Question

The rate constant for a first order reaction is 60 sec\(^{-1}\). How much time will it take for the reactant to reduce to 1/16th of its initial concentration?

Hint:

Use \( t=\frac{2.303}{k}\log\frac{[A]_0}{[A]} \) with ratio 16, or note that 1/16 is four half-lives.

Answer 1

Step 1 (Formula): For a first order reaction, \[ t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} \]
Step 2 (Given): \( k = 60\ s^{-1} \); the reactant reduces to \( 1/16 \) of its initial value, so \( \dfrac{[A]_0}{[A]} = 16 \).
Step 3 (Substitution): \[ t = \frac{2.303}{60}\log 16 \]
Step 4 (Arithmetic): \( \log 16 = 1.204 \), so \( t = \dfrac{2.303 \times 1.204}{60} = \dfrac{2.773}{60} \).
Step 5: \( t = 0.0462\ s \).
\[ \boxed{t \approx 0.046\ \text{s}} \]