For a second order reaction , `t_(75%) = x t _(50%)` find the value of x

Explanation

To solve the problem, we need to find the relationship between \( t_{75\%} \) and \( t_{50\%} \) for a second-order reaction.

### Step-by-Step Solution:

1. **Understand the Second-Order Reaction Kinetics**:
For a second-order reaction, the integrated rate law is given by:
\[ \frac{1}{[A]} - \frac{1}{[A_0]} = kt \]
where \([A]\) is the concentration at time \( t \), \([A_0]\) is the initial concentration, and \( k \) is the rate constant.

2. **Determine \( t_{50\%} \)**:
At \( t_{50\%} \), 50% of the reactant has reacted, meaning:
\[ [A] = \frac{1}{2}[A_0] \]
Substituting into the integrated rate law:
\[ \frac{1}{\frac{1}{2}[A_0]} - \frac{1}{[A_0]} = k t_{50\%} \]
Simplifying gives:
\[ \frac{2}{[A_0]} - \frac{1}{[A_0]} = k t_{50\%} \]
\[ \frac{1}{[A_0]} = k t_{50\%} \]
Therefore, we can express \( t_{50\%} \) as:
\[ t_{50\%} = \frac{1}{k[A_0]} \]

3. **Determine \( t_{75\%} \)**:
At \( t_{75\%} \), 75% of the reactant has reacted, meaning:
\[ [A] = \frac{1}{4}[A_0] \]
Substituting into the integrated rate law:
\[ \frac{1}{\frac{1}{4}[A_0]} - \frac{1}{[A_0]} = k t_{75\%} \]
Simplifying gives:
\[ \frac{4}{[A_0]} - \frac{1}{[A_0]} = k t_{75\%} \]
\[ \frac{3}{[A_0]} = k t_{75\%} \]
Therefore, we can express \( t_{75\%} \) as:
\[ t_{75\%} = \frac{3}{k[A_0]} \]

4. **Relate \( t_{75\%} \) and \( t_{50\%} \)**:
Now we can find the ratio \( \frac{t_{75\%}}{t_{50\%}} \):
\[ \frac{t_{75\%}}{t_{50\%}} = \frac{\frac{3}{k[A_0]}}{\frac{1}{k[A_0]}} = 3 \]
Thus, we have:
\[ t_{75\%} = 3 t_{50\%} \]

5. **Conclusion**:
From the relationship we derived, we find that:
\[ x = 3 \]

### Final Answer:
The value of \( x \) is \( 3 \).