Solid ammonium carbamate dissociated according to the given reaction
`NH_2COONH_4(s) hArr 2NH_3(g) +CO(g)`
Total pressure of the gases in equilibrium is 5 atm. Hence `K_p`.

Options

A. 18.5
B. 16.4
C. `1//5`
D. 12.5

Answer

To find the equilibrium constant \( K_p \) for the dissociation of solid ammonium carbamate, we start with the given reaction:

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

### Step 1: Understand the Reaction
In this reaction, solid ammonium carbamate dissociates into two moles of ammonia gas and one mole of carbon monoxide gas. The total pressure of the gases at equilibrium is given as 5 atm.

### Step 2: Set Up the Equilibrium Expression
The equilibrium constant \( K_p \) is defined in terms of the partial pressures of the gaseous products:

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

Since the pressure of solids does not appear in the expression, we can simplify this to:

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

### Step 3: Define Partial Pressures
Let’s denote the partial pressure of ammonia as \( P_{\text{NH}_3} = 2P \) (since there are 2 moles of ammonia) and the partial pressure of carbon monoxide as \( P_{\text{CO}} = P \).

### Step 4: Total Pressure Equation
The total pressure at equilibrium is the sum of the partial pressures of the gases:

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

Given that the total pressure is 5 atm, we can set up the equation:

\[ 3P = 5 \quad \Rightarrow \quad P = \frac{5}{3} \text{ atm} \]

### Step 5: Calculate Partial Pressures
Now we can find the partial pressures:

\[ P_{\text{NH}_3} = 2P = 2 \times \frac{5}{3} = \frac{10}{3} \text{ atm} \]
\[ P_{\text{CO}} = P = \frac{5}{3} \text{ atm} \]

### Step 6: Substitute into the \( K_p \) Expression
Now we can substitute these values into the expression for \( K_p \):

\[ K_p = \left(\frac{10}{3}\right)^2 \cdot \left(\frac{5}{3}\right) \]

Calculating this:

\[ K_p = \frac{100}{9} \cdot \frac{5}{3} = \frac{500}{27} \]

### Step 7: Final Calculation
Now we can compute the numerical value:

\[ K_p \approx 18.5 \]

### Conclusion
Thus, the value of \( K_p \) is approximately 18.5.

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