Two first order reaction have half-life in the ratio `3:2`. Calculate the ratio of time intervals `t_(1):t_(2)`. The time `t_(1)` and `t_(2)` are the time period for `25%` and `75%` completion for the first and second reaction respectively:

Options

A. `0.311:1`
B. `0.420:1`
C. `0.273:1`
D. `0.119:1`

Explanation

To solve the problem, we need to calculate the time intervals \( t_1 \) and \( t_2 \) for two first-order reactions, given that their half-lives are in the ratio \( 3:2 \).

### Step 1: Understand the relationship between half-life and rate constant
For a first-order reaction, the half-life (\( t_{1/2} \)) is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant. Since the half-lives are in the ratio \( 3:2 \), we can express this as:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \frac{3}{2} \]
Let \( t_{1/2,1} = 3k \) and \( t_{1/2,2} = 2k \) for some constant \( k \).

### Step 2: Calculate \( t_1 \) for the first reaction
The time \( t_1 \) for 25% completion of the first reaction can be calculated using the formula:
\[ t = \frac{2.303}{k} \log \left( \frac{A}{A - x} \right) \]
where \( A \) is the initial concentration and \( x \) is the amount reacted. For 25% completion:
\[ A = 100\% \quad \text{and} \quad x = 25\% \]
Thus:
\[ t_1 = \frac{2.303}{k_1} \log \left( \frac{100}{100 - 25} \right) = \frac{2.303}{k_1} \log \left( \frac{100}{75} \right) \]

### Step 3: Calculate \( t_2 \) for the second reaction
Similarly, for the second reaction, the time \( t_2 \) for 75% completion is:
\[ t_2 = \frac{2.303}{k_2} \log \left( \frac{100}{100 - 75} \right) = \frac{2.303}{k_2} \log \left( \frac{100}{25} \right) \]

### Step 4: Relate the rate constants \( k_1 \) and \( k_2 \)
From the half-life relationship:
\[ k_1 = \frac{0.693}{t_{1/2,1}} = \frac{0.693}{3k} \quad \text{and} \quad k_2 = \frac{0.693}{t_{1/2,2}} = \frac{0.693}{2k} \]
Thus:
\[ \frac{k_1}{k_2} = \frac{2}{3} \]

### Step 5: Substitute \( k_1 \) and \( k_2 \) into \( t_1 \) and \( t_2 \)
Now substituting \( k_1 \) and \( k_2 \) into \( t_1 \) and \( t_2 \):
\[ t_1 = \frac{2.303}{\frac{0.693}{3k}} \log \left( \frac{100}{75} \right) = \frac{2.303 \cdot 3k}{0.693} \log \left( \frac{100}{75} \right) \]
\[ t_2 = \frac{2.303}{\frac{0.693}{2k}} \log \left( \frac{100}{25} \right) = \frac{2.303 \cdot 2k}{0.693} \log \left( \frac{100}{25} \right) \]

### Step 6: Calculate the ratio \( \frac{t_1}{t_2} \)
Now, we can find the ratio \( \frac{t_1}{t_2} \):
\[ \frac{t_1}{t_2} = \frac{3 \log \left( \frac{100}{75} \right)}{2 \log \left( \frac{100}{25} \right)} \]

### Step 7: Simplify the logarithmic terms
Calculating the logarithmic terms:
\[ \log \left( \frac{100}{75} \right) = \log \left( \frac{4}{3} \right) \quad \text{and} \quad \log \left( \frac{100}{25} \right) = \log(4) \]
Thus:
\[ \frac{t_1}{t_2} = \frac{3 \log \left( \frac{4}{3} \right)}{2 \log(4)} \]

### Step 8: Final Calculation
Using the values of logarithms:
\[ \log(4) \approx 0.602 \quad \text{and} \quad \log \left( \frac{4}{3} \right) \approx 0.125 \]
So:
\[ \frac{t_1}{t_2} = \frac{3 \cdot 0.125}{2 \cdot 0.602} \approx \frac{0.375}{1.204} \approx 0.311 \]

### Final Result
Thus, the ratio \( t_1 : t_2 \) is approximately \( 0.311 : 1 \).