The value of `t_(0.875)/(t_(0.50)` for `n^(th)` order reaction is

Options

A. `2^((2n-2))`
B. `2^((2n-2)-1)`
C. `(8^(n-1)-1)/(2^(n-1)-1)`
D. None of these

Answer

To solve the question regarding the value of \( \frac{t_{0.875}}{t_{0.50}} \) for an \( n^{th} \) order reaction, we will follow these steps:

### Step 1: Understand the Integrated Rate Law for \( n^{th} \) Order Reactions

The integrated rate law for an \( n^{th} \) order reaction can be expressed as:

\[ \frac{1}{C} - \frac{1}{C_0} = kt^{(n-1)} \]

Where:
- \( C \) is the concentration at time \( t \)
- \( C_0 \) is the initial concentration
- \( k \) is the rate constant
- \( t \) is the time

### Step 2: Set Up the Equations for \( t_{0.875} \) and \( t_{0.50} \)

For \( t_{0.875} \) (which corresponds to 87.5% completion), we can express the concentration as:

\[ C_{0.875} = C_0 \times (1 - 0.875) = C_0 \times 0.125 \]

Substituting into the integrated rate law:

\[ \frac{1}{0.125 C_0} - \frac{1}{C_0} = k t_{0.875}^{(n-1)} \]

This simplifies to:

\[ \frac{1}{0.125} - 1 = k t_{0.875}^{(n-1)} \]

Calculating \( \frac{1}{0.125} \):

\[ 8 - 1 = k t_{0.875}^{(n-1)} \implies 7 = k t_{0.875}^{(n-1)} \]

### Step 3: Repeat for \( t_{0.50} \)

For \( t_{0.50} \) (which corresponds to 50% completion):

\[ C_{0.50} = C_0 \times (1 - 0.50) = C_0 \times 0.50 \]

Using the integrated rate law again:

\[ \frac{1}{0.50 C_0} - \frac{1}{C_0} = k t_{0.50}^{(n-1)} \]

This simplifies to:

\[ \frac{1}{0.50} - 1 = k t_{0.50}^{(n-1)} \]

Calculating \( \frac{1}{0.50} \):

\[ 2 - 1 = k t_{0.50}^{(n-1)} \implies 1 = k t_{0.50}^{(n-1)} \]

### Step 4: Formulate the Ratio \( \frac{t_{0.875}}{t_{0.50}} \)

Now we have two equations:

1. \( 7 = k t_{0.875}^{(n-1)} \)
2. \( 1 = k t_{0.50}^{(n-1)} \)

To find the ratio \( \frac{t_{0.875}}{t_{0.50}} \):

\[ \frac{t_{0.875}^{(n-1)}}{t_{0.50}^{(n-1)}} = \frac{7}{1} \]

Taking the \( (n-1)^{th} \) root on both sides:

\[ \frac{t_{0.875}}{t_{0.50}} = \left(7\right)^{\frac{1}{n-1}} \]

### Conclusion

Thus, the value of \( \frac{t_{0.875}}{t_{0.50}} \) for an \( n^{th} \) order reaction is:

\[ \frac{t_{0.875}}{t_{0.50}} = 7^{\frac{1}{n-1}} \]