is \(10^{-2}\ \mathrm{M}\)?
Half-life of a first order reaction is \(10\) minutes. What is the rate of reaction after \(20\) minutes, if the initial concentration is \(10^{-2}\ \mathrm{M}\)?
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For a first-order reaction:
\[
\boxed{t_{1/2}=\frac{0.693}{k}}
\]
After each half-life, the concentration becomes half of its previous value.
Updated On: Jul 9, 2026
- \(1.73\times10^{-4}\ \mathrm{M\,min^{-1}}\)
- \(1.73\times10^{-2}\ \mathrm{M\,min^{-1}}\)
- \(3.46\times10^{-4}\ \mathrm{M\,min^{-1}}\)
- \(4.19\times10^{-5}\ \mathrm{M\,min^{-1}}\) \bigskip
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The Correct Option is A
Solution and Explanation
Concept:
For a first-order reaction,
\[
t_{1/2}=\frac{0.693}{k}
\]
and
\[
\text{Rate}=k[A]
\]
Step 1: Calculate the rate constant. \[ k=\frac{0.693}{10} =0.0693\ \mathrm{min^{-1}} \]
Step 2: Find the concentration after \(20\) minutes. Since \(20\) minutes corresponds to two half-lives, \[ [A] = 10^{-2}\times\left(\frac12\right)^2 = 2.5\times10^{-3}\ \mathrm{M} \]
Step 3: Calculate the reaction rate. \[ \text{Rate} = k[A] = 0.0693\times2.5\times10^{-3} = 1.73\times10^{-4}\ \mathrm{M\,min^{-1}} \]
Step 4: Final conclusion. \[ \boxed{1.73\times10^{-4}\ \mathrm{M\,min^{-1}}} \] Hence, the correct option is \(\boxed{(A)}\).
Step 1: Calculate the rate constant. \[ k=\frac{0.693}{10} =0.0693\ \mathrm{min^{-1}} \]
Step 2: Find the concentration after \(20\) minutes. Since \(20\) minutes corresponds to two half-lives, \[ [A] = 10^{-2}\times\left(\frac12\right)^2 = 2.5\times10^{-3}\ \mathrm{M} \]
Step 3: Calculate the reaction rate. \[ \text{Rate} = k[A] = 0.0693\times2.5\times10^{-3} = 1.73\times10^{-4}\ \mathrm{M\,min^{-1}} \]
Step 4: Final conclusion. \[ \boxed{1.73\times10^{-4}\ \mathrm{M\,min^{-1}}} \] Hence, the correct option is \(\boxed{(A)}\).
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