For n th order reaction `(t_(1//2))/(t_(3//4))` depends on `(n ne1)` :

Options

A. initial concentration only
B. "n" only
C. initial concentration of "n" both
D. Some times "n" and some times initial concentration

Answer

To solve the problem of finding the ratio \( \frac{t_{1/2}}{t_{3/4}} \) for an nth order reaction, we can follow these steps:

### Step 1: Understand the nth Order Reaction
For an nth order reaction, the rate of reaction is given by:
\[ -\frac{dA}{dt} \propto A^n \]
This means that the rate of change of concentration \( A \) with respect to time \( t \) is proportional to the concentration raised to the power of \( n \).

### Step 2: Integrate the Rate Equation
Integrating the rate equation, we have:
\[ \int_{A_0}^{A_t} \frac{dA}{A^n} = -k \int_0^t dt \]
This leads to:
\[ \frac{A_0^{1-n} - A_t^{1-n}}{1-n} = kt \]
Rearranging gives:
\[ A_0^{1-n} - A_t^{1-n} = k(1-n)t \]

### Step 3: Calculate \( t_{1/2} \)
For half-life \( t_{1/2} \), we set \( A_t = \frac{A_0}{2} \):
\[ A_0^{1-n} - \left(\frac{A_0}{2}\right)^{1-n} = k(1-n)t_{1/2} \]
This simplifies to:
\[ A_0^{1-n} - \frac{A_0^{1-n}}{2^{1-n}} = k(1-n)t_{1/2} \]
Factoring out \( A_0^{1-n} \):
\[ A_0^{1-n} \left(1 - \frac{1}{2^{1-n}}\right) = k(1-n)t_{1/2} \]
Thus, we have:
\[ t_{1/2} = \frac{A_0^{1-n} \left(1 - \frac{1}{2^{1-n}}\right)}{k(1-n)} \]

### Step 4: Calculate \( t_{3/4} \)
For \( t_{3/4} \), we set \( A_t = \frac{A_0}{4} \):
\[ A_0^{1-n} - \left(\frac{A_0}{4}\right)^{1-n} = k(1-n)t_{3/4} \]
This simplifies to:
\[ A_0^{1-n} - \frac{A_0^{1-n}}{4^{1-n}} = k(1-n)t_{3/4} \]
Factoring out \( A_0^{1-n} \):
\[ A_0^{1-n} \left(1 - \frac{1}{4^{1-n}}\right) = k(1-n)t_{3/4} \]
Thus, we have:
\[ t_{3/4} = \frac{A_0^{1-n} \left(1 - \frac{1}{4^{1-n}}\right)}{k(1-n)} \]

### Step 5: Find the Ratio \( \frac{t_{1/2}}{t_{3/4}} \)
Now, we can find the ratio:
\[ \frac{t_{1/2}}{t_{3/4}} = \frac{\frac{A_0^{1-n} \left(1 - \frac{1}{2^{1-n}}\right)}{k(1-n)}}{\frac{A_0^{1-n} \left(1 - \frac{1}{4^{1-n}}\right)}{k(1-n)}} \]
This simplifies to:
\[ \frac{t_{1/2}}{t_{3/4}} = \frac{1 - \frac{1}{2^{1-n}}}{1 - \frac{1}{4^{1-n}}} \]

### Conclusion
Thus, the ratio \( \frac{t_{1/2}}{t_{3/4}} \) depends only on \( n \) and does not depend on the initial concentration \( A_0 \).

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