For the following equilibrium `:`
`NH_(2)CO_(2)NH_(4)(s)hArr 2NH_(3)(g)+CO_(2)(s)`
`K_(p)` is found to be 0.5 at 4400 K . Hence, partial pressure of `NH_(3)` and `CO_(2)` are respectively `:`

Options

A. 2.0,1.0 atm
B. 1.0,2.0 atm
C. 1.0,0.5 atm
D. 0.5,1.0 atm

Explanation

To find the partial pressures of \( NH_3 \) and \( CO_2 \) at equilibrium for the reaction:

\[ NH_2CO_2NH_4(s) \rightleftharpoons 2NH_3(g) + CO_2(g) \]

with \( K_p = 0.5 \) at \( 4400 \, K \), we can follow these steps:

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

The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the gaseous products:

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

Here, \( P_{NH_3} \) and \( P_{CO_2} \) are the partial pressures of ammonia and carbon dioxide, respectively. The solid \( NH_2CO_2NH_4 \) does not appear in the expression since the concentration of solids is constant.

### Step 2: Define variables for partial pressures

Let:
- \( P_{CO_2} = x \) (the partial pressure of \( CO_2 \))
- \( P_{NH_3} = 2x \) (since 2 moles of \( NH_3 \) are produced for every mole of \( CO_2 \))

### Step 3: Substitute into the \( K_p \) expression

Substituting the expressions for the partial pressures into the \( K_p \) equation gives:

\[ K_p = \frac{(2x)^2 \cdot x}{1} = 4x^3 \]

### Step 4: Set up the equation with the known \( K_p \)

Now we can set this equal to the given \( K_p \):

\[ 4x^3 = 0.5 \]

### Step 5: Solve for \( x \)

To find \( x \), we rearrange the equation:

\[ x^3 = \frac{0.5}{4} = 0.125 \]

Taking the cube root of both sides:

\[ x = \sqrt[3]{0.125} = 0.5 \]

### Step 6: Calculate the partial pressures

Now we can find the partial pressures:

- \( P_{CO_2} = x = 0.5 \, \text{atm} \)
- \( P_{NH_3} = 2x = 2 \times 0.5 = 1.0 \, \text{atm} \)

### Final Answer

The partial pressures of \( NH_3 \) and \( CO_2 \) are:

- \( P_{NH_3} = 1.0 \, \text{atm} \)
- \( P_{CO_2} = 0.5 \, \text{atm} \)