For an ideal gas reaction
`2A+BhArrC+D`
the value of `K_(p)` will be:

Options

A. `K_(p)=(n_(C )n_(D))/(n_(A)^(2)n_(B)).V/(RT^(2))`
B. `K_(p)=(n_(C )n_(D))/(n_(A)^(2)n_(B)).V/(RT)`
C. `K_(p)=(n_(C )n_(D))/(n_(A)^(2)n_(B)).(RT)/V`
D. `K_(p)=(n_(C )n_(D))/(4n_(A)^(2)n_(B)).V/(RT)`

Explanation

To find the value of \( K_p \) for the reaction \( 2A + B \rightleftharpoons C + D \), we can follow these steps:

### Step 1: Write the expression for \( K_p \)

For a general reaction of the form \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K_p \) is given by:

\[ K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} \]

For our specific reaction \( 2A + B \rightleftharpoons C + D \), this becomes:

\[ K_p = \frac{(P_C)^1 (P_D)^1}{(P_A)^2 (P_B)^1} \]

### Step 2: Relate partial pressures to concentrations

Using the ideal gas law, \( PV = nRT \), we can express the partial pressures in terms of concentrations. The partial pressure \( P \) of a gas can be expressed as:

\[ P = \frac{n}{V}RT \]

where \( n \) is the number of moles, \( V \) is the volume, and \( R \) is the ideal gas constant. Therefore, we can write:

\[ P_A = \frac{n_A}{V}RT, \quad P_B = \frac{n_B}{V}RT, \quad P_C = \frac{n_C}{V}RT, \quad P_D = \frac{n_D}{V}RT \]

### Step 3: Substitute the expressions into the \( K_p \) formula

Substituting these expressions into the \( K_p \) formula gives:

\[ K_p = \frac{\left(\frac{n_C}{V}RT\right)^1 \left(\frac{n_D}{V}RT\right)^1}{\left(\frac{n_A}{V}RT\right)^2 \left(\frac{n_B}{V}RT\right)^1} \]

### Step 4: Simplify the expression

This simplifies to:

\[ K_p = \frac{(n_C)(n_D)(RT)^2}{(n_A)^2(n_B)(V^2)(RT)^2} \]

The \( (RT)^2 \) terms cancel out, leading to:

\[ K_p = \frac{n_C n_D}{n_A^2 n_B} \cdot \frac{1}{V^2} \]

### Step 5: Final expression for \( K_p \)

Thus, we can express \( K_p \) as:

\[ K_p = \frac{n_C n_D}{n_A^2 n_B} \cdot \frac{1}{V^2} \]

### Conclusion

The final expression for \( K_p \) for the reaction \( 2A + B \rightleftharpoons C + D \) is:

\[ K_p = \frac{n_C n_D}{n_A^2 n_B} \cdot \frac{1}{V^2} \]