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 will follow these steps:

### Step 1: Write the balanced chemical equation
The dissociation of solid ammonium carbamate can be represented as:
\[ \text{NH}_2\text{COONH}_4(s) \rightleftharpoons 2\text{NH}_3(g) + \text{CO}(g) \]

### Step 2: Identify the total pressure at equilibrium
The total pressure of the gases at equilibrium is given as:
\[ P_t = 5 \text{ atm} \]

### Step 3: Determine the mole fractions of the gases
From the balanced equation, we see that:
- For every 1 mole of ammonium carbamate that dissociates, 2 moles of ammonia (\( \text{NH}_3 \)) and 1 mole of carbon monoxide (\( \text{CO} \)) are produced.
- Therefore, at equilibrium, the total number of moles of gas is:
\[ n_{\text{total}} = 2 + 1 = 3 \text{ moles} \]

### Step 4: Calculate the mole fractions
- The mole fraction of ammonia (\( \chi_{\text{NH}_3} \)):
\[ \chi_{\text{NH}_3} = \frac{2}{3} \]
- The mole fraction of carbon monoxide (\( \chi_{\text{CO}} \)):
\[ \chi_{\text{CO}} = \frac{1}{3} \]

### Step 5: Calculate the partial pressures using Dalton's Law
Using Dalton's Law, the partial pressures can be calculated as follows:
- Partial pressure of ammonia (\( P_{\text{NH}_3} \)):
\[ P_{\text{NH}_3} = \chi_{\text{NH}_3} \times P_t = \frac{2}{3} \times 5 \text{ atm} = \frac{10}{3} \text{ atm} \]
- Partial pressure of carbon monoxide (\( P_{\text{CO}} \)):
\[ P_{\text{CO}} = \chi_{\text{CO}} \times P_t = \frac{1}{3} \times 5 \text{ atm} = \frac{5}{3} \text{ atm} \]

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{(P_{\text{NH}_3})^2 \times (P_{\text{CO}})}{1} = (P_{\text{NH}_3})^2 \times P_{\text{CO}} \]

### Step 7: Substitute the values into the \( K_p \) expression
Substituting the calculated partial pressures:
\[ K_p = \left(\frac{10}{3}\right)^2 \times \left(\frac{5}{3}\right) \]
Calculating this:
\[ K_p = \frac{100}{9} \times \frac{5}{3} = \frac{500}{27} \]

### Step 8: Calculate the numerical value of \( K_p \)
Calculating \( K_p \):
\[ K_p \approx 18.52 \]

### Final Answer
Thus, the equilibrium constant \( K_p \) is approximately:
\[ K_p \approx 18.52 \]