Question

Ratio of time required for 99% completion vs 90% completion of a first order reaction is:

• A. $\frac{99}{90}$
• B. 2
• C. $\frac{10}{9}$
• D. 1

Hint:

Key points:

• Time for 99\% completion is double that for 90\%
• Ratio is independent of rate constant $k$
• Comes from $\log 100 = 2 \times \log 10$
• Characteristic of first-order reactions only

Answer 1

The Correct Option is B

Step 1: Recall first-order kinetics
For a first-order reaction: \[ t = \frac{2.303}{k} \log \frac{a}{a-x} \] where: - $t$ = time - $k$ = rate constant - $a$ = initial concentration - $x$ = reacted concentration
Step 2: Calculate time for 90% completion
For 90% completion ($x = 0.9a$): \[ t_{90} = \frac{2.303}{k} \log \frac{a}{a-0.9a} = \frac{2.303}{k} \log 10 = \frac{2.303}{k} \]
Step 3: Calculate time for 99% completion
For 99% completion ($x = 0.99a$): \[ t_{99} = \frac{2.303}{k} \log \frac{a}{a-0.99a} = \frac{2.303}{k} \log 100 = \frac{2 \times 2.303}{k} \]
Step 4: Compute the ratio
\[ \frac{t_{99}}{t_{90}} = \frac{\frac{2 \times 2.303}{k}}{\frac{2.303}{k}} = 2 \]
Step 5: Verify with options
The ratio equals 2, corresponding to option (b).