Question

When heated, ammonium carbamate decomposes as follows: $$ \mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftarrows 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) $$ At a certain temperature, the equilibrium pressure of the system is 0.318 atm. Calculate \(K_{p}\) for the reaction.

Short answer

\(K_p = 0.00476\)

Step-by-step solution

Understanding the Reaction

The decomposition of ammonium carbamate into ammonia and carbon dioxide results in the formation of gases from a solid. The equation is: \[ \mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g) + \mathrm{CO}_{2}(g) \] Since solids do not appear in the expression for the equilibrium constant \(K_p\), only the gases are considered.

Expression for \(K_p\)

To express the equilibrium constant \(K_p\) for this reaction, use the partial pressures of the gases involved. The expression for \(K_p\) is given by: \[ K_p = \frac{(P_{\mathrm{NH}_3})^2 \cdot P_{\mathrm{CO}_2}}{1} \] Here, \(P_{\mathrm{NH}_3}\) and \(P_{\mathrm{CO}_2}\) are the partial pressures of ammonia and carbon dioxide, respectively.

Relating Total Pressure to Partial Pressures

At equilibrium, the total pressure is 0.318 atm, and it includes the partial pressures of both ammonia and carbon dioxide. Assume \(x\) to be the change in pressure due to decomposition producing gases: - Ammonia: \(2x\)- Carbon dioxide: \(x\) Thus, the expression for total pressure is: \[ 2x + x = 0.318 \] Solving gives \(3x = 0.318\), so \(x = 0.106\) atm.

Calculate Partial Pressures

Use the value of \(x\) to find individual partial pressures: - \(P_{\mathrm{NH}_3} = 2x = 2 \times 0.106 = 0.212\) atm - \(P_{\mathrm{CO}_2} = x = 0.106\) atm.

Calculate \(K_p\)

Plug the partial pressures back into the expression for \(K_p\): \[ K_p = (0.212)^2 \times 0.106 = 0.00476 \] Therefore, the equilibrium constant \(K_p\) is calculated to be 0.00476 for the given reaction at this temperature.

Key concepts

Ammonium Carbamate Decomposition

Ammonium carbamate is a compound made of ammonium ions and carbamate ions. When heated, it readily decomposes into ammonia (\(\mathrm{NH}_{3}\)) and carbon dioxide (\(\mathrm{CO}_{2}\)).
The decomposition reaction is represented as follows: \[ \mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftarrows 2 \mathrm{NH}_{3}(g) + \mathrm{CO}_{2}(g) \] This reaction involves a solid reacting to form gaseous products. A fascinating aspect of this reaction is how it shifts from a single-phase (solid) to a gas-phase equilibrium when heated.
This means that the decomposition can be monitored by looking at the pressures of the gases produced.

Equilibrium Constant

The equilibrium constant (\(K_p\)) is a value that expresses the ratio of the concentrations of products to reactants for a reaction at equilibrium.
In the context of gas-phase reactions, like the decomposition of ammonium carbamate, \(K_p\) is determined using the partial pressures of the gaseous components.
\(K_p\) provides important information about the position of equilibrium and the extent of the reaction. To express \(K_p\) for the decomposition reaction, we only consider the gases. Solids are not included in the expression for equilibrium constants.
Thus, the \(K_p\) expression for our reaction is: \[ K_p = \frac{(P_{\mathrm{NH}_3})^2 \times P_{\mathrm{CO}_2}}{1} \] This formula essentially describes how the partial pressures are related in equilibrium.

Partial Pressures

Partial pressure is the pressure that a single gas in a mixture would exert if it occupied the entire volume by itself.
For gas reactions, understanding partial pressures helps in calculating equilibrium constants like \(K_p\).
Each gas in the mixture contributes to the total pressure depending on its mole fraction and individual properties. In the reaction of ammonium carbamate, as decomposition occurs, the total pressure measured arises from the partial pressures of ammonia and carbon dioxide.
By considering the reaction stoichiometry, we can determine the contribution of each gas to the total pressure and solve for their partial pressures.
This is crucial for accurately determining the equilibrium constant.

Chemical Equilibrium

Chemical equilibrium describes a state in which the rate of the forward reaction equals the rate of the reverse reaction.
At equilibrium, the concentrations of reactants and products remain constant over time. For ammonium carbamate decomposition, even as its solid form continues to break down to form gases, some of the gaseous molecules recombine to reform the solid.
Equilibrium is reached when these happenings balance out.
At this point, the measurable pressures of ammonia and carbon dioxide become constant.
This constancy allows us to calculate important values such as \(K_p\). The concept of chemical equilibrium is pivotal in understanding how reactions proceed and the factors influencing their dynamics.