Question

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 \(50.0-\mathrm{L}\) 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. \(715 \mathrm{~g} \mathrm{CaCO}_{3}, 813 \mathrm{~g} \mathrm{CaO}, P_{\mathrm{CO}_{2}}=0.211 \mathrm{~atm}\)

Short answer

For each given mixture, the amount of calcium oxide will change as follows: - (a) Decrease - (b) Remain the same - (c) Remain the same - (d) Increase

Step-by-step solution

The reaction is as follows: \(\ce{CaCO3}(s) \rightleftharpoons \ce{CaO}(s) + \ce{CO2}(g)\) At \(900^{\circ} \mathrm{C}\), \(K_{\mathrm{p}}=1.04\) We have four scenarios (a, b, c, d) with different amounts of \(\ce{CaCO3}\), \(\ce{CaO}\), and \(\ce{CO2}\). #Step 2: Calculate the initial Reaction Quotient (Q) for each scenario#

Since the equation involves the partial pressure of a gas, we'll use the equation: \(Q = \dfrac{P_{\ce{CO2}}}{1}\) As solid concentrations are considered to be constant and generally omitted from the reaction quotient. Calculate Q for each scenario using the given partial pressure of \(P_{\ce{CO2}}\). (a) Q = \(\dfrac{2.55}{1} = 2.55\) (b) Q = \(\dfrac{1.04}{1} = 1.04\) (c) Q = \(\dfrac{1.04}{1} = 1.04\) (d) Q = \(\dfrac{0.211}{1} = 0.211\) #Step 3: Compare Q with Kp to determine the reaction direction#

To determine whether the amount of calcium oxide will increase, decrease, or remain the same, we need to see how Q compares to Kp for each of the scenario. (a) Q (2.55) is greater than Kp (1.04): The reaction moves to the left (reverse direction, amount of CaO will decrease) (b) Q (1.04) is equal to Kp (1.04): The reaction is at equilibrium (amount of CaO will remain the same) (c) Q (1.04) is equal to Kp (1.04): The reaction is at equilibrium (amount of CaO will remain the same) (d) Q (0.211) is less than Kp (1.04): The reaction moves to the right (forward direction, amount of CaO will increase) #Step 4: State the final answer#

For each given mixture, the amount of calcium oxide will change as follows: - (a) Decrease - (b) Remain the same - (c) Remain the same - (d) Increase

Key concepts

Reaction Quotient

The Reaction Quotient, represented as \( Q \), is a measure used to determine the direction in which a chemical reaction is proceeding concerning its equilibrium state. It compares the current state of a reaction to its position at equilibrium. In our exercise, for the reaction \( \ce{CaCO3(s) \rightleftharpoons CaO(s) + CO2(g)} \), \( Q \) is derived based on the partial pressure of \( \ce{CO2} \), which is the sole gaseous component that appears in our simplified expression for \( Q \).

• Calculate \( Q \) using the equation: \( Q = \dfrac{P_{\ce{CO2}}}{1} \). Here, only the partial pressure of \( \ce{CO2} \) is used because both \( \ce{CaCO3} \) and \( \ce{CaO} \) are solids, whose concentrations remain constant and hence do not appear in the \( Q \) expression.
• Comparing \( Q \) to the Equilibrium Constant \( K_p \) can provide valuable insight into the state of the reaction.

In our provided scenarios, we calculated \( Q \) values based on the initial partial pressures provided, and compared these against \( K_p = 1.04 \). This comparison helps determine if the reaction will proceed forward to produce more \( \ce{CO2} \), backward to consume \( \ce{CO2} \), or if it is already in equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle helps predict how a chemical equilibrium will respond to disturbances or changes in conditions. According to this principle, if an external change is applied to a system at equilibrium, the system will adjust in a way that counteracts this change and tries to re-establish equilibrium.
When applied to our context of the exercise:

• If \( Q > K_p \): The reaction will shift towards the left, or the reverse direction. This means that some of the \( \ce{CO2} \) gas will react with \( \ce{CaO} \) to form more \( \ce{CaCO3} \), thus decreasing the amount of \( \ce{CaO} \).
• If \( Q < K_p \): The reaction will shift towards the right, or the forward direction, producing more gaseous \( \ce{CO2} \), resulting in an increase in \( \ce{CaO} \).
• If \( Q = K_p \): The system is already at equilibrium, therefore no further changes in amounts of reactants or products will occur. \( \ce{CaO} \) remains unchanged.

By understanding these shifts, you can predict how the amounts of substances will change as the system attempts to restore equilibrium.

Equilibrium Constant

The Equilibrium Constant, denoted as \( K \), is a fundamental concept that quantifies the ratio of the concentrations of products to reactants at equilibrium. For gaseous reactions, the symbol \( K_p \) is used, representing equilibrium in terms of partial pressures. The magnitude of \( K_p \) gives insight into the dominant direction of the reaction at equilibrium.
In our scenario:

• The reaction \( \ce{CaCO3(s) \rightleftharpoons CaO(s) + CO2(g)} \) has an equilibrium constant \( K_p = 1.04 \) at \( 900^{\circ} \text{C} \).
• If \( K_p \) is greater than 1, the reaction favors products at equilibrium. If smaller, it favors reactants. Here, \( K_p \approx 1.04 \) suggests a balanced production of products and reactants at this specific temperature.

By understanding \( K_p \), you can interpret whether a reaction heavily favors reactants or products. It also allows you to see how the system might shift when disrupted, a key aspect when evaluating reaction mixtures as discussed in this context. In practice, knowing the value of \( K_p \) helps predict the blend of reactants and products and how to manipulate conditions to achieve desired outcomes.