Question

A first order reaction proceeds to 90% completion. What will be the approximate time taken for 90% completion in relation to \(t_{1/2}\) of the reaction?

• A. 5.02 times of \(t_{1/2}\)
• B. 4.54 times of \(t_{1/2}\)
• C. 5 times of \(t_{1/2}\)
• D. 3.32 times of \(t_{1/2}\)

Hint:

For first-order reactions, the time taken for a certain percentage of completion can be calculated using the relation \(\frac{\ln(\text{initial concentration})}{\ln(\text{final concentration})}\).

Answer 1

The Correct Option is D

For a first-order reaction, the integrated rate law is:
\[ \ln \left(\frac{[A_0]}{[A_t]}\right) = kt \] where:
\([A_0]\) is the initial concentration,
\([A_t]\) is the concentration at time \(t\),
\(k\) is the rate constant, and \(t\) is the time.
For 90% completion, \([A_t] = 0.10[A_0]\). Thus:
\[ \ln \left(\frac{[A_0]}{0.10[A_0]}\right) = kt \] Simplifying:
\[ \ln(10) = kt \] For the half-life \(t_{1/2}\) of a first-order reaction:
\[ t_{1/2} = \frac{\ln(B)}{k} \] Thus, the time for 90% completion in terms of \(t_{1/2}\) is:
\[ t = \frac{\ln(10)}{\ln(B)} \cdot t_{1/2} \] Approximating \(\ln(10) \approx 2.3026\) and \(\ln(B) \approx 0.6931\), we get:
\[ t \approx 3.32 \cdot t_{1/2} \]