Problem 44 - Chemistry and Chemical Reactivity

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, \(K_{\mathrm{p}} ?\)

Step-by-Step Solution

Answer

The equilibrium constant \(K_{\mathrm{p}}\) is \(2.32 \times 10^{-4}\).

Step 1: Analyze the Reaction

The given reaction is: \(\left(\mathrm{NH}_{4}\right)\left(\mathrm{H}_{2}\mathrm{NCO}_{2}\right)(\mathrm{s}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})\). When the solid ammonium carbamate sublimes, it dissociates into ammonia gas \((\mathrm{NH}_{3})\) and carbon dioxide gas \((\mathrm{CO}_{2})\). The reaction is at equilibrium at 25°C with a total pressure of 0.116 atm.

Step 2: Identify Partial Pressures

Let \(x\) be the partial pressure of \(\mathrm{CO}_{2}\) at equilibrium. Since the stoichiometry of \(\mathrm{NH}_{3}\) to \(\mathrm{CO}_{2}\) is 2:1, the partial pressure of \(\mathrm{NH}_{3}\) will be \(2x\). The total pressure of the gas, given as 0.116 atm, is the sum of the partial pressures: \(2x + x = 3x = 0.116 \text{ atm}\).

Step 3: Solve for Partial Pressures

From the equation \(3x = 0.116 \text{ atm}\), solve for \(x\):\[x = \frac{0.116}{3} = 0.0387 \text{ atm}\]Thus, the partial pressure of \(\mathrm{CO}_{2}\) is \(0.0387\) atm, and for \(\mathrm{NH}_{3}\), it is \(2x = 0.0774\) atm.

Step 4: Write the Expression for Kp

The equilibrium constant \(K_{\mathrm{p}}\) for the reaction in terms of partial pressures is given by:\[K_{\mathrm{p}} = \left(P_{\mathrm{NH}_{3}}\right)^2 \cdot P_{\mathrm{CO}_{2}}\]This comes from the balanced chemical equation: \(K_{\mathrm{p}} = (P_{\mathrm{NH}_{3}})^2 \times P_{\mathrm{CO}_{2}}\).

Step 5: Calculate Kp Using Partial Pressures

Substitute the calculated partial pressures into the \(K_{\mathrm{p}}\) expression:\[K_{\mathrm{p}} = (0.0774)^2 \times 0.0387 = 0.000232 = 2.32 \times 10^{-4}\]Therefore, the equilibrium constant \(K_{\mathrm{p}}\) is \(2.32 \times 10^{-4}\).

Key Concepts

Sublimation

Sublimation is a fascinating phase transition where a substance directly changes from a solid to a gas without first becoming a liquid.
This process occurs under specific conditions of temperature and pressure. Generally, it happens when the solid has lower intermolecular forces, allowing it to transition more easily into a gas.
In the exercise, ammonium carbamate is the solid that sublimes. It directly turns into ammonia and carbon dioxide gases.

• Ammonium carbamate gradually breaks down as it sublimes.
• No liquid phase is observed in this transition, making it distinct from melting.
• Sublimation is relatively rare compared to other phase changes like melting and boiling.

Understanding sublimation aids in comprehending how certain solids can bypass the liquid state under suitable conditions, significantly affecting how we calculate equilibrium in gaseous systems like in this problem.

Partial Pressures

In chemical equilibrium involving gases, each gas component exerts a certain pressure, known as its partial pressure.
Partial pressure is the pressure that a gas would exert if it occupied the whole volume by itself at the same temperature. So what happens in our exercise?
The total pressure given for the system was 0.116 atm, which is a sum of partial pressures of ammonia and carbon dioxide.

• The partial pressure of a gas is directly related to its mole fraction in a mixture.
• In our case, the stoichiometry indicates that the partial pressure of ammonia will be twice that of carbon dioxide due to its coefficients.
• Using the equation for total pressure, we can mathematically derive each gas's partial pressure.

This understanding is fundamental for determining equilibrium constants when dealing with gaseous reactions. Calculating partial pressures helps us understand how gas mixtures steadily state themselves in equilibrium.

Chemical Equilibrium

Chemical equilibrium marks a state where the rates of the forward and backward reactions are equal.
At this stage, the composition of reactants and products remains constant over time.In the context of our exercise, establishing chemical equilibrium for the sublimation reaction is crucial.
The equilibrium constant, denoted as \(K_{\mathrm{p}}\), is pivotal.

• It reflects the ratio of the product of the partial pressures of the products to the reactants, each raised to their respective stoichiometric coefficients.
• A small value for \(K_{\mathrm{p}}\), like \(2.32 \times 10^{-4}\), suggests that the reaction strongly favors the reactants; the products are present in minimal amounts.
• Understanding this helps predict how changes in conditions like temperature or pressure affect the equilibrium position.

Ultimately, chemical equilibrium illustrates how dynamic processes in chemistry can reach a state of balance, helping explain why certain reactions favor specific directions under given conditions.