\( \mathrm{~A} 25.0 \mathrm{~g}\) 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 \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)(\mathrm{s})=2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) .\)

Short Answer

Expert verified
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 by step solution

01

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.
02

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}): \[\text{Moles of } \mathrm{CO}_2 = \frac{0.0174 \mathrm{~g}}{44.01 \mathrm{~g/mol}}.\]
03

Calculate Concentration of CO2

Divide the moles of \(\mathrm{CO}_2\) by the volume of the flask to find its concentration at equilibrium: \[ [\mathrm{CO}_2] = \frac{\text{Moles of } \mathrm{CO}_2}{0.250 \text{ L}}. \]
04

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: \[ [\mathrm{NH}_3] = 2 \times [\mathrm{CO}_2]. \]
05

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 \(\mathrm{NH}_3\).
06

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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

The Kc 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 Kc 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, Kc, which is a unique value describing the system's state of balance under the given conditions.

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Most popular questions from this chapter

Explain why the following equilibria are heterogencous and write the reaction quoticnt \(Q\) for each one. (a) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})=\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})\) (b) \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{OH}_{2} \mathrm{O}\) (s) \(\rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{~s})+10 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (c) \(2 \mathrm{KCO}_{3}\) (s) \(\Rightarrow 2 \mathrm{KCl}\) (s) \(+3 \mathrm{O}_{2}\) (g)

Use Le Chatelier's principle to predict the effect that the change given in the first column of the following table has on the quantity in the second column for the following equilibrium system: $$ \begin{gathered} 5 \mathrm{CO}(\mathrm{g})+\mathrm{I}_{2} \mathrm{O}_{5}(\mathrm{~s}) \rightleftharpoons \mathrm{I}_{2}(\mathrm{~g})+5 \mathrm{CO}_{2}(\mathrm{~g}) \\\ \Delta H^{\circ}=-1175 \mathrm{~kJ} \end{gathered} $$

A reaction mixture consisting of \(2.00 \mathrm{~mol} \mathrm{CO}\) and \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) is placed into a \(10.0-\mathrm{L}\) reaction vessel and heated to \(1200 \mathrm{~K}\). At cquilibrium, \(0.478 \mathrm{~mol} \mathrm{CH}_{4}\) was present in the system. Determine the value of \(K_{c}\) for the reaction \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

A 0.500-L reaction vessel at \(700 \mathrm{~K}\) contains \(1.20 \times 10^{-3} \mathrm{~mol} \mathrm{SO}_{2}(\mathrm{~g}), 5.0 \times 10^{-4} \mathrm{~mol} \mathrm{O}_{2}(\mathrm{~g})\), and \(1.0 \times 10^{-4} \mathrm{~mol} \mathrm{SO}_{3}(\mathrm{~g}) .\) At \(700 \mathrm{~K}, K_{c}=1.7 \times 10^{6}\) for the cquilibrium \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}) .\) (a) Calculate the reaction quotient \(Q_{e}\) (b) Will more \(\mathrm{SO}_{3}(\mathrm{~g})\) tend to form?

A mixture consisting of \(1.1 \mathrm{mmol} \mathrm{SO}_{2}\) and \(2.2 \mathrm{mmol} \mathrm{O}_{2}\) in a \(250-\mathrm{mL}\) container was heated to \(500 \mathrm{~K}\) and allowed to reach equilibrium. Will more sulfur trioxide be formed if that equilibrium mixture is cooled to \(25^{\circ} \mathrm{C}\) ? For the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\) \(\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}), \mathrm{K}=2.5 \times 10^{10}\) at \(500 \mathrm{~K}\) and \(4.0 \times 10^{24}\) at \(25^{\circ} \mathrm{C}\).

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