The rate constant ,k, of a second -order reaction ,`A to ` Products is given by
`k= 1/t {x/(a(a-x))}`
The ratio `t_(3//4)//t_(1//2)` is eual to

Options

A. `1//5`
B. `2//1`
C. `3//1`
D. `5//1`

Answer

To solve the problem, we need to find the ratio \( \frac{t_{3/4}}{t_{1/2}} \) for a second-order reaction \( A \rightarrow \text{Products} \) using the given formula for the rate constant \( k \):

\[ k = \frac{1}{t} \cdot \frac{x}{a(a-x)} \]

### Step 1: Understand the Variables
- \( a \): Initial concentration of reactant \( A \).
- \( x \): Amount of reactant \( A \) that has reacted.
- \( t \): Time taken for the reaction to reach a certain extent.

### Step 2: Define \( t_{3/4} \) and \( t_{1/2} \)
- \( t_{3/4} \): Time taken for \( 75\% \) of \( A \) to react, which means \( x = \frac{3}{4}a \) and \( a - x = \frac{1}{4}a \).
- \( t_{1/2} \): Time taken for \( 50\% \) of \( A \) to react, which means \( x = \frac{1}{2}a \) and \( a - x = \frac{1}{2}a \).

### Step 3: Calculate \( t_{3/4} \)
Using the formula for \( k \):

\[ k = \frac{1}{t_{3/4}} \cdot \frac{\frac{3}{4}a}{a\left(a - \frac{3}{4}a\right)} = \frac{1}{t_{3/4}} \cdot \frac{\frac{3}{4}a}{a \cdot \frac{1}{4}a} = \frac{1}{t_{3/4}} \cdot \frac{3}{4} \cdot \frac{4}{a} = \frac{3}{t_{3/4}a} \]

Rearranging gives:

\[ t_{3/4} = \frac{3}{k} \]

### Step 4: Calculate \( t_{1/2} \)
Using the same formula for \( k \):

\[ k = \frac{1}{t_{1/2}} \cdot \frac{\frac{1}{2}a}{a\left(a - \frac{1}{2}a\right)} = \frac{1}{t_{1/2}} \cdot \frac{\frac{1}{2}a}{a \cdot \frac{1}{2}a} = \frac{1}{t_{1/2}} \cdot \frac{1}{2} \cdot \frac{2}{a} = \frac{1}{t_{1/2}a} \]

Rearranging gives:

\[ t_{1/2} = \frac{1}{k} \]

### Step 5: Find the Ratio \( \frac{t_{3/4}}{t_{1/2}} \)
Now we can find the ratio:

\[ \frac{t_{3/4}}{t_{1/2}} = \frac{\frac{3}{k}}{\frac{1}{k}} = 3 \]

### Final Answer
Thus, the ratio \( \frac{t_{3/4}}{t_{1/2}} \) is equal to \( 3 \).

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