Question

In a first order reaction at a given temperature the time required to complete $99\%$ of the reaction ($T_1$) is related to time required for $90\%$ completion ($T_2$) as

• A. $T_1 = T_2$
• B. $T_1 = 4T_2$
• C. $T_1 = 3T_2$
• D. $T_1 = 2T_2$

Hint:

For first-order reactions, the time taken is proportional to the number of $10$-fold reductions. $90\%$ completion is a $1$-log reduction, $99\%$ is a $2$-log reduction. Hence, $T_{99\%} = 2 \times T_{90\%}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For a first-order reaction, the time required for a specific percentage of completion is independent of the initial concentration.
Step 2: Key Formula or Approach:
Integrated rate law: \( t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]}\right) \).
Step 3: Detailed Explanation:
For 99% completion (\( T_1 \)):
Amount left \( [A] = 100 - 99 = 1% \).
\( T_1 = \frac{2.303}{k} \log\left(\frac{100}{1}\right) = \frac{2.303}{k} \times 2 \).
For 90% completion (\( T_2 \)):
Amount left \( [A] = 100 - 90 = 10% \).
\( T_2 = \frac{2.303}{k} \log\left(\frac{100}{10}\right) = \frac{2.303}{k} \times 1 \).
Dividing the two:
\( \frac{T_1}{T_2} = \frac{2}{1} \implies T_1 = 2T_2 \).
Step 4: Final Answer:
The relation is T\textsubscript{1} = 2T\textsubscript{2}.