Question:
The rate constant for a first order reaction is \(60\text{ s}^{-1}\). How much time will it take to reduce the concentration of the reactant to \(1/20^{\text{th}}\) of its initial value?
The rate constant for a first order reaction is \(60\text{ s}^{-1}\). How much time will it take to reduce the concentration of the reactant to \(1/20^{\text{th}}\) of its initial value?
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Use $t=\frac{2.303}{k}\log\frac{[A]_0}{[A]}$ with $[A]_0/[A]=20$.
Updated On: Oct 1, 2026
- \(0.0529\) s
- \(0.0852\) s
- \(0.0499\) s
- \(0.0357\) s
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Write the integrated law
\[ t=\frac{2.303}{k}\log\frac{[A]_0}{[A]} \]
Step 2: Substitute
The concentration falls to \(\frac{1}{20}\), so the ratio is \(20\) and \(\log20=1.301\).
\[ t=\frac{2.303\times1.301}{60}=\frac{2.996}{60}=0.0499\text{ s} \]
Step 3: Result
Option (C), \(0.0499\) s.
Final Answer:
The time is \(0.0499\) s, option (C). \[ \boxed{\text{(C)}} \]
\[ t=\frac{2.303}{k}\log\frac{[A]_0}{[A]} \]
Step 2: Substitute
The concentration falls to \(\frac{1}{20}\), so the ratio is \(20\) and \(\log20=1.301\).
\[ t=\frac{2.303\times1.301}{60}=\frac{2.996}{60}=0.0499\text{ s} \]
Step 3: Result
Option (C), \(0.0499\) s.
Final Answer:
The time is \(0.0499\) s, option (C). \[ \boxed{\text{(C)}} \]
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