For a first order reaction at `27^(@)C` the rato of time required for 75% completion of 25% completion of reaction is

Options

A. 3
B. 2.303
C. 4.8
D. 0.477

Answer

To solve the problem of finding the ratio of time required for 75% completion of a first-order reaction to the time required for 25% completion of the same reaction, we will follow these steps:

### Step 1: Understand the first-order reaction kinetics
For a first-order reaction, the rate constant \( k \) is related to the time taken for a certain percentage of the reaction to complete. The integrated rate law for a first-order reaction is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{A_0}{A} \right) \]
where:
- \( t \) = time taken for the reaction to reach a certain completion
- \( A_0 \) = initial concentration
- \( A \) = concentration at time \( t \)

### Step 2: Calculate time for 75% completion
For 75% completion:
- \( A_0 = 100 \) (initial concentration)
- \( A = 100 - 75 = 25 \) (remaining concentration)

Substituting into the formula:
\[ t_1 = \frac{2.303}{k} \log \left( \frac{100}{25} \right) \]
Calculating the logarithm:
\[ \log \left( \frac{100}{25} \right) = \log(4) \approx 0.602 \]
Thus,
\[ t_1 = \frac{2.303}{k} \cdot 0.602 \]

### Step 3: Calculate time for 25% completion
For 25% completion:
- \( A_0 = 100 \)
- \( A = 100 - 25 = 75 \)

Substituting into the formula:
\[ t_2 = \frac{2.303}{k} \log \left( \frac{100}{75} \right) \]
Calculating the logarithm:
\[ \log \left( \frac{100}{75} \right) = \log \left( \frac{4}{3} \right) \approx 0.125 \]
Thus,
\[ t_2 = \frac{2.303}{k} \cdot 0.125 \]

### Step 4: Find the ratio of \( t_1 \) to \( t_2 \)
Now, we can find the ratio of the times:
\[ \frac{t_1}{t_2} = \frac{\frac{2.303}{k} \cdot 0.602}{\frac{2.303}{k} \cdot 0.125} \]
The \( \frac{2.303}{k} \) terms cancel out:
\[ \frac{t_1}{t_2} = \frac{0.602}{0.125} \]
Calculating the ratio:
\[ \frac{t_1}{t_2} \approx 4.816 \approx 4.8 \]

### Conclusion
The ratio of the time required for 75% completion to the time required for 25% completion of the reaction is approximately **4.8**.