Two first order freaction have half-lives in the ratio `8:1` Calculate the ratio of time intervals `t_(1):t_(2)` and `t_(1)` and `t_(2)` are the time period for `((1)/(4))^(th)` and `((3)/(4))^(th)` completion.

Options

A. `1:0.301`
B. `0.125:0.602`
C. `1:0.602`
D. none of these

Answer

To solve the problem, we need to calculate the time intervals \( t_1 \) and \( t_2 \) for the completion of a first-order reaction given their half-lives in the ratio \( 8:1 \). Here’s a step-by-step breakdown of the solution:

### 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.

Given that the half-lives of the two reactions are in the ratio \( 8:1 \), we can express this as:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \frac{8}{1} \]

### Step 2: Express the rate constants in terms of half-lives
From the half-life formula, we can write:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \frac{0.693/k_1}{0.693/k_2} = \frac{k_2}{k_1} \]
Thus, we have:
\[ \frac{k_2}{k_1} = \frac{8}{1} \implies k_2 = 8k_1 \]

### Step 3: Calculate \( t_1 \) for one-fourth completion
For \( t_1 \), which corresponds to one-fourth completion, we have:
- Initial concentration = \( C_0 \)
- Remaining concentration after one-fourth completion = \( \frac{3}{4}C_0 \)

Using the first-order kinetics equation:
\[ t = \frac{2.303}{k} \log \left( \frac{C_0}{C} \right) \]
For \( t_1 \):
\[ t_1 = \frac{2.303}{k_1} \log \left( \frac{C_0}{\frac{3}{4}C_0} \right) = \frac{2.303}{k_1} \log \left( \frac{4}{3} \right) \]

### Step 4: Calculate \( t_2 \) for three-fourth completion
For \( t_2 \), which corresponds to three-fourth completion, we have:
- Remaining concentration after three-fourth completion = \( \frac{1}{4}C_0 \)

Using the same first-order kinetics equation:
\[ t_2 = \frac{2.303}{k_2} \log \left( \frac{C_0}{\frac{1}{4}C_0} \right) = \frac{2.303}{k_2} \log \left( 4 \right) \]

### Step 5: Calculate the ratio \( \frac{t_1}{t_2} \)
Now we can find the ratio:
\[ \frac{t_1}{t_2} = \frac{\frac{2.303}{k_1} \log \left( \frac{4}{3} \right)}{\frac{2.303}{k_2} \log \left( 4 \right)} = \frac{k_2}{k_1} \cdot \frac{\log \left( \frac{4}{3} \right)}{\log \left( 4 \right)} \]
Substituting \( \frac{k_2}{k_1} = 8 \):
\[ \frac{t_1}{t_2} = 8 \cdot \frac{\log \left( \frac{4}{3} \right)}{\log \left( 4 \right)} \]

### Step 6: Final calculation
Using logarithmic values:
- \( \log(4) = 2 \log(2) \)
- \( \log\left(\frac{4}{3}\right) = \log(4) - \log(3) \)

Thus, we can compute the final ratio:
\[ \frac{t_1}{t_2} = 8 \cdot \frac{\log(4) - \log(3)}{\log(4)} \]

### Conclusion
By calculating the above expression, we will arrive at the final ratio of \( t_1 : t_2 \).