`NH_(4)COONH_(2)(s)hArr2NH_(3)(g)+CO_(2)(g)` If equilibrium pressure is 3 atm for the above reaction, then `K_(p)` for the reaction is

Options

A. 4
B. 27
C. `4/27`
D. `1/27`

Answer

To find the equilibrium constant \( K_p \) for the reaction

\[ \text{NH}_4\text{COONH}_2(s) \rightleftharpoons 2\text{NH}_3(g) + \text{CO}_2(g) \]

given that the equilibrium pressure is 3 atm, we can follow these steps:

### Step 1: Identify the Reaction Components
The reaction shows that solid ammonium carbamate (\( \text{NH}_4\text{COONH}_2 \)) decomposes into two moles of ammonia gas (\( \text{NH}_3 \)) and one mole of carbon dioxide gas (\( \text{CO}_2 \)).

### Step 2: Write the Expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the gaseous products:

\[ K_p = \frac{(P_{\text{NH}_3})^2 \cdot (P_{\text{CO}_2})}{(P_{\text{NH}_4\text{COONH}_2})} \]

Since \( \text{NH}_4\text{COONH}_2 \) is a solid, it does not appear in the expression for \( K_p \).

### Step 3: Define the Partial Pressures
Let \( P \) be the partial pressure of \( \text{CO}_2 \). According to the stoichiometry of the reaction, the partial pressure of \( \text{NH}_3 \) will be \( 2P \) because there are two moles of \( \text{NH}_3 \) produced for every mole of \( \text{CO}_2 \).

### Step 4: Relate Total Pressure to Partial Pressures
The total pressure at equilibrium is given as 3 atm:

\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{CO}_2} = 2P + P = 3P \]

Since the total pressure is 3 atm, we can set up the equation:

\[ 3P = 3 \text{ atm} \]

From this, we find:

\[ P = 1 \text{ atm} \]

### Step 5: Substitute Partial Pressures into \( K_p \) Expression
Now we can substitute \( P \) into the expression for \( K_p \):

\[ P_{\text{NH}_3} = 2P = 2 \times 1 \text{ atm} = 2 \text{ atm} \]
\[ P_{\text{CO}_2} = P = 1 \text{ atm} \]

Substituting these values into the \( K_p \) expression:

\[ K_p = \frac{(2 \text{ atm})^2 \cdot (1 \text{ atm})}{1} = 4 \text{ atm}^2 \]

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

\[ K_p = 4 \text{ atm}^2 \]