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

Options

A. `3.0`
B. 2.303
C. 4.8
D. 0.47

Explanation

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

### Step 1: Understand the first-order reaction formula
For a first-order reaction, the time required for a certain percentage of completion can be calculated using the formula:
\[ T = \frac{2.303}{k} \log \left( \frac{a}{a - x} \right) \]
where:
- \( T \) is the time taken for the reaction to reach completion,
- \( k \) is the rate constant,
- \( a \) is the initial concentration,
- \( x \) is the amount reacted.

### Step 2: Calculate time for 25% completion
For 25% completion, \( x = 0.25a \):
\[ T_{25\%} = \frac{2.303}{k} \log \left( \frac{a}{a - 0.25a} \right) \]
\[ T_{25\%} = \frac{2.303}{k} \log \left( \frac{a}{0.75a} \right) \]
\[ T_{25\%} = \frac{2.303}{k} \log \left( \frac{1}{0.75} \right) \]
Using the property of logarithms:
\[ \log \left( \frac{1}{0.75} \right) = -\log(0.75) \]
Thus,
\[ T_{25\%} = \frac{2.303}{k} (-\log(0.75)) \]

### Step 3: Calculate time for 75% completion
For 75% completion, \( x = 0.75a \):
\[ T_{75\%} = \frac{2.303}{k} \log \left( \frac{a}{a - 0.75a} \right) \]
\[ T_{75\%} = \frac{2.303}{k} \log \left( \frac{a}{0.25a} \right) \]
\[ T_{75\%} = \frac{2.303}{k} \log \left( \frac{1}{0.25} \right) \]
Using the property of logarithms:
\[ \log \left( \frac{1}{0.25} \right) = -\log(0.25) \]
Thus,
\[ T_{75\%} = \frac{2.303}{k} (-\log(0.25)) \]

### Step 4: Find the ratio of times
Now, we need to find the ratio of \( T_{75\%} \) to \( T_{25\%} \):
\[ \text{Ratio} = \frac{T_{75\%}}{T_{25\%}} = \frac{\frac{2.303}{k} (-\log(0.25))}{\frac{2.303}{k} (-\log(0.75))} \]
The \( \frac{2.303}{k} \) cancels out:
\[ \text{Ratio} = \frac{\log(0.25)}{\log(0.75)} \]

### Step 5: Calculate the logarithms
Using logarithm values:
- \( \log(0.25) = \log\left(\frac{1}{4}\right) = -2 \log(2) \)
- \( \log(0.75) = \log\left(\frac{3}{4}\right) = \log(3) - \log(4) = \log(3) - 2\log(2) \)

Thus, the ratio can be approximated numerically or left in this logarithmic form.

### Final Answer
The ratio of time required for 75% completion to 25% completion of the reaction is:
\[ \text{Ratio} = \frac{\log(0.25)}{\log(0.75)} \]