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Chapter 14: Problem 134

Calcium carbonate, \(\mathrm{CaCO}_{3}\), decomposes when heated to give calcium oxide, \(\mathrm{CaO},\) and carbon dioxide, \(\mathrm{CO}_{2}\) $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ \(K_{p}\) for this reaction at \(900^{\circ} \mathrm{C}\) is \(1.040 .\) What would be the yield of carbon dioxide (in grams) when \(1.000 \mathrm{~g}\) of \(\mathrm{CaCO}_{3}\) and \(1.000 \mathrm{~g} \mathrm{CaO}\) are heated to \(900^{\circ} \mathrm{C}\) in a \(1.000-\mathrm{L}\) vessel. (Ignore the volume occupied by the solids.) What would be the effect of adding a similar quantity of carbon dioxide to this equilibrium mixture? What would happen if the quantity of calcium carbonate were doubled?

Short Answer

Expert verified
0.475 g of CO2 is produced. Adding CO2 decreases yield; doubling CaCO3 has no effect.

Step by step solution

01

Write the Initial Reaction and Equilibrium Expression

The given reaction is \( \mathrm{CaCO}_3(s) \rightleftharpoons \mathrm{CaO}(s) + \mathrm{CO}_2(g) \). This equilibrium involves the partial pressure of \( \mathrm{CO}_2 \) only, because solids do not appear in the equilibrium expression. Thus, \( K_p = P_{CO_2} \). Since \( K_p = 1.040 \) at \( 900^{\circ} \mathrm{C} \), we have \( P_{CO_2} = 1.040 \ \text{atm} \).
02

Determine Moles of CaCO3

The molar mass of \( \mathrm{CaCO}_3 \) is approximately \( 100.1 \ \text{g/mol} \). From the given mass, calculate moles: \[\text{Moles of } \mathrm{CaCO}_3 = \frac{1.000 \ \text{g}}{100.1 \ \text{g/mol}} = 0.010 \ \text{mol}.\]
03

Calculate Moles of CO2 at Equilibrium

At equilibrium, the partial pressure of \( \mathrm{CO}_2 \) is 1.040 atm in a 1.000 L vessel. Use the ideal gas law: \[P V = nRT \Rightarrow\ n = \frac{PV}{RT}\]Where \(R = 0.0821\ \text{L atm K}^{-1} \text{mol}^{-1}\) and \(T = 1173 \ \text{K} \).\[n = \frac{1.040 \times 1.000}{0.0821 \times 1173} \approx 0.0108 \ \text{mol}.\]
04

Convert Moles of CO2 to Grams

Calculate the mass of \( \mathrm{CO}_2 \) formed:\[\text{Molar mass of } \mathrm{CO}_2 = 44.0 \ \text{g/mol}\]\[\text{Mass of } \mathrm{CO}_2 = 0.0108 \times 44.0 \approx 0.475 \ \text{g}.\]
05

Analyze Additional CO2 Effect

Adding more \( \mathrm{CO}_2 \) shifts the equilibrium toward the left, reducing the decomposition of \( \mathrm{CaCO}_3 \), and resulting in less \( \mathrm{CO}_2 \) being released back into the gas phase, thus decreasing the yield of \( \mathrm{CO}_2 \) in grams.
06

Analyze Doubling CaCO3 Effect

Doubling the amount of \( \mathrm{CaCO}_3 \) does not affect the equilibrium position or pressure since solids do not appear in the equilibrium expression for \( K_p \). Thus, the yield of \( \mathrm{CO}_2 \) in grams remains unchanged.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle serves as a guide to predict how a change in conditions affects a chemical equilibrium. This principle states that if a system at equilibrium is disturbed, it will adjust in a way that counteracts the disturbance and re-establishes equilibrium.
For instance, in the decomposition of calcium carbonate, if additional carbon dioxide is added to the system, the equilibrium will shift towards the left. This means the reaction will favor the formation of calcium carbonate and decrease the amount of free carbon dioxide. This adjustment helps reduce the increase in carbon dioxide, thereby balancing the system again.
Conversely, if carbon dioxide was removed from the system, the equilibrium would shift to the right, producing more carbon dioxide by decomposing more calcium carbonate. Understanding this can help predict how changing the conditions can influence the concentration of substances in equilibrium.
Equilibrium Constant (Kp)
The equilibrium constant for a reaction at a given temperature is a way to express the ratio of products to reactants at equilibrium. For reactions involving gases, the equilibrium constant is often given as \( K_p \), which is defined using partial pressures.
In the given reaction, \( K_p \) is determined by the partial pressure of carbon dioxide because it is the only gaseous component that plays a role in the equilibrium expression. For the decomposition of calcium carbonate at 900掳C, \( K_p = 1.040 \), which corresponds directly to the partial pressure of \( \mathrm{CO}_2 \) in atmospheres.
This means that at equilibrium, the partial pressure of carbon dioxide is 1.040 atm. This constant is crucial in calculating how many moles of carbon dioxide are present and ultimately determining the mass of \( \mathrm{CO}_2 \) formed under these conditions. Using \( K_p \) allows chemists to predict the behavior of reactions involving gases.
Ideal Gas Law
The Ideal Gas Law is a fundamental equation in chemistry that relates the parameters of a gas: pressure (\( P \)), volume (\( V \)), temperature (\( T \)), and the number of moles (\( n \)). The equation is written as \( PV = nRT \), where \( R \) is the gas constant.
This law is particularly useful in calculating the behavior of gases under different conditions. In the calcium carbonate decomposition process, the Ideal Gas Law helps calculate the number of moles of carbon dioxide at equilibrium. Given the partial pressure (1.040 atm), volume (1.000 L), and temperature (1173 K), it's possible to solve for the number of moles of \( \mathrm{CO}_2 \), which turned out to be approximately 0.0108 mol.
The Ideal Gas Law simplifies the complex interactions in gases to a straightforward relation, making it easier to manage calculations regarding gases in chemical reactions. It is key to determining quantities like the mass of gas produced or consumed when considering changes in conditions.

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Most popular questions from this chapter

During an experiment with the Haber process, a researcher put \(1 \mathrm{~mol} \mathrm{~N}_{2}\) and \(1 \mathrm{~mol} \mathrm{H}_{2}\) into a reaction vessel to observe the equilibrium formation of ammonia, \(\mathrm{NH}_{3}\). $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ When these reactants come to equilibrium, assume that \(x\) mol \(\mathrm{H}_{2}\) react. How many moles of ammonia form?

A 2.0-L reaction flask initially contains \(0.010 \mathrm{~mol}\) \(\begin{array}{lllll}\mathrm{CO}, & 0.80 \mathrm{~mol} \mathrm{H}_{2} \text { , and } & 0.50 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH} & \text { (methanol). }\end{array}\) If this mixture is brought in contact with a zinc oxidechromium(III) oxide catalyst, the equilibrium $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ is attained. The equilibrium constant \(K_{c}\) for this reaction at \(300^{\circ} \mathrm{C}\) is \(1.1 \times 10^{-2}\). What is the direction of reaction (forward or reverse) as the mixture attains equilibrium?

The equilibrium-constant expression for a reaction is $$ K_{c}=\frac{\left[\mathrm{NO}_{2}\right]^{4}\left[\mathrm{O}_{2}\right]}{\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]^{2}} $$ What is the equilibrium-constant expression when the equation for this reaction is halved and then reversed?

List four ways in which the yield of ammonia in the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) ; \Delta H^{\circ}<0 $$ can be improved for a given amount of \(\mathrm{H}_{2}\). Explain the principle behind each way.

You mix equal moles of \(\mathrm{N}_{2}, \mathrm{O}_{2}\), and NO and place them into a container fitted with a movable piston. Then you heat the mixture to \(2127^{\circ} \mathrm{C}\), compressing the mixture with the piston to \(10.0 \mathrm{~atm}\), where the following reaction may occur: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ \(K_{p}\) for this reaction at \(2127^{\circ} \mathrm{C}\) is \(0.0025 .\) Is the mixture of gases initially at equilibrium at this temperature and pressure, or does it undergo reaction to the left or to the right? Do you have enough information to answer this question? Explain your answer.
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