The value of `K_(p)` for the reaction, `A(g) +2B(g) hArr C(g) ` is `25atm^(-2)` at a certain temperature. The value of `K_(p)` for the reaction , `(1)/(2) C(g) hArr (1)/(2) A(g)+B(g)` at the same temperature would be `:`

Options

A. `25 atm^(-1)`
B. `(1)/(25) atm^(-1)`
C. `(1)/(5) atm`
D. `5 atm`

Answer

To find the value of \( K_p \) for the reaction

\[ \frac{1}{2} C(g) \rightleftharpoons \frac{1}{2} A(g) + B(g) \]

given that the \( K_p \) for the reaction

\[ A(g) + 2B(g) \rightleftharpoons C(g) \]

is \( 25 \, \text{atm}^{-2} \), we can follow these steps:

### Step 1: Write down the given equilibrium constant
The equilibrium constant \( K_p \) for the reaction

\[ A(g) + 2B(g) \rightleftharpoons C(g) \]

is given as

\[ K_p = 25 \, \text{atm}^{-2} \]

### Step 2: Reverse the reaction
When we reverse the reaction, we get:

\[ C(g) \rightleftharpoons A(g) + 2B(g) \]

According to the rules of equilibrium constants, when a reaction is reversed, the new equilibrium constant \( K_p' \) is the reciprocal of the original:

\[ K_p' = \frac{1}{K_p} = \frac{1}{25 \, \text{atm}^{-2}} = 0.04 \, \text{atm}^{2} \]

### Step 3: Adjust the stoichiometry
Now, we need to manipulate the reaction to match the desired equation:

\[ \frac{1}{2} C(g) \rightleftharpoons \frac{1}{2} A(g) + B(g) \]

To do this, we multiply the entire reversed reaction by \( \frac{1}{2} \):

\[ \frac{1}{2} C(g) \rightleftharpoons \frac{1}{2} A(g) + B(g) \]

### Step 4: Calculate the new equilibrium constant
When we multiply the reaction by a factor \( n \), the new equilibrium constant \( K_p'' \) is given by:

\[ K_p'' = (K_p')^n \]

In this case, \( n = \frac{1}{2} \):

\[ K_p'' = (0.04 \, \text{atm}^{2})^{\frac{1}{2}} = \sqrt{0.04} = 0.2 \, \text{atm} \]

### Final Answer
Thus, the value of \( K_p \) for the reaction

\[ \frac{1}{2} C(g) \rightleftharpoons \frac{1}{2} A(g) + B(g) \]

is

\[ 0.2 \, \text{atm} \]