For a second order reaction `dx//dt =k(a-x)^(2)`. Its half life periof is

Options

A. `(1)/(a.k)`
B. `(0.693)/(k)`
C. `(a)/(k)`
D. `(0.693)/(ak)`

Answer

To find the half-life period for the second-order reaction given by the rate equation \( \frac{dx}{dt} = k(a - x)^2 \), we can follow these steps:

### Step 1: Understand the Reaction Order
The given reaction is a second-order reaction because the rate of change of concentration \( x \) is proportional to the square of the remaining concentration \( (a - x) \).

### Step 2: Write the Integrated Rate Law for a Second-Order Reaction
For a second-order reaction, the integrated rate law can be expressed as:
\[ kt = \frac{1}{(a - x)} - \frac{1}{a} \]
where \( a \) is the initial concentration and \( x \) is the concentration at time \( t \).

### Step 3: Determine the Half-Life Expression
At half-life \( t_{1/2} \), the concentration \( x \) will be half of the initial concentration \( a \):
\[ x = \frac{a}{2} \]
Substituting \( x = \frac{a}{2} \) into the integrated rate law gives:
\[ kt_{1/2} = \frac{1}{(a - \frac{a}{2})} - \frac{1}{a} \]

### Step 4: Simplify the Equation
Now, simplify the right-hand side:
\[ kt_{1/2} = \frac{1}{\frac{a}{2}} - \frac{1}{a} \]
\[ kt_{1/2} = \frac{2}{a} - \frac{1}{a} \]
\[ kt_{1/2} = \frac{2 - 1}{a} = \frac{1}{a} \]

### Step 5: Solve for Half-Life
Now, rearranging the equation to find \( t_{1/2} \):
\[ t_{1/2} = \frac{1}{k} \cdot a \]

### Final Result
Thus, the half-life period \( t_{1/2} \) for the second-order reaction is:
\[ t_{1/2} = \frac{1}{k} \cdot a \]