The time `t_1` and `t_2` are time period for `1/3`rd and `2/3`rd completion of two first order reactions If the half lives are in the ratio 10:3. then the ratio of `t_1` and `t_2` will be

Options

A. 1.23
B. 0.81
C. 0.11
D. 2.7

Answer

To solve the problem, we need to find the ratio of the time periods \( t_1 \) and \( t_2 \) for the completion of \( \frac{1}{3} \) and \( \frac{2}{3} \) of two first-order reactions, respectively, given that their half-lives are in the ratio of \( 10:3 \).

### Step-by-Step Solution:

1. **Understanding First-Order Reactions**:
For a first-order reaction, the relationship between the time \( t \), the rate constant \( k \), and the concentration is given by:
\[ t = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \]
where \( [A_0] \) is the initial concentration and \( [A] \) is the concentration at time \( t \).

2. **Expressing Rate Constant in Terms of Half-Life**:
The rate constant \( k \) for a first-order reaction can be expressed in terms of the half-life \( t_{1/2} \):
\[ k = \frac{0.693}{t_{1/2}} \]

3. **Finding \( t_1 \) for \( \frac{1}{3} \) Completion**:
For \( t_1 \) (where \( \frac{1}{3} \) of the reaction is completed):
- Initial concentration \( [A_0] = A \)
- Final concentration \( [A] = A - \frac{1}{3}A = \frac{2}{3}A \)
\[ t_1 = \frac{2.303}{k} \log \left( \frac{A}{\frac{2}{3}A} \right) = \frac{2.303}{k} \log \left( \frac{3}{2} \right) \]

4. **Finding \( t_2 \) for \( \frac{2}{3} \) Completion**:
For \( t_2 \) (where \( \frac{2}{3} \) of the reaction is completed):
- Final concentration \( [A] = A - \frac{2}{3}A = \frac{1}{3}A \)
\[ t_2 = \frac{2.303}{k} \log \left( \frac{A}{\frac{1}{3}A} \right) = \frac{2.303}{k} \log \left( 3 \right) \]

5. **Finding the Ratio \( \frac{t_1}{t_2} \)**:
Now, we can find the ratio of \( t_1 \) and \( t_2 \):
\[ \frac{t_1}{t_2} = \frac{\frac{2.303}{k} \log \left( \frac{3}{2} \right)}{\frac{2.303}{k} \log \left( 3 \right)} = \frac{\log \left( \frac{3}{2} \right)}{\log \left( 3 \right)} \]

6. **Using the Given Ratio of Half-Lives**:
The half-lives are given in the ratio \( 10:3 \). Let \( t_{1/2,1} = 10x \) and \( t_{1/2,2} = 3x \).
Therefore, the rate constants are:
\[ k_1 = \frac{0.693}{10x}, \quad k_2 = \frac{0.693}{3x} \]

7. **Substituting into the Ratio**:
Now substituting the values of \( k_1 \) and \( k_2 \) into the ratio:
\[ \frac{t_1}{t_2} = \frac{10}{3} \cdot \frac{\log \left( \frac{3}{2} \right)}{\log \left( 3 \right)} \]

8. **Calculating the Logs**:
Using approximate values:
- \( \log(3) \approx 0.477 \)
- \( \log(2) \approx 0.301 \)
Thus, \( \log \left( \frac{3}{2} \right) = \log(3) - \log(2) \approx 0.477 - 0.301 = 0.176 \).

9. **Final Calculation**:
\[ \frac{t_1}{t_2} = \frac{10}{3} \cdot \frac{0.176}{0.477} \approx \frac{10}{3} \cdot 0.369 \approx 1.23 \]

### Conclusion:
The ratio \( \frac{t_1}{t_2} \) is approximately \( 1.23 \).