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 \( K_p \) 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 original reaction and its \( K_p \)

The original reaction is:

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

with

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

### Step 2: Reverse the reaction

When we reverse a reaction, the equilibrium constant for the new reaction is the reciprocal of the original equilibrium constant. Thus, for the reversed reaction:

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

the new \( K_p \) is:

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

### Step 3: Adjust the reaction by multiplying by \( \frac{1}{2} \)

Now, we need to multiply the entire reversed reaction by \( \frac{1}{2} \):

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

When we multiply a reaction by a factor, we raise the equilibrium constant to the power of that factor. Therefore, we have:

\[ K_p'' = (K_p')^{\frac{1}{2}} = \left(\frac{1}{25}\right)^{\frac{1}{2}} = \frac{1}{\sqrt{25}} = \frac{1}{5} \, \text{atm} \]

### 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

\[ \frac{1}{5} \, \text{atm} \]

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