Question

Ammonium carbamate \(\left(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}\right)\) decomposes as follows: $$ \mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftarrows 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) $$ Starting with only the solid, it is found that when the system reaches equilibrium at \(40^{\circ} \mathrm{C},\) the total gas pressure \(\left(\mathrm{NH}_{3}\right.\) and \(\mathrm{CO}_{2}\) ) is 0.363 atm. Calculate the equilibrium constant \(K_{P}\).

Short answer

The equilibrium constant \(K_P\) is approximately 0.0586.

Step-by-step solution

Understand the Reaction

The decomposition of ammonium carbamate is described by the equation:\[ \mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftarrows 2 \mathrm{NH}_{3}(g) + \mathrm{CO}_{2}(g) \]This indicates that one mole of solid decomposes to produce two moles of ammonia gas \((\mathrm{NH}_3)\) and one mole of carbon dioxide gas \((\mathrm{CO}_2)\).

Initial Setup and Definition of Terms

Start with solid ammonium carbamate. The gases will form at equilibrium, with ammonia being \(2x\) and carbon dioxide as \(x\) due to stoichiometry. The total pressure of these gases at equilibrium is given as 0.363 atm.

Express Total Pressure

From the reaction stoichiometry and given total pressure, express the total pressure of gases in terms of \(x\):\[ P_{\text{total}} = P_{\text{NH}_3} + P_{\text{CO}_2} \]\[ 0.363 = 2x + x = 3x \]Solve for \(x\):\[ x = \frac{0.363}{3} = 0.121\text{ atm} \]

Calculate Partial Pressures

Calculate the partial pressures at equilibrium:\[ P_{\text{NH}_3} = 2x = 2 \times 0.121 = 0.242 \text{ atm} \]\[ P_{\text{CO}_2} = x = 0.121 \text{ atm} \]

Calculate Equilibrium Constant \(K_P\)

The equilibrium constant \(K_P\) for the reaction is expressed in terms of partial pressures:\[ K_P = (P_{\text{NH}_3})^2 \times (P_{\text{CO}_2}) \]Substitute the values:\[ K_P = (0.242)^2 \times 0.121 \]\[ K_P = 0.0585648 \]

Analyze Results

Round the value of \(K_P\) to obtain a more concise result:\[ K_P \approx 0.0586 \]

Key concepts

Ammonium Carbamate Decomposition

Ammonium carbamate is an interesting compound that decomposes in a reversible reaction. When this solid decomposes, it produces two gases, ammonia (\( \mathrm{NH}_3\)) and carbon dioxide (\(\mathrm{CO}_2\)).
At the heart of this decomposition process is the transition from a solid state to gaseous products.
This highlights how chemical properties can change state under different conditions. In this scenario, when heat is applied, the energy provided allows the tightly packed molecules in the solid to break free and form gaseous products.

• The decomposition occurs as: \(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)\)
• Solid decomposition typically results in an increase in the total pressure due to the generation of gaseous molecules.
• The reaction proceeds until it reaches equilibrium, where no net change in the concentration of products and reactants happens.

Gas Partial Pressure

When ammonium carbamate decomposes, it produces gases that exert pressure in the container. This is known as gas partial pressure. In any gaseous chemical reaction, the pressure exerted by each individual gas is crucial for understanding the system’s behavior.
Partial pressure represents the contribution each gas makes to the total pressure in a mixture.
For our reaction, ammonia and carbon dioxide are the gases involved. Their pressures add up to the total pressure observed at equilibrium.

• In the given reaction, at equilibrium, the total pressure is \(0.363 \) atm.
• Partial pressures of gases are derived using their stoichiometric coefficients from the chemical equation.
• The relationship between partial pressures and total pressure can be expressed as: \( P_{\text{total}} = P_{\text{NH}_3} + P_{\text{CO}_2} \)
• Here, the calculations show \( P_{\text{NH}_3} \) is \(0.242 \) atm and \( P_{\text{CO}_2}\) is \(0.121 \) atm.

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry. It occurs when a chemical reaction's forward and backward reactions occur at the same rate. In the decomposition of ammonium carbamate, this is the point where the production of ammonia and carbon dioxide balances with their recombination into solid ammonium carbamate.
At equilibrium, the concentrations of reactants and products remain constant, though not necessarily equal.
The equilibrium constant, denoted as \(K_P\) for gas-phase reactions, provides insight into the position of the equilibrium.

• \(K_P\) is calculated using the partial pressures of products raised to the power of their coefficients from the balanced equation.
• For our reaction: \(K_P = (P_{\text{NH}_3})^2 \times (P_{\text{CO}_2})\)
• A higher \(K_P\) value means a greater concentration of products at equilibrium.
• In our case, \(K_P \) is approximately \(0.0586\), indicating the product's dominance in this reversible reaction.

Reaction Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves using balanced equations to calculate molar proportions and is critical for understanding the decomposition of ammonium carbamate.
In our specific reaction, stoichiometry provides the foundation for determining the amounts of ammonia and carbon dioxide produced.

• The balanced equation \(\mathrm{NH}_{4} \mathrm{CO}_{2} \mathrm{NH}_{2}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)\) shows the ratio of moles forming; 2 moles of \(\mathrm{NH}_3\) for every mole of \(\mathrm{CO}_{2}\).
• This ratio is used to express the gas pressures in terms of a single unknown, often represented as \(x\).
• Total pressure \(0.363\) atm results from the combination of \(2x\) for ammonia and \(x\) for carbon dioxide, leading to \(3x = 0.363\).
• Solving gives \(x = 0.121\), allowing us to find the individual partial pressures essential for equilibrium constant calculations.