Question

Show that the time required for 99.9% completion in a first-order reaction is 10 times of half-life (t$_{1/2}$) of the reaction.

Hint:

For first-order reactions, the time for 99.9% completion is approximately 10 times the half-life (t$_{1/2}$).

Answer 1

The integrated rate law for a first-order reaction is: \[ \ln \left( \frac{[R_0]}{[R]} \right) = kt \] Where:
- \( [R_0] \) is the initial concentration,
- \( [R] \) is the concentration at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.


Step 1: For 99.9% completion, \( [R] = 0.1 \times [R_0] \), so: \[ \ln \left( \frac{[R_0]}{0.1 [R_0]} \right) = kt_{99.9%} \] \[ \ln 10 = kt_{99.9%} \] \[ t_{99.9%} = \frac{2.303}{k} \log 10 = \frac{2.303}{k} \]

Step 2: For half-life \( t_{1/2} \), \( [R] = 0.5 [R_0] \), so: \[ \ln \left( \frac{[R_0]}{0.5 [R_0]} \right) = kt_{1/2} \] \[ \ln 2 = kt_{1/2} \] \[ t_{1/2} = \frac{2.303}{k} \log 2 = \frac{2.303}{k} \times 0.3010 = \frac{2.303}{k} \times 0.3010 \]

Step 3: Now, dividing the two expressions: \[ \frac{t_{99.9%}}{t_{1/2}} = \frac{\frac{2.303}{k}}{\frac{2.303 \times 0.3010}{k}} = \frac{1}{0.3010} = 10 \] Final Answer: The time required for 99.9% completion is 10 times the half-life of the reaction.