Ammonium carbamate can dissociate into gases at 25°C according to the reaction: NH₂COONH₄(s) ⇌ 2 NH₃(g) + CO₂(g)
If sufficient ammonium carbamate is sealed in a flask,the total pressure will be 0.117 atm at equilibrium.What is the value of K_p at 25°C?

Question

Quizplus · Accepted Answer

The equilibrium expression for the dissociation of ammonium carbamate into gases is given by $K_p = \frac{P_{NH_3}^2 \cdot P_{CO_2}}{P_{solid}}$. Since the pressure of the solid is constant and not included in the $K_p$ expression, we can simplify the expression to $K_p = P_{NH_3}^2 \cdot P_{CO_2}$. At equilibrium, the total pressure ($P_{total}$) is the sum of the partial pressures of $NH_3$ and $CO_2$. Given that $2NH_3$ and $1CO_2$ are produced, for every mole of ammonium carbamate that dissociates, the ratio of $NH_3$ to $CO_2$ will be 2:1. Therefore, if $x$ is the partial pressure of $CO_2$, then $2x$ will be the partial pressure of $NH_3$, making $P_{total} = 3x = 0.117$ atm. Solving for $x$ gives $x = 0.039$ atm. Substituting these values into the $K_p$ expression gives $K_p = (2x)^2 \cdot x = (2 \cdot 0.039)^2 \cdot 0.039 = 2.37 \times 10^{-4}$.

Ammonium Carbamate Can Dissociate into Gases at 25°C According to the Reaction