Step 1: Understanding the Concept:
This problem applies the mathematical relationship between different percentage completion times in first-order kinetics.
We can solve this either by calculating the rate constant $k$ first or by using the ratio relationship derived in the previous question.
Step 2: Key Formula or Approach:
Method 1 (Ratio Method):
$t_{99%} = \frac{2.303}{k} \log \frac{100}{1} = \frac{2.303}{k} \times 2$
$t_{99.9%} = \frac{2.303}{k} \log \frac{100}{0.1} = \frac{2.303}{k} \times 3$
Therefore, $\frac{t_{99.9%}}{t_{99%}} = \frac{3}{2} = 1.5$.
Step 3: Detailed Explanation:
Given:
$t_{99%} = 32$ minutes.
We need to find $t_{99.9%}$.
Using the ratio relationship:
\[ t_{99.9%} = 1.5 \times t_{99%} \]
\[ t_{99.9%} = 1.5 \times 32 \]
\[ t_{99.9%} = \frac{3}{2} \times 32 = 3 \times 16 = 48 \text{ minutes.} \]
Alternative Calculation (finding $k$):
$32 = \frac{2.303}{k} \times 2 \implies \frac{2.303}{k} = 16$.
Now, $t_{99.9%} = 16 \times \log 1000 = 16 \times 3 = 48$ minutes.
Both methods yield the same result. This highlights the logarithmic nature of decay in first-order reactions: every factor of 10 reduction in concentration takes the same amount of time ($t_{90%}$).
Going from 100% to 1% is two factors of 10 (32 mins), so each factor of 10 (90% completion) takes 16 mins.
To reach 99.9% (remaining 0.1%) is three factors of 10, thus $16 \times 3 = 48$ mins.
Step 4: Final Answer:
The reaction will be 99.9% complete in 48 minutes.