Solid ammonium carbomate, `NH_4CO_2NH_2(s),` dissociates into ammonia and carbon dioxide when it evaporates as shown by
`NH_4CO_2NH_2(s)hArr2NH_3(g)+CO_2(g)`
At `25^@C`, the total pressure of the gases in equilibrium with the solid is 0.116 atm.If 0.1 atm of `CO_2` is introduced after equilibrium is reached then :

Options

A. Final pressure of `CO_2` will be less than 0.1 atm
B. Final pressure of `CO_2` will be more than 0.1 atm
C. Pressure of `NH_3` will decrease due to addition of `CO_2`
D. Pressure of `NH_3` will increase due to addition of `CO_2`

Explanation

To solve the problem, we need to analyze the dissociation of solid ammonium carbonate and the effect of adding carbon dioxide on the equilibrium of the reaction.

### Step-by-Step Solution:

1. **Write the Equilibrium Reaction:**
The dissociation of solid ammonium carbonate can be represented as:
\[ \text{NH}_4\text{CO}_2\text{NH}_2(s) \rightleftharpoons 2\text{NH}_3(g) + \text{CO}_2(g) \]

2. **Initial Conditions:**
At 25°C, the total pressure of the gases in equilibrium with the solid is given as 0.116 atm. This pressure is the sum of the partial pressures of ammonia (\(P_{\text{NH}_3}\)) and carbon dioxide (\(P_{\text{CO}_2}\)):
\[ P_{\text{NH}_3} + P_{\text{CO}_2} = 0.116 \text{ atm} \]

3. **Determine Partial Pressures:**
Let’s assume the partial pressure of \(CO_2\) at equilibrium is \(x\) atm. Then, the partial pressure of \(NH_3\) will be:
\[ P_{\text{NH}_3} = 0.116 - x \]
Since the stoichiometry of the reaction shows that for every mole of \(CO_2\), there are 2 moles of \(NH_3\), we can express the relationship as:
\[ P_{\text{NH}_3} = 2P_{\text{CO}_2} \]
Substituting \(P_{\text{CO}_2} = x\):
\[ 0.116 - x = 2x \]
Solving for \(x\):
\[ 0.116 = 3x \implies x = \frac{0.116}{3} \approx 0.03867 \text{ atm} \]
Thus, \(P_{\text{CO}_2} \approx 0.03867 \text{ atm}\) and:
\[ P_{\text{NH}_3} = 0.116 - 0.03867 \approx 0.07733 \text{ atm} \]

4. **Effect of Adding \(CO_2\):**
Now, if we introduce an additional \(0.1 \text{ atm}\) of \(CO_2\), the new total pressure of \(CO_2\) will be:
\[ P'_{\text{CO}_2} = 0.03867 + 0.1 = 0.13867 \text{ atm} \]

5. **Applying Le Chatelier's Principle:**
According to Le Chatelier's principle, adding more \(CO_2\) will shift the equilibrium to the left to counteract the change, thus favoring the formation of solid ammonium carbonate and reducing the amount of \(NH_3\) present in the system.

6. **Final Pressures:**
As a result, the pressure of \(NH_3\) will decrease due to the shift in equilibrium, while the pressure of \(CO_2\) will not remain at \(0.13867 \text{ atm}\) but will be lower than this value due to the shift in equilibrium.

### Conclusion:
- The final pressure of \(CO_2\) will be less than \(0.13867 \text{ atm}\).
- The pressure of \(NH_3\) will decrease.