For the decomposition reaction
`NH_2COONH_4(s) hArr 2NH_3(g) +CO_2(g)`
the `K_p =2.9 xx 10^(-5) "atm"^3`. The total pressure of gases at equilibrium when 1 mole of `NH_2COONH_4` (s) was taken to start with would be

Options

A. 0.0194 atm
B. 0.0388 atm
C. 0.0582 atm
D. 0.0766 atm

Answer

To solve the problem for the decomposition reaction

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

with \( K_p = 2.9 \times 10^{-5} \, \text{atm}^3 \), we need to find the total pressure of gases at equilibrium when 1 mole of \( \text{NH}_2\text{COONH}_4 \) is taken to start with.

### Step 1: Set up the initial conditions
Initially, we have 1 mole of solid \( \text{NH}_2\text{COONH}_4 \) and no gases. Therefore, the initial amounts are:
- \( [\text{NH}_2\text{COONH}_4] = 1 \, \text{mol} \)
- \( [\text{NH}_3] = 0 \, \text{mol} \)
- \( [\text{CO}_2] = 0 \, \text{mol} \)

### Step 2: Define the change at equilibrium
Let \( x \) be the amount of \( \text{NH}_2\text{COONH}_4 \) that decomposes at equilibrium. According to the stoichiometry of the reaction:
- \( \text{NH}_2\text{COONH}_4 \) produces 2 moles of \( \text{NH}_3 \) and 1 mole of \( \text{CO}_2 \).
- At equilibrium:
- Moles of \( \text{NH}_2\text{COONH}_4 = 1 - x \)
- Moles of \( \text{NH}_3 = 2x \)
- Moles of \( \text{CO}_2 = x \)

### Step 3: Calculate total moles at equilibrium
The total moles of gas at equilibrium is:
\[ \text{Total moles} = 2x + x = 3x \]

### Step 4: Express \( K_p \) in terms of \( P \)
The expression for \( K_p \) is given by:
\[ K_p = \frac{(P_{\text{NH}_3})^2 (P_{\text{CO}_2})}{1} \]
Where \( P_{\text{NH}_3} \) and \( P_{\text{CO}_2} \) are the partial pressures of ammonia and carbon dioxide, respectively.

Using the ideal gas law, we can express the partial pressures in terms of total pressure \( P \):
- \( P_{\text{NH}_3} = \frac{2x}{3x} P = \frac{2}{3} P \)
- \( P_{\text{CO}_2} = \frac{x}{3x} P = \frac{1}{3} P \)

### Step 5: Substitute into the \( K_p \) expression
Substituting these into the \( K_p \) expression:
\[ K_p = \left(\frac{2}{3} P\right)^2 \left(\frac{1}{3} P\right) = \frac{4}{9} P^2 \cdot \frac{1}{3} P = \frac{4}{27} P^3 \]

### Step 6: Set up the equation with the given \( K_p \)
Now, we can set this equal to the given \( K_p \):
\[ \frac{4}{27} P^3 = 2.9 \times 10^{-5} \]

### Step 7: Solve for \( P^3 \)
Rearranging gives:
\[ P^3 = \frac{2.9 \times 10^{-5} \times 27}{4} \]
Calculating the right side:
\[ P^3 = \frac{2.9 \times 27}{4} \times 10^{-5} = \frac{78.3}{4} \times 10^{-5} = 19.575 \times 10^{-5} = 1.9575 \times 10^{-4} \]

### Step 8: Calculate \( P \)
Taking the cube root:
\[ P = (1.9575 \times 10^{-4})^{1/3} \approx 0.058 \, \text{atm} \]

### Final Answer
The total pressure of gases at equilibrium is approximately \( 0.058 \, \text{atm} \).

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