Problem 82 - Chemistry The Central Science

Question

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a \(10.0-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO},\) and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO},\) and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (c) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO},\) and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)

Step-by-Step Solution

Answer

For the given mixtures: (a) The amount of \(CaCO_3\) will increase. (b) The amount of \(CaCO_3\) will decrease. (c) The amount of \(CaCO_3\) will remain the same.

Step 1: Calculate the moles of given reactants and products for each case

Use the molar mass of CaCO3, CaO, and CO2 to determine the moles of each in every case. Then, divide the moles by the volume to obtain their respective concentrations. Remember that solids do not affect the equilibrium situation, so solid products and reactants can be ignored when calculating Q. (a) Moles and Concentrations: \(n_{CaCO3} = \frac{15 g}{100.09 \frac{g}{mol}} = 0.15 mol\) \(C_{CaCO3} = \frac{0.15 mol}{10.0 L} = 0.015 M\) \(n_{CaO} = \frac{15 g}{56.08 \frac{g}{mol}} = 0.267 mol\) \(C_{CaO} = \frac{0.267 mol}{10.0 L} = 0.0267 M\) \(n_{CO2} = \frac{4.25 g}{44.01 \frac{g}{mol}} = 0.0965 mol\) \(C_{CO2} = \frac{0.0965 mol}{10.0 L} = 0.00965 M\) Repeat the same process for cases (b) and (c).

Step 2: Calculate the reaction quotient (Q) for each case

According to the balanced chemical equation: \(Q_c = \frac{[CO_2]}{[CaCO_3][CaO]} \) To calculate Q for each case, substitute the values calculated in Step 1: (a) \(Q_{c(a)} = \frac{0.00965}{(0.015)(0.0267)} = 0.0239\) Repeat the same calculation for cases (b) and (c).

Step 3: Compare Q and Kc for each case to determine the direction of the reaction

In each case, compare Q to the given equilibrium constant, Kc. If \(Q < K_c\), the system will shift forward to reach equilibrium and thus, the amount of CaCO_3 will decrease. If \(Q > K_c\), the system will shift backward to reach equilibrium and the amount of CaCO_3 will increase. If \(Q = K_c\), the system is at equilibrium, and the amount of CaCO_3 will remain the same. (a) Since \(Q_{c(a)} = 0.0239 > K_c = 0.0108\), the reaction will shift backward, and the amount of \(CaCO_3\) will increase. Now, compare Q and Kc for cases (b) and (c).

Step 4: Give the conclusion for each case

Summarize the results for each case: (a) The amount of \(CaCO_3\) will increase. (b) Calculate and compare Q and Kc to determine whether the amount of \(CaCO_3\) will increase, decrease, or remain the same. (c) Calculate and compare Q and Kc to determine if the amount of \(CaCO_3\) will increase, decrease, or remain the same.

Key Concepts

Reaction Quotient

The reaction quotient, denoted as \( Q \), is a central concept in understanding chemical equilibrium. It is a mathematical expression that quantifies the ratio of the concentrations of products to reactants at any non-equilibrium condition of a reaction. The formula for the reaction quotient is similar to that of the equilibrium constant \( K \), but it uses concentrations at a particular moment in time rather than at equilibrium. For the reaction \( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)\), the reaction quotient \( Q_c \) is calculated as:

• \( Q_c = \frac{[\text{CO}_2]}{[\text{CaCO}_3][\text{CaO}]} \)

It's important to note that the concentrations of solid reactants and products do not appear in the equation. Hence, the equation simplifies to the concentration of \( \text{CO}_2 \) because \( \text{CaCO}_3 \) and \( \text{CaO} \) are solids. Calculating \( Q \) helps in predicting how a system will proceed to reach equilibrium by comparing it with the equilibrium constant \( K_c \).
- If \( Q < K_c \), the reaction will shift forward to produce more products.- If \( Q > K_c \), the reaction will shift backward, favoring the formation of reactants.- If \( Q = K_c \), the system is already at equilibrium and no shift occurs.

Equilibrium Constant

The equilibrium constant, \( K_c \), is a crucial concept in chemical equilibrium. It provides a ratio that represents the balance between product and reactant concentrations when a reaction is at equilibrium. For the reaction \( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \) at \( 900^{\circ} C \), \( K_c \) is given as 0.0108. In this reaction, \( K_c \) only considers the concentration of \( \text{CO}_2 \) because \( \text{CaCO}_3 \) and \( \text{CaO} \) are solids, and solids are not included in the expression for \( K_c \):

• \( K_c = \frac{[\text{CO}_2]}{[\text{CaCO}_3][\text{CaO}]} \)

This constant does not alter with changes in concentration or pressure, as long as temperature remains constant. By comparing \( Q \) and \( K_c \), one can determine the direction in which a reaction will naturally proceed. It's a snapshot of the energy dynamics of a reaction at a given temperature, showing a balance when all opposing rates are the same.
Understanding \( K_c \) is imperative for predicting reaction behaviors and manipulating reaction conditions to yield desired results, making it fundamental in industrial and laboratory chemical processes.

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemistry that describes how a system at equilibrium responds to disturbances or stresses. This principle asserts that if a system at equilibrium is exposed to external changes such as concentration, pressure (involving gases), or temperature, the system will adjust itself to counteract the effect of the disturbance and achieve a new equilibrium state.
For the reaction \( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \), when evaluated at a fixed temperature of \( 900^{\circ} C \), applying Le Chatelier's Principle can help predict the system’s response to changes:

• If additional \( \text{CO}_2 \) is added to the system, the equilibrium will shift towards the left, promoting the formation of \( \text{CaCO}_3 \) to reduce \( \text{CO}_2 \) concentration.
• Conversely, removing \( \text{CO}_2 \) will shift the equilibrium to the right, encouraging the breakdown of \( \text{CaCO}_3 \) to produce more \( \text{CO}_2 \).
• Changes in pressure would have little direct effect on this particular equilibrium due to the involvement of solids which are not typically influenced by pressure changes.

Le Chatelier's Principle thus serves as a predictive tool, allowing chemists to manipulate conditions and optimize reactions in both theoretical studies and practical applications. By appreciating how a chemical system responds to different stressors, chemists and engineers can tailor reactions for improved yield or efficiency in diverse contexts.