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?

• A. \(0.0529\) s
• B. \(0.0852\) s
• C. \(0.0499\) s
• D. \(0.0357\) s

Hint:

Use $t=\frac{2.303}{k}\log\frac{[A]_0}{[A]}$ with $[A]_0/[A]=20$.

Answer 1

The Correct Option is C

Step 1: Use the natural log form
$\ln\frac{[A]_0}{[A]}=kt$, so $t=\frac{\ln20}{60}$.
$\ln20=2.996$, so $t=0.0499$ s.

Final Answer:
Option (C). \[ \boxed{\text{(C)}} \]