Solid ammonium carbamate `NH_(2)COONH_(4)` was taken in excess in closed container according to the following reaction
`NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g)`. If equilibrium pressure is `4` atm, it's equilibrium constnat `K_(P)` is ?

Answer

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

\[ NH_2COONH_4(s) \rightleftharpoons 2NH_3(g) + CO_2(g) \]

given that the equilibrium pressure is \( 4 \) atm, we can follow these steps:

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

The equilibrium constant \( K_p \) for the reaction is given by the formula:

\[ K_p = \frac{(P_{NH_3})^2 \cdot (P_{CO_2})}{1} \]

where \( P_{NH_3} \) is the partial pressure of ammonia and \( P_{CO_2} \) is the partial pressure of carbon dioxide. Since the solid does not appear in the expression, we only consider the gases.

### Step 2: Determine the partial pressures at equilibrium

Let’s denote the total pressure at equilibrium as \( P \). Given that the equilibrium pressure is \( 4 \) atm, we have:

\[ P = 4 \text{ atm} \]

### Step 3: Calculate the mole fractions of the gases

From the stoichiometry of the reaction, we know that for every mole of \( NH_2COONH_4 \) that decomposes, we produce \( 2 \) moles of \( NH_3 \) and \( 1 \) mole of \( CO_2 \). Therefore, if \( \alpha \) moles of \( NH_2COONH_4 \) dissociate, we will have:

- Moles of \( NH_3 = 2\alpha \)
- Moles of \( CO_2 = \alpha \)

The total moles of gas at equilibrium will be:

\[ \text{Total moles} = 2\alpha + \alpha = 3\alpha \]

### Step 4: Calculate the partial pressures

The partial pressures can be expressed in terms of the total pressure:

- \( P_{NH_3} = \frac{2\alpha}{3\alpha} \cdot P = \frac{2}{3}P \)
- \( P_{CO_2} = \frac{\alpha}{3\alpha} \cdot P = \frac{1}{3}P \)

Substituting \( P = 4 \) atm:

- \( P_{NH_3} = \frac{2}{3} \cdot 4 = \frac{8}{3} \text{ atm} \)
- \( P_{CO_2} = \frac{1}{3} \cdot 4 = \frac{4}{3} \text{ atm} \)

### Step 5: Substitute into the \( K_p \) expression

Now we can substitute the partial pressures into the \( K_p \) expression:

\[ K_p = \left( \frac{8}{3} \right)^2 \cdot \left( \frac{4}{3} \right) \]

Calculating this gives:

\[ K_p = \frac{64}{9} \cdot \frac{4}{3} = \frac{256}{27} \]

### Step 6: Calculate the numerical value

Now we can compute the numerical value of \( K_p \):

\[ K_p = \frac{256}{27} \approx 9.48 \]

Thus, the equilibrium constant \( K_p \) is approximately \( 9.48 \).

### Final Answer

\[ K_p \approx 9.48 \]

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