Question

For a first order reaction, time required for 99% completion is \(x\) times the time required for 90% completion. Find \(x\).

Hint:

For first order reactions, \(t_{99\%} = 2 \times t_{90\%}\) because \(\log 100 = 2 \log 10\). In general, \(t_{99.9\%} = 3 \times t_{90\%}\), etc.

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
For a first-order reaction, the rate depends only on the concentration of one reactant. The integrated rate law allows us to calculate the time required for any specific percentage of completion.
Step 2: Key Formula or Approach:
Integrated rate law for first order:
\[ t = \frac{2.303}{k} \log \left( \frac{a}{a - x} \right) \]
Where \(a\) is initial concentration and \(a-x\) is concentration at time \(t\).
: Detailed Explanation:
1. For \(90%\) completion (\(a-x = 10% \text{ of } a = 0.1a\)):
\[ t_{90%} = \frac{2.303}{k} \log \left( \frac{a}{0.1a} \right) = \frac{2.303}{k} \log(10) = \frac{2.303}{k} \dots (i) \]
2. For \(99%\) completion (\(a-x = 1% \text{ of } a = 0.01a\)):
\[ t_{99%} = \frac{2.303}{k} \log \left( \frac{a}{0.01a} \right) = \frac{2.303}{k} \log(100) = \frac{2.303}{k} \times 2 \dots (ii) \]
3. Finding the ratio \(x\):
\[ \frac{t_{99%}}{t_{90%}} = \frac{2 \times \frac{2.303}{k}}{\frac{2.303}{k}} = 2 \]
Thus, \(t_{99%} = 2 \times t_{90%}\).
Step 3: Final Answer:
The value of \(x\) is \(2\).