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) \(15.0 \mathrm{~g} \mathrm{CaCO}, 15.0 \mathrm{~g} \mathrm{CaO}\), and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}=25.0 \mathrm{~g} \mathrm{CaO}\), and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (c) \(30.5 \mathrm{~g} \mathrm{CaCO}, 25.5 \mathrm{~g} \mathrm{CaO}\), and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)

Short answer

For the three mixtures: (a) The amount of CaCO3 will decrease as the system approaches equilibrium. (b) The amount of CaCO3 will increase as the system approaches equilibrium. (c) The amount of CaCO3 will increase as the system approaches equilibrium.

Step-by-step solution

We must first convert the given masses of CaCO3, CaO, and CO2 into moles using their respective molar masses (100.09 g/mol for CaCO3, 56.08 g/mol for CaO, and 44.01 g/mol for CO2). Then, we can calculate their initial concentrations by dividing the number of moles by the volume of the container (10.0 L). #Step 2: Calculate the Initial Reaction Quotient, Qc#

The reaction quotient, Qc, is calculated using the equation: \[Q_c = \frac{[CO_2]}{1}\] Note that solids are not included in reaction quotients. #Step 3: Compare Qc to Kc for Each Mixture#

Once we have the initial Qc values for each mixture, we can compare them to the given Kc value of 0.0108: - If Qc > Kc, the reaction will shift towards the reactants, and the amount of CaCO3 will increase. - If Qc < Kc, the reaction will shift towards the products, and the amount of CaCO3 will decrease. - If Qc = Kc, the system is already at equilibrium, and the amount of CaCO3 will remain the same. Now, let's apply these steps to each mixture: (a) \(15.0 \mathrm{~g} \mathrm{CaCO}, 15.0 \mathrm{~g} \mathrm{CaO}\), and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) #Step 1a: Calculate Initial Concentrations#

Moles of CO2 = (4.25 g) / (44.01 g/mol) = 0.09658 mol Initial concentration of CO2 = (0.09658 mol) / (10.0 L) = 0.009658 M #Step 2a: Calculate the Initial Reaction Quotient, Qc#

Qc = [CO2] = 0.009658 #Step 3a: Compare Qc to Kc for Mixture (a)#

Qc (0.009658) < Kc (0.0108), so the reaction will shift towards the products, and the amount of CaCO3 will decrease. (b) \(2.50 \mathrm{~g} \mathrm{CaCO}=25.0 \mathrm{~g} \mathrm{CaO}\), and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) #Step 1b: Calculate Initial Concentrations#

Moles of CO2 = (5.66 g) / (44.01 g/mol) = 0.1286 mol Initial concentration of CO2 = (0.1286 mol) / (10.0 L) = 0.01286 M #Step 2b: Calculate the Initial Reaction Quotient, Qc#

Qc = [CO2] = 0.01286 #Step 3b: Compare Qc to Kc for Mixture (b)#

Qc (0.01286) > Kc (0.0108), so the reaction will shift towards the reactants, and the amount of CaCO3 will increase. (c) \(30.5 \mathrm{~g} \mathrm{CaCO}, 25.5 \mathrm{~g} \mathrm{CaO}\), and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\) #Step 1c: Calculate Initial Concentrations#

Moles of CO2 = (6.48 g) / (44.01 g/mol) = 0.1472 mol Initial concentration of CO2 = (0.1472 mol) / (10.0 L) = 0.01472 M #Step 2c: Calculate the Initial Reaction Quotient, Qc#

Qc = [CO2] = 0.01472 #Step 3c: Compare Qc to Kc for Mixture (c)#

Qc (0.01472) > Kc (0.0108), so the reaction will shift towards the reactants, and the amount of CaCO3 will increase.

Key concepts

Reaction Quotient

In chemical equilibrium, the reaction quotient, denoted as \( Q_c \), helps us determine the direction in which a reaction will proceed to achieve equilibrium. It is essentially a snapshot of the current state of the reaction. Particularly, it compares the current concentrations of the reaction products and reactants to those when the reaction is at equilibrium.

For the reaction \(\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)\), \( Q_c \) is calculated as \( Q_c = \frac{[\text{CO}_2]}{1}\) because the concentration of solids, namely \(\text{CaCO}_3\) and \(\text{CaO}\), do not alter the reaction quotient.

This approach is vital because:

• If \( Q_c > K_c \), the reaction will shift towards the formation of reactants to reach equilibrium.
• If \( Q_c < K_c \), the reaction shifts towards forming more products.
• If \( Q_c = K_c \), the system is at equilibrium and needs no shift.

This quantitative measure gives you the power to predict the direction of the reaction’s shift, thus anticipating changes in concentration over time.

Equilibrium Constant

The equilibrium constant, symbolized as \( K_c \), represents the ratio of the concentrations of products to reactants at equilibrium for a given reaction. In the equation \( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \), \( K_c \) is focused solely on the concentration of \( \text{CO}_2 \), making it \( K_c = 0.0108 \) at \(900{ }^\circ\text{C}\).

This value is pivotal because:

• It is a characteristic of a specific reaction at a definite temperature, so it differs with temperature changes.
• \( K_c \) helps in predicting the extent to which a reaction will occur before reaching equilibrium.
• The constant tells us about the composition of the equilibrium mixture; a larger \( K_c \) implies more product formation.

Understanding \( K_c \)’s role is critical to anticipate and verify reactions’ behavior under varying conditions, giving you insight into chemical processes.

Le Chatelier's Principle

Le Chatelier's Principle provides insight into how a system at equilibrium reacts to disturbances. It states that if a dynamic equilibrium is disturbed by changing the conditions, the system will respond to counteract the change and re-establish equilibrium.

In the practical context of our reaction: \( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \), this principle tells us that:

• If the concentration of \( \text{CO}_2 \) increases, the system will shift towards the formation of \( \text{CaCO}_3 \) to reduce the disturbance.
• If the concentration decreases, more \( \text{CaCO}_3 \) will decompose to replace the \( \text{CO}_2 \).

Considerations like temperature, pressure, and concentration shifts all influence equilibrium in predictable ways thanks to Le Chatelier's Principle. This makes it a versatile tool for predicting how chemical systems will react to environmental and experimental changes.