Given that for a reaction of nth order, the integrated rate equation is:
`K=1/(t(n-1))[1/C^(n-1)-1/C_(0)^(n-1)]`, where `C` and `C_(0)` are the concentration of reactant at time `t` and initially respectively. The `t_(3//4)` and `t_(1//2)` are related as `t_(3//4)` is time required for C to become `C_(1//4`) :

Options

A. `t_((3)/(4))=t_((1)/(2))][2^(n-1)+1]`
B. `t_((3)/(4))=t_((1)/(2))[2^(n-1)-1]`
C. `t_((3)/(4))=t_((1)/(2)[2^(n+1)-1]`
D. `t_((3)/(4))=t_((1)/(2))[2^(n+1)+1]`

Answer

To solve the problem, we will derive the relationship between \( t_{3/4} \) and \( t_{1/2} \) using the integrated rate equation for an nth order reaction.

### Step-by-Step Solution:

1. **Understand the Integrated Rate Equation**:
The integrated rate equation for an nth order reaction is given by:
\[ k = \frac{1}{t^{(n-1)}} \left( \frac{1}{C^{(n-1)}} - \frac{1}{C_0^{(n-1)}} \right) \]

2. **Define Concentrations**:
- For \( t_{3/4} \): The concentration at this time is \( C = \frac{C_0}{4} \).
- For \( t_{1/2} \): The concentration at this time is \( C = \frac{C_0}{2} \).

3. **Substituting for \( t_{3/4} \)**:
Substitute \( C = \frac{C_0}{4} \) into the integrated rate equation:
\[ t_{3/4} = \frac{1}{k(n-1)} \left( \frac{1}{\left(\frac{C_0}{4}\right)^{(n-1)}} - \frac{1}{C_0^{(n-1)}} \right) \]
Simplifying this gives:
\[ t_{3/4} = \frac{1}{k(n-1)} \left( \frac{4^{(n-1)}}{C_0^{(n-1)}} - \frac{1}{C_0^{(n-1)}} \right) = \frac{1}{k(n-1)} \left( \frac{4^{(n-1)} - 1}{C_0^{(n-1)}} \right) \]

4. **Substituting for \( t_{1/2} \)**:
Substitute \( C = \frac{C_0}{2} \) into the integrated rate equation:
\[ t_{1/2} = \frac{1}{k(n-1)} \left( \frac{1}{\left(\frac{C_0}{2}\right)^{(n-1)}} - \frac{1}{C_0^{(n-1)}} \right) \]
Simplifying this gives:
\[ t_{1/2} = \frac{1}{k(n-1)} \left( \frac{2^{(n-1)}}{C_0^{(n-1)}} - \frac{1}{C_0^{(n-1)}} \right) = \frac{1}{k(n-1)} \left( \frac{2^{(n-1)} - 1}{C_0^{(n-1)}} \right) \]

5. **Finding the Ratio \( \frac{t_{3/4}}{t_{1/2}} \)**:
Now, we can find the ratio:
\[ \frac{t_{3/4}}{t_{1/2}} = \frac{\frac{1}{k(n-1)} \left( \frac{4^{(n-1)} - 1}{C_0^{(n-1)}} \right)}{\frac{1}{k(n-1)} \left( \frac{2^{(n-1)} - 1}{C_0^{(n-1)}} \right)} \]
The \( \frac{1}{k(n-1)} \) and \( \frac{1}{C_0^{(n-1)}} \) terms cancel out:
\[ \frac{t_{3/4}}{t_{1/2}} = \frac{4^{(n-1)} - 1}{2^{(n-1)} - 1} \]

6. **Simplifying the Expression**:
Notice that \( 4^{(n-1)} = (2^2)^{(n-1)} = 2^{2(n-1)} \):
\[ \frac{t_{3/4}}{t_{1/2}} = \frac{2^{2(n-1)} - 1}{2^{(n-1)} - 1} \]
This can be factored as:
\[ \frac{t_{3/4}}{t_{1/2}} = \frac{(2^{(n-1)} - 1)(2^{(n-1)} + 1)}{2^{(n-1)} - 1} = 2^{(n-1)} + 1 \]

7. **Final Result**:
Thus, we arrive at the relationship:
\[ t_{3/4} = t_{1/2} \cdot (2^{(n-1)} + 1) \]