For the equlibrium
`NH_(2)CONH_(4)(s) rarr 2NH_(3)(g) + CO_(2)(g)`
`p_(CO_(2)) = 1 ` atm at `100^(@)C`. What will be the equilibrium constant at `100^(@)C` in atm.

Answer

To solve the equilibrium constant for the reaction

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

given that the partial pressure of \( \text{CO}_2 \) is 1 atm at \( 100^\circ C \), we can follow these steps:

### Step 1: Write the expression for the equilibrium constant (K)
The equilibrium constant \( K \) for the reaction can be expressed in terms of the partial pressures of the gaseous products. Since \( \text{NH}_2\text{CONH}_4 \) is a solid, it does not appear in the equilibrium expression. Therefore, the equilibrium constant \( K \) is given by:

\[ K = \frac{P_{\text{NH}_3}^2 \cdot P_{\text{CO}_2}}{1} \]

### Step 2: Identify the known values
From the problem statement, we know:
- \( P_{\text{CO}_2} = 1 \) atm
- We need to find \( P_{\text{NH}_3} \).

### Step 3: Relate the pressures of the products
For every mole of \( \text{NH}_2\text{CONH}_4 \) that dissociates, it produces 2 moles of \( \text{NH}_3 \) and 1 mole of \( \text{CO}_2 \). If we assume \( x \) moles of \( \text{NH}_2\text{CONH}_4 \) dissociate, then:
- \( P_{\text{CO}_2} = x \)
- \( P_{\text{NH}_3} = 2x \)

Given \( P_{\text{CO}_2} = 1 \) atm, we have:

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

Thus, the pressure of \( \text{NH}_3 \) is:

\[ P_{\text{NH}_3} = 2x = 2 \times 1 = 2 \text{ atm} \]

### Step 4: Substitute the values into the equilibrium constant expression
Now, substituting the values into the equilibrium constant expression:

\[ K = \frac{(2)^2 \cdot (1)}{1} = \frac{4 \cdot 1}{1} = 4 \]

### Conclusion
The equilibrium constant \( K \) at \( 100^\circ C \) is:

\[ K = 4 \]