The decomposition of solid ammonium carbamate, `(NH_(4))(NH_(2)CO_(2))`, to gaseous ammonia and carbon dioxide is an endothermic reaction.
`9NH_(4))(NH_(2)CO_(2)))(s)hArr2NH_(3)(g)+CO_(2)(g)`
(a) When solid `(NH_(4))(NH_(2)CO_(2))` is introduced into and evacuated flask at `25^(@)C`, the total pressure of gas at equilibrium is `0.3` atm. What is the value of `K_(p) "at" 25^(@)C`?

Explanation

To solve the problem step by step, we will analyze the equilibrium reaction and calculate the equilibrium constant \( K_p \).

### Step 1: Write the balanced chemical equation
The decomposition of solid ammonium carbamate is given by the equation:
\[ 9 \text{NH}_4\text{NH}_2\text{CO}_2 (s) \rightleftharpoons 2 \text{NH}_3 (g) + \text{CO}_2 (g) \]

### Step 2: Understand the equilibrium conditions
At equilibrium, the total pressure of the gases is given as \( P_{\text{total}} = 0.3 \) atm. We need to express the partial pressures of the gases involved.

### Step 3: Define variables for partial pressures
Let \( P \) be the partial pressure of \( \text{CO}_2 \) at equilibrium. Since there are 2 moles of \( \text{NH}_3 \) produced for every mole of \( \text{CO}_2 \), the partial pressure of \( \text{NH}_3 \) will be \( 2P \).

### Step 4: Set up the total pressure equation
The total pressure at equilibrium can be expressed as:
\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{CO}_2} \]
Substituting the expressions for the partial pressures:
\[ P_{\text{total}} = 2P + P = 3P \]
Given that \( P_{\text{total}} = 0.3 \) atm, we can write:
\[ 3P = 0.3 \]
From this, we can solve for \( P \):
\[ P = \frac{0.3}{3} = 0.1 \text{ atm} \]

### Step 5: Calculate the partial pressures
Now we can find the partial pressures of \( \text{NH}_3 \) and \( \text{CO}_2 \):
- The partial pressure of \( \text{NH}_3 \):
\[ P_{\text{NH}_3} = 2P = 2 \times 0.1 = 0.2 \text{ atm} \]
- The partial pressure of \( \text{CO}_2 \):
\[ P_{\text{CO}_2} = P = 0.1 \text{ atm} \]

### Step 6: Write the expression for \( K_p \)
The equilibrium constant \( K_p \) is defined as:
\[ K_p = \frac{(P_{\text{NH}_3})^2 \cdot (P_{\text{CO}_2})}{1} \]
Substituting the values we found:
\[ K_p = (0.2)^2 \cdot (0.1) \]

### Step 7: Calculate \( K_p \)
Now, we calculate \( K_p \):
\[ K_p = 0.04 \cdot 0.1 = 0.004 \text{ atm}^3 \]
or
\[ K_p = 4 \times 10^{-3} \text{ atm}^3 \]

### Final Answer
Thus, the value of \( K_p \) at \( 25^\circ C \) is:
\[ K_p = 4 \times 10^{-3} \text{ atm}^3 \]

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