At \(900^{\circ} \mathrm{C}, K_{\mathrm{p}}=1.04\) for the reaction $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ At a low temperature, dry ice (solid \(\mathrm{CO}_{2}\) ), calcium oxide, and calcium carbonate are introduced into a 50.0-L reaction chamber. The temperature is raised to \(900^{\circ} \mathrm{C}\), resulting in the dry ice converting to gaseous \(\mathrm{CO}_{2} .\) For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at \(900{ }^{\circ} \mathrm{C}\) ? a. \(655 \mathrm{~g} \mathrm{CaCO}_{3}, 95.0 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=2.55 \mathrm{~atm}\) b. \(780 \mathrm{~g} \mathrm{CaCO}_{3}, 1.00 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{~atm}\) c. \(0.14 \mathrm{~g} \mathrm{CaCO}_{3}, 5000 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{~atm}\) d. \(715 \mathrm{~g} \mathrm{CaCO}_{3}, 813 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=0.211 \mathrm{~atm}\)

Step by step solution

01

Calculate the initial pressure of CO2

The initial pressure of CO2 is already given (\(P_{\mathrm{CO}_{2}}=2.55\) atm).

02

Calculate the reaction quotient, Qp

The reaction quotient, \(Q_{\mathrm{p}}\), for the given reaction can be calculated using the expression \(Q_{\mathrm{p}} = P_{\mathrm{CO}_{2}}\). In this case, \(Q_{\mathrm{p}}=2.55\) atm.

03

Compare Qp with Kp

For this mixture, we have \(Q_{\mathrm{p}} = 2.55\) and \(K_{\mathrm{p}} = 1.04\). Since \(Q_{\mathrm{p}} > K_{\mathrm{p}}\), the reaction will shift to the left, meaning the initial amount of calcium oxide will decrease. b. \(780 \mathrm{~g} \mathrm{CaCO}_{3}, 1.00 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{~atm}\)

04

Calculate the initial pressure of CO2

The initial pressure of CO2 is already given (\(P_{\mathrm{CO}_{2}}=1.04\) atm).

05

Calculate the reaction quotient, Qp

In this case, the reaction quotient, \(Q_{\mathrm{p}}\), is equal to the initial pressure of CO2 (\(Q_{\mathrm{p}} = P_{\mathrm{CO}_{2}} = 1.04\) atm).

06

Compare Qp with Kp

For this mixture, we have \(Q_{\mathrm{p}} = 1.04\) and \(K_{\mathrm{p}} = 1.04\). Since \(Q_{\mathrm{p}} = K_{\mathrm{p}}\), the reaction is already at equilibrium, and the initial amount of calcium oxide will remain the same. c. \(0.14 \mathrm{~g} \mathrm{CaCO}_{3}, 5000 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=1.04 \mathrm{~atm}\)

07

Calculate the initial pressure of CO2

The initial pressure of CO2 is already given (\(P_{\mathrm{CO}_{2}}=1.04\) atm).

08

Calculate the reaction quotient, Qp

In this case, the reaction quotient, \(Q_{\mathrm{p}}\), is equal to the initial pressure of CO2 (\(Q_{\mathrm{p}} = P_{\mathrm{CO}_{2}} = 1.04\) atm).

09

Compare Qp with Kp

For this mixture, we have \(Q_{\mathrm{p}} = 1.04\) and \(K_{\mathrm{p}} = 1.04\). Since \(Q_{\mathrm{p}} = K_{\mathrm{p}}\), the reaction is already at equilibrium, and the initial amount of calcium oxide will remain the same. d. \(715 \mathrm{~g} \mathrm{CaCO}_{3}, 813 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=0.211 \mathrm{~atm}\)

10

Calculate the initial pressure of CO2

The initial pressure of CO2 is already given (\(P_{\mathrm{CO}_{2}}=0.211\) atm).

11

Calculate the reaction quotient, Qp

In this case, the reaction quotient, \(Q_{\mathrm{p}}\), is equal to the initial pressure of CO2 (\(Q_{\mathrm{p}} = P_{\mathrm{CO}_{2}} = 0.211\) atm).

12

Compare Qp with Kp

For this mixture, we have \(Q_{\mathrm{p}} = 0.211\) and \(K_{\mathrm{p}} = 1.04\). Since \(Q_{\mathrm{p}} < K_{\mathrm{p}}\), the reaction will shift to the right, meaning the initial amount of calcium oxide will increase.

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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 is a key concept in understanding how a chemical system at equilibrium responds to changes in concentration, temperature, and pressure. According to this principle, if an external change is applied to a system at equilibrium, the system adjusts in such a way as to minimize that change.

For example, consider a sealed container where nitrogen gas reacts with hydrogen gas to form ammonia. If we were to increase the pressure of the system, Le Chatelier's Principle predicts that the reaction will shift towards forming more ammonia since it has fewer gas molecules and thus reduces the pressure.

In the context of the textbook exercise, when the pressure of CO₂ is altered, the system will shift either to the right or to the left to counteract this change and restore equilibrium. This will result in either an increase or decrease in the amount of calcium oxide present. Hence, understanding this principle allows us to predict the direction of a shift in equilibrium in response to changes in experimental conditions.

Reaction Quotient (Qp)

The reaction quotient, usually denoted as Q_p, helps determine the direction in which a reaction must shift to reach equilibrium. It is a ratio that compares the current pressures of the gaseous reactants and products in a reaction at any point in time, not just at equilibrium.

For a generic reaction where a gas A reacts with gas B to form gas C, the reaction quotient Q_p is given by the expression \( Q_{\text{p}} = \frac{P_{\text{C}}}{P_{\text{A}} \times P_{\text{B}}} \), assuming the reaction is \(aA + bB \rightleftharpoons cC\).As seen in the textbook exercise, the reaction quotient for the decomposition of calcium carbonate is simply the pressure of CO₂ because calcium carbonate and calcium oxide are solids and do not appear in the expression for Q_p. By comparing Q_p with the equilibrium constant, K_p, students can predict which direction the reaction will proceed to achieve equilibrium.

Equilibrium Constant (Kp)

The equilibrium constant, represented as K_p, is a number that characterizes the chemical equilibrium of a reaction. It's unique for a given reaction at a specific temperature. If the reaction involves gases, the equilibrium constant in terms of partial pressures is used, denoted as K_p.

The value of K_p is calculated using the equilibrium partial pressures of the reactants and products. For the reaction in the textbook exercise, the equilibrium constant K_p is expressed as \( K_{\text{p}} = P_{\text{CO}_2} \) because the solid reactants and products are not included in the expression. When K_p is known, it can be used together with Q_p to predict the direction of the shift needed to reach equilibrium. A comparison of Q_p to K_p lets us know if the reaction at a given moment is product-favored (\(Q_{\text{p}} < K_{\text{p}}\)), reactant-favored (\(Q_{\text{p}} > K_{\text{p}}\)), or at equilibrium (\(Q_{\text{p}} = K_{\text{p}}\)). This understanding is crucial for solving equilibrium problems and is effectively demonstrated in the textbook exercise.