The ratio of `t_(0.75)` and `t_(0.5)` for first order reaction :-

Options

A. `4:3`
B. `3:2`
C. `2:1`
D. `1:2`

Explanation

To find the ratio of \( t_{0.75} \) and \( t_{0.5} \) for a first-order reaction, we can follow these steps:

### Step 1: Understand the formula for first-order reaction time
For a first-order reaction, the time \( t \) can be expressed using the formula:
\[ t = \frac{2.303}{k} \log \left( \frac{A}{A - x} \right) \]
where:
- \( t \) = time taken for the reaction
- \( k \) = rate constant
- \( A \) = initial concentration of the reactant
- \( x \) = amount of reactant that has reacted

### Step 2: Calculate \( t_{0.75} \)
For \( t_{0.75} \), we need to find the time taken when 75% of the reactant has reacted. Thus, \( x = 0.75A \):
\[ t_{0.75} = \frac{2.303}{k} \log \left( \frac{A}{A - 0.75A} \right) = \frac{2.303}{k} \log \left( \frac{A}{0.25A} \right) = \frac{2.303}{k} \log(4) \]

### Step 3: Calculate \( t_{0.5} \)
For \( t_{0.5} \), we need to find the time taken when 50% of the reactant has reacted. Thus, \( x = 0.5A \):
\[ t_{0.5} = \frac{2.303}{k} \log \left( \frac{A}{A - 0.5A} \right) = \frac{2.303}{k} \log \left( \frac{A}{0.5A} \right) = \frac{2.303}{k} \log(2) \]

### Step 4: Find the ratio \( \frac{t_{0.75}}{t_{0.5}} \)
Now, we can find the ratio of \( t_{0.75} \) to \( t_{0.5} \):
\[ \frac{t_{0.75}}{t_{0.5}} = \frac{\frac{2.303}{k} \log(4)}{\frac{2.303}{k} \log(2)} \]
The \( \frac{2.303}{k} \) cancels out:
\[ \frac{t_{0.75}}{t_{0.5}} = \frac{\log(4)}{\log(2)} \]

### Step 5: Simplify the ratio
We know that \( \log(4) = \log(2^2) = 2 \log(2) \):
\[ \frac{t_{0.75}}{t_{0.5}} = \frac{2 \log(2)}{\log(2)} = 2 \]

### Final Answer
Thus, the ratio of \( t_{0.75} \) to \( t_{0.5} \) for a first-order reaction is:
\[ \frac{t_{0.75}}{t_{0.5}} = 2:1 \]