Problem 43 - Chemical Principles: The Quest for Insight

Question

\( \mathrm{~A} 25.0 \mathrm{~g}\) sample of ammonium carbamate, \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)\), was placed in an evacuated \(0.250-\mathrm{L}\). flask and kept at \(25^{\circ} \mathrm{C}\). At equilibrium, the flask contained \(17.4 \mathrm{mg}\) of \(\mathrm{CO}_{2}\). What is the value of \(K_{c}\) for the decomposition of ammonium carbamate into ammonia and carbon dioxide? The reaction is \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)(\mathrm{s})=2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) .\)

Step-by-Step Solution

Answer

Calculate the moles of CO2 from its mass, use that to find the equilibrium concentrations of CO2 and NH3, and then use those values to calculate Kc with the equilibrium expression for the reaction.

Step 1: Write the Balanced Chemical Equation and Express Kc

First, write the balanced equation for the decomposition of ammonium carbamate: \[\mathrm{NH}_4(\mathrm{NH}_2\mathrm{CO}_2)(\mathrm{s}) \longrightarrow 2\mathrm{NH}_3(\mathrm{g}) + \mathrm{CO}_2(\mathrm{g}).\] Then express the equilibrium constant, \(K_c\), in terms of the concentrations of the gases at equilibrium: \[K_c = \frac{[\mathrm{NH}_3]^2[\mathrm{CO}_2]}{1}\] since the concentration of a pure solid, ammonium carbamate, is constant and can be omitted.

Step 2: Convert Mass of CO2 to Moles

Calculate the number of moles of \(\mathrm{CO}_2\) at equilibrium using its mass: \(17.4 \mathrm{mg}\) or \(0.0174 \mathrm{g}\) and the molar mass of \(\mathrm{CO}_2\) (44.01 \mathrm{g/mol}): \[\text{Moles of } \mathrm{CO}_2 = \frac{0.0174 \mathrm{~g}}{44.01 \mathrm{~g/mol}}.\]

Step 3: Calculate Concentration of CO2

Divide the moles of \(\mathrm{CO}_2\) by the volume of the flask to find its concentration at equilibrium: \[ [\mathrm{CO}_2] = \frac{\text{Moles of } \mathrm{CO}_2}{0.250 \text{ L}}. \]

Step 4: Relate Moles of CO2 to Moles of NH3 and Calculate Concentrations

Using the stoichiometry from the balanced equation, 1 mole of \(\mathrm{CO}_2\) is produced for every 2 moles of \(\mathrm{NH}_3\) present. Thus the moles of \(\mathrm{NH}_3\) will be twice that of \(\mathrm{CO}_2\), and its concentration will be: \[ [\mathrm{NH}_3] = 2 \times [\mathrm{CO}_2]. \]

Step 5: Substitute the Concentrations into the Expression for Kc

Substitute the concentrations \([\mathrm{NH}_3]\) and \([\mathrm{CO}_2]\) into the expression for \(K_c\) to find its value: \[ K_c = \frac{([\mathrm{NH}_3])^2 \times [\mathrm{CO}_2]}{1}. \] Make sure to square the concentration of \(\mathrm{NH}_3\).

Step 6: Perform the Calculations to Find Kc

Insert the calculated concentrations into the equilibrium expression and solve for \(K_c\). Remember to square the concentration of \(\mathrm{NH}_3\) when you substitute it into the equation.

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a key concept in the field of chemical reactions and occurs when the rates of the forward and reverse reactions are equal, leading to no overall change in the amounts of the reactants and products over time. It's essential to understand that being at equilibrium does not mean that the reactants and products are present in equal amounts, but rather that their quantities are constant because the reaction rates are balanced.

In the context of the ammonium carbamate decomposition exercise, the equilibrium state is reached when ammonium carbamate decomposes into ammonia and carbon dioxide at the same rate at which they recombine to form ammonium carbamate. At this point, the amounts of ammonia and carbon dioxide become constant, indicating that the system has attained chemical equilibrium.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It's based on the balanced chemical equation, which shows the ratios in which substances react and are produced. For instance, the decomposition reaction in the exercise shows that one molecule of ammonium carbamate yields two molecules of ammonia (NH₃) and one molecule of carbon dioxide (CO₂).

Analyzing the Stoichiometry of Decomposition

It's critical to understand these ratios to calculate the moles and concentrations of the gases formed. In our exercise, the stoichiometry of the decomposition reaction is used to deduce that two moles of ammonia are generated for each mole of carbon dioxide produced.

Molar Concentration

Molar concentration, also known as molarity, is a measurement of the concentration of a solute in a solution, expressed as the number of moles of solute per liter of solution (mol/L). It is a fundamental concept in chemical equilibrium as it allows us to quantify the concentrations of reactants and products.

During the equilibrium constant calculation for the ammonium carbamate decomposition, molar concentration is used to describe the amounts of ammonia and carbon dioxide in the system. To find the molar concentration of CO₂, we divided the moles of CO₂ by the volume of the flask, obtaining its equilibrium concentration in mol/L.

Equilibrium Constant Calculation

The equilibrium constant, represented by K_c, quantifies the concentrations of the products and reactants at chemical equilibrium. A higher K_c value suggests a greater extent of reaction towards products, while a lower K_c indicates a reaction that favors the reactants.

The K_c is calculated by substituting the equilibrium concentrations of reactants and products into the reaction's equilibrium constant expression. The exercise demonstrates this calculation step-by-step for the decomposition of ammonium carbamate. It's also important to note that for a reaction at a given temperature, the value of K_c is constant.

Ammonium Carbamate Decomposition

Ammonium carbamate is a compound that can decompose into ammonia and carbon dioxide. This decomposition reaction is important in the study of equilibrium as it illustrates the concepts of equilibrium, stoichiometry, and concentration in a practical setting.

In our exercise, we analyze the equilibrium of ammonium carbamate's decomposition reaction, showing the practical application of chemical equilibrium concepts. By applying stoichiometry and molar concentration calculations, we can solve for the equilibrium constant, K_c, which is a unique value describing the system's state of balance under the given conditions.