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 integrated rate equation is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.

Step 2: Detailed Explanation:
1. For $99\%$ completion ($T_1$):
Remaining concentration $[A]_t = 100 - 99 = 1\%$.
\[ k = \frac{2.303}{T_1} \log \frac{100}{1} = \frac{2.303}{T_1} \log(10^2) = \frac{2.303 \times 2}{T_1} \]
2. For $90\%$ completion ($T_2$):
Remaining concentration $[A]_t = 100 - 90 = 10\%$.
\[ k = \frac{2.303}{T_2} \log \frac{100}{10} = \frac{2.303}{T_2} \log(10) = \frac{2.303 \times 1}{T_2} \]
3. Since $k$ is constant for the same reaction at the same temperature:
\[ \frac{2.303 \times 2}{T_1} = \frac{2.303}{T_2} \]
\[ \frac{2}{T_1} = \frac{1}{T_2} \implies T_1 = 2T_2 \]

Step 3: Final Answer:
The relationship is $T_1 = 2T_2$.