A first order reaction is completed by 75% in time t, completed by 99.9% in time t'. Ratio `(t')/(t)` will be equal to

Explanation

To solve the problem, we need to find the ratio \( \frac{t'}{t} \) for a first-order reaction that is 75% completed in time \( t \) and 99.9% completed in time \( t' \).

### Step-by-Step Solution:

1. **Understand the First-Order Reaction Formula**:
The time taken for a first-order reaction can be expressed using the formula:
\[ t = \frac{2.303}{k} \log \left( \frac{A_0}{A} \right) \]
where:
- \( t \) = time taken for the reaction
- \( k \) = rate constant
- \( A_0 \) = initial concentration of the reactant
- \( A \) = concentration of the reactant at time \( t \)

2. **Set Up the Equations**:
- For 75% completion, 25% of the reactant remains. Thus, \( A = 0.25 A_0 \):
\[ t = \frac{2.303}{k} \log \left( \frac{A_0}{0.25 A_0} \right) = \frac{2.303}{k} \log \left( \frac{1}{0.25} \right) = \frac{2.303}{k} \log (4) \]
- For 99.9% completion, 0.1% of the reactant remains. Thus, \( A = 0.001 A_0 \):
\[ t' = \frac{2.303}{k} \log \left( \frac{A_0}{0.001 A_0} \right) = \frac{2.303}{k} \log \left( \frac{1}{0.001} \right) = \frac{2.303}{k} \log (1000) \]

3. **Calculate the Ratio \( \frac{t'}{t} \)**:
Now we can find the ratio of \( t' \) to \( t \):
\[ \frac{t'}{t} = \frac{\frac{2.303}{k} \log (1000)}{\frac{2.303}{k} \log (4)} = \frac{\log (1000)}{\log (4)} \]
We know that:
- \( \log (1000) = 3 \) (since \( 1000 = 10^3 \))
- \( \log (4) = 2 \log (2) \approx 2 \times 0.301 = 0.602 \)

Therefore:
\[ \frac{t'}{t} = \frac{3}{\log (4)} \approx \frac{3}{0.602} \approx 4.975 \approx 5 \]

### Final Result:
The ratio \( \frac{t'}{t} \) is approximately equal to 5.

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