Question

\- When solid ammonium carbamate sublimes, it dissociates completely into ammonia and carbon dioxide according to the following equation: $$ \left(\mathrm{NH}_{4}\right)\left(\mathrm{H}_{2} \mathrm{NCO}_{2}\right)(\mathrm{s}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) $$ At \(25^{\circ} \mathrm{C},\) experiment shows that the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium constant, \(K_{p} ?\)

Short answer

The equilibrium constant \( K_p \) is approximately \( 8.63 \times 10^{-6} \).

Step-by-step solution

Understand the Chemical Equation

The given equilibrium reaction is: \[ (\mathrm{NH}_{4})(\mathrm{H}_{2} \mathrm{NCO}_{2})(\mathrm{s}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g}) + \mathrm{CO}_{2}(\mathrm{g}) \]Solid ammonium carbamate dissociates into ammonia and carbon dioxide gases. The equilibrium constant \( K_p \) expresses the ratio of the partial pressures of products to the reactants. Since the reactant is a solid, it does not appear in the expression for \( K_p \).

Write the Expression for Kp

In this reaction, the equilibrium constant, \( K_p \), is related to the partial pressures of the gaseous products:\[ K_p = P_{\mathrm{NH}_3}^2 \cdot P_{\mathrm{CO}_2} \]Since solid concentrations are not included, we focus only on \( \mathrm{NH}_3 \) and \( \mathrm{CO}_2 \) pressures.

Calculate the Partial Pressures

At equilibrium, the total pressure of the gases is 0.116 atm. Let \( P_{\mathrm{CO}_2} = x \) and \( P_{\mathrm{NH}_3} = 2x \) since two moles of \( \mathrm{NH}_3 \) form for each mole of \( \mathrm{CO}_2 \). Thus, the total pressure is:\[ P_{\text{total}} = P_{\mathrm{NH}_3} + P_{\mathrm{CO}_2} = 2x + x = 3x \]Since \( P_{\text{total}} = 0.116 \text{ atm} \), solve for \( x \):\[ 3x = 0.116 \quad \Rightarrow \quad x = \frac{0.116}{3} = 0.0387 \text{ atm} \]

Substitute the Partial Pressures into Kp

Now substitute the partial pressures into the expression for \( K_p \):\[ P_{\mathrm{CO}_2} = 0.0387 \text{ atm}, \quad P_{\mathrm{NH}_3} = 2 \times 0.0387 = 0.0774 \text{ atm} \]\[ K_p = (0.0774)^2 \cdot 0.0387 \]

Calculate Kp Value

Compute the value of \( K_p \):\[ K_p = 0.0774^2 \cdot 0.0387 = 0.000223 \cdot 0.0387 = 8.63 \times 10^{-6} \]Hence, the equilibrium constant \( K_p \) is approximately \( 8.63 \times 10^{-6} \).

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K \), is a crucial concept in the study of chemical equilibrium. It gives a numerical value to the ratio of concentrations or partial pressures of products and reactants at equilibrium in a reversible chemical reaction.
This value is specific for a given reaction at a particular temperature. For reactions involving gases, the equilibrium constant is expressed in terms of partial pressures, symbolized as \( K_p \).

Here is how you determine \( K_p \):

• Identify the products and reactants in your balanced equation.
• Write the expression for \( K_p \), which includes only gaseous species (solids and liquids are not included).
• The expression is the product of the partial pressures of the products raised to their respective coefficients, divided by the reactants raised to their coefficients.

For the sublimation of ammonium carbamate, only gaseous products are considered. This gives the expression \( K_p = P_{\mathrm{NH}_3}^2 \cdot P_{\mathrm{CO}_2} \). Knowing the \( K_p \) of a reaction allows chemists to predict the position of equilibrium and how the balance can shift with changes in pressure or temperature.

Partial Pressure

In a mixture of gases, each component exerts its own pressure as if it were alone in the container. This is known as partial pressure. The sum of the partial pressures of all gases equals the total pressure exerted by the gas mixture.
When dealing with equilibrium involving gases, understanding partial pressure is critical. It allows us to relate pressures of individual gases to the overall reaction conditions.

How to calculate partial pressure:

• Express each gas's pressure as a part of the total pressure using the mole ratio from the balanced equation.
• In the sublimation example, the total pressure at equilibrium is known, and the partial pressures of the products can be expressed in terms of a variable, \( x \).
• The total pressure is then solved by summing the individual partial pressures, leading to equations like \( P_{\text{total}} = 2x + x = 3x \).

Correctly solving for \( x \) gives us the individual partial pressures, essential for determining the equilibrium constant.

Sublimation

Sublimation is the process where a solid changes directly into a gas without passing through the liquid state. This endothermic process can occur under specific environmental conditions, such as temperature and pressure.
Sublimation is common in certain compounds, such as ammonium carbamate, as shown in the provided problem. Understanding this phase transition is important because it affects the availability of gaseous reactants or products.

Key Points about Sublimation:

• In chemical equations, sublimating solids don't appear in the equilibrium constant expression because they do not affect the pressure of gases.
• The gaseous products resulting from sublimation will have their partial pressures considered in equilibrium calculations, like in the expression \( K_p = P_{\mathrm{NH}_3}^2 \cdot P_{\mathrm{CO}_2} \).

This process is relevant in industries where purification and refinement are essential, and it provides insight into phase transitions that are not intuitive like melting and boiling.

Gaseous Products

When discussing chemical reactions and equilibria, gaseous products play a significant role in determining the system's behavior. In our example, the decomposition of ammonium carbamate results in two gaseous products: ammonia \((\mathrm{NH}_3)\) and carbon dioxide \((\mathrm{CO}_2)\).
Gaseous products are especially important as they directly impact calculations of equilibrium constants and reaction conditions like pressure and temperature.

Considerations for Gaseous Products:

• The formation of gases from solids or liquids usually increases the system's pressure. This increase is directly used to calculate equilibrium constants.
• In reactions at equilibrium, the partial pressures of gases are used to derive an expression for \( K_p \), which helps to predict where the equilibrium will lie.
• The behavior of gases often obeys the ideal gas law, but real gas behavior at different conditions should also be considered.

In conclusion, understanding the role of gaseous products in a chemical reaction is pivotal for mastering the principles of equilibrium and their application in real-world chemical processes.