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Chapter 14: Problem 85

At \(850^{\circ} \mathrm{C}\) and 1.000 atm pressure, a gaseous mixture of carbon monoxide and carbon dioxide in equilibrium with solid carbon is \(90.55 \%\) CO by mass. $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Calculate \(K_{c}\) for this reaction at \(850^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The equilibrium constant \(K_c\) is calculated using mole percentages and stoichiometry of the reaction.

Step by step solution

01

Understand the Problem

We need to calculate the equilibrium constant, \(K_c\), for the given reaction where carbon monoxide and carbon dioxide are in equilibrium with solid carbon at 850掳C. The gaseous mixture is 90.55% CO by mass.
02

Write Down the Balanced Equation

The reaction given is \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\). It shows solid carbon reacting with carbon dioxide gas to form carbon monoxide gas.
03

Define Molar Masses and Mass Percentages

Calculate the molar masses: \( \text{Molar mass of CO} = 28.01\, \text{g/mol} \) and \( \text{Molar mass of CO}_2 = 44.01\, \text{g/mol} \). Given that 90.55% of the mixture is CO by mass, we can write the masses as fractions (0.9055 CO and 0.0945 CO鈧�).
04

Set Up the Mass-to-Mole Conversion

Let \( x \) be the total mass of the mixture in grams. The mass of CO is \( 0.9055x\) grams and CO鈧� is \( 0.0945x \) grams. Convert these masses to moles: \( \text{moles of CO} = \frac{0.9055x}{28.01}\) and \( \text{moles of CO}_2 = \frac{0.0945x}{44.01}\).
05

Find Mole Fractions at Equilibrium

At equilibrium, use the stoichiometry to find the change in moles. Since CO is formed from CO鈧�, the increase in moles for CO is double the decrease in moles of CO鈧�. Therefore, \(n_{CO} = 2(x)\) and \(n_{CO_{2}} = x\). In this case, equate the expression for moles of CO we calculated earlier to solve for \( x \), the moles at equilibrium.
06

Calculate the Equilibrium Constant \(K_c\)

\( K_c = \frac{[CO]^2}{[CO_2]} \). Use the equilibrium concentrations calculated (from moles divided by volume V), assuming V = 1 L for simplification, and plug the values into the formula to find \( K_c \).
07

Solve for \( K_c \) Value

Finally, substitute the values into the equilibrium expression and calculate \( K_c \). With values known, express the CO concentration squared and divide by the CO鈧� concentration to obtain \( K_c \).

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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 state where the concentrations of reactants and products remain constant over time. This happens because the forward and reverse reactions occur at the same rate. In our problem, we have an equilibrium between carbon monoxide (CO), carbon dioxide (CO鈧�), and solid carbon. Even though the solid carbon does not appear in the equilibrium expression, it plays a crucial role by participating in the reaction. Understanding equilibrium helps predict how the concentrations of the involved gases will affect the reaction, especially when calculating the equilibrium constant, which is a measure of the extent of a reaction at equilibrium.
Reaction Stoichiometry
Reaction stoichiometry involves the quantitative relationships between the substances involved in a chemical reaction, based on the balanced chemical equation. For the reaction given: \ \[\mathrm{C}(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\] \, we see that one mole of CO鈧� can produce two moles of CO. This stoichiometric relationship is important when calculating how the amounts of reactions and products change during the reaction. Stoichiometry helps us keep track of moles during the equilibrium as changes in one component affect others proportionally, according to the balanced equation.
Mole Concept
The mole concept is fundamental in chemistry, allowing us to count entities at the molecular scale. A mole represents Avogadro's number of particles, often atoms or molecules. In the reaction under study, the mole concept is used to convert mass percentages into moles, enabling us to apply stoichiometry. The conversion is crucial to determine the number of moles of CO and CO鈧� present in the equilibrium mixture, which we need to calculate the equilibrium constant \(K_c\). It highlights the direct relationship between the mass of substances and the number of moles they contain.
Mass Percentage
Mass percentage expresses the concentration of a component in a mixture as a percentage of the total mass. In the problem, the mixture contains 90.55% CO by mass. This means that out of 100 grams of the gaseous mixture, 90.55 grams is CO. To use this information in calculations, we must convert these percentages into usable forms like mass or moles. This allows us to determine the composition of the equilibrium mixture quantitatively and to perform further computational steps, such as determining concentrations.
Molar Mass
Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). For gases involved in this reaction, we have the molar mass of CO as 28.01 g/mol and CO鈧� as 44.01 g/mol. Knowing the molar mass of each gas is essential when converting mass to moles, as shown in the calculations. By dividing the mass by the molar mass, we can find out how many moles of each substance are present in our reaction. It is a key parameter needed to bridge the gap between macroscopic measurements (like mass) and the microscopic quantities (moles) used in chemistry.
Gaseous Reactions
Gaseous reactions involve substances in the gas phase, and they feature unique characteristics like pressure, volume, and temperature effects. In our reaction, the equilibrium mixture involves gaseous CO and CO鈧�. These reactions are highly dependent on conditions such as temperature and pressure, both of which affect the equilibrium constant \(K_c\). Understanding gaseous reactions is crucial for predicting how changes in conditions may shift the equilibrium position. The concepts of partial pressures and the ideal gas law often aid calculations related to gaseous reactions.

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Most popular questions from this chapter

Calculate the composition of the gaseous mixture obtained when \(1.25 \mathrm{~mol}\) of carbon dioxide is exposed to hot carbon at \(800^{\circ} \mathrm{C}\) in a 1.25 - \(\mathrm{L}\) vessel. The equilibrium constant \(K_{c}\) at \(800^{\circ} \mathrm{C}\) is 14.0 for the reaction $$ \mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g) $$

A researcher put \(0.400 \mathrm{~mol} \mathrm{PCl}_{3}\) and \(0.600 \mathrm{~mol}\) \(\mathrm{Cl}_{2}\) into a 5.00-L vessel at a given temperature to produce phosphorus pentachloride, \(\mathrm{PCl}_{5}:\) $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ What will be the composition of this gaseous mixture at equilibrium? \(K_{c}=25.6\) at the temperature of this experiment.

Iodine, \(\mathrm{I}_{2}\), is a blue-black solid, but it easily vaporizes to give a violet vapor. At high temperatures, this molecular substance dissociates to atoms: $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ An absent-minded professor measured equilibrium constants for this dissociation at two temperatures, \(700^{\circ} \mathrm{C}\) and \(800^{\circ} \mathrm{C}\). He obtained the \(K_{p}\) values 0.01106 and \(0.001745,\) but he couldn't remember which value went with what temperature. Can you please help him assign temperatures to the equilibrium constants? State your argument carefully.

Ammonium hydrogen sulfide, \(\mathrm{NH}_{4} \mathrm{HS}\), is unstable at room temperature and decomposes: $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ You have placed some solid ammonium hydrogen sulfide in a closed flask. Which of the following would produce less hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S},\) which is a poisonous gas? a. Removing some \(\mathrm{NH}_{3}\) from the flask b. Adding some \(\mathrm{NH}_{3}\) to the flask c. Removing some of the \(\mathrm{NH}_{4} \mathrm{HS}\) d. Increasing the pressure in the flask by adding helium gas Explain each of your answers.

The equilibrium constant \(K_{c}\) for the equation $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ at \(321^{\circ} \mathrm{C}\) is 1.03 . What is the value of \(K_{c}\) for the following equation? $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$
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