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. \(655 \mathrm{g} \mathrm{CaCO}_{3}, 95.0 \mathrm{g}\) 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

a. The initial amount of CaO will decrease. b. The initial amount of CaO will remain the same. c. The initial amount of CaO will increase. d. The initial amount of CaO will increase.

Step-by-step solution

Case a: Moles of CaCO3, CaO, and CO2

Use the molecular weights to convert 655g CaCO3, 95.0g CaO, and CO2 pressure (P_CO2) of 2.55 atm to moles.

Case b: Moles of CaCO3, CaO, and CO2

Use the molecular weights to convert 780g CaCO3, 1.00g CaO, and CO2 pressure (P_CO2) of 1.04 atm to moles.

Case c: Moles of CaCO3, CaO, and CO2

Use the molecular weights to convert 0.14g CaCO3, 5000g CaO, and CO2 pressure (P_CO2) of 1.04 atm to moles.

Case d: Moles of CaCO3, CaO, and CO2

Use the molecular weights to convert 715g CaCO3, 813g CaO, and CO2 pressure (P_CO2) of 0.211 atm to moles. #Step 2: Calculate the Reaction Quotient (Qp)# Now we will calculate the reaction quotient (Qp) for each case, using the formula Qp = P_CO2.

Case a: Calculate Qp

Calculate Qp for Case a using P_CO2 of 2.55 atm.

Case b: Calculate Qp

Calculate Qp for Case b using P_CO2 of 1.04 atm.

Case c: Calculate Qp

Calculate Qp for Case c using P_CO2 of 1.04 atm.

Case d: Calculate Qp

Calculate Qp for Case d using P_CO2 of 0.211 atm. #Step 3: Compare Qp to Kp for Each Case and Determine System's Behavior# Now, we will compare Qp to the given Kp value (1.04) for each case to determine if the reaction will shift right (forward), left (reverse), or remain unchanged.

Case a: Compare Qp to Kp

Compare Qp to Kp for Case a to determine if the initial amount of CaO will increase, decrease, or remain the same as the system moves towards equilibrium.

Case b: Compare Qp to Kp

Compare Qp to Kp for Case b to determine if the initial amount of CaO will increase, decrease, or remain the same as the system moves towards equilibrium.

Case c: Compare Qp to Kp

Compare Qp to Kp for Case c to determine if the initial amount of CaO will increase, decrease, or remain the same as the system moves towards equilibrium.

Case d: Compare Qp to Kp

Compare Qp to Kp for Case d to determine if the initial amount of CaO will increase, decrease, or remain the same as the system moves towards equilibrium.

Key concepts

Reaction Quotient (Qp)

The reaction quotient, denoted as \( Q_p \), helps to predict the direction in which a chemical reaction will proceed. It is very similar to the equilibrium constant \( K_p \), but it uses the initial concentrations or partial pressures to calculate the ratio of products to reactants. In the context of a gas-phase reaction like:\[ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \]we focus on the gaseous product, \( \mathrm{CO}_2 \), since solids do not appear in the expression. Therefore, the expression for \( Q_p \) is simply the partial pressure of \( CO_2 \):

• \( Q_p = P_{\mathrm{CO}_2} \)

Knowing \( Q_p \) allows us to determine if the system is at equilibrium or which direction it will shift to reach equilibrium.When \( Q_p < K_p \), the system shifts to the right (forward reaction) to produce more products. If \( Q_p > K_p \), the system shifts to the left (reverse reaction) to produce more reactants. Lastly, when \( Q_p = K_p \), the system is already at equilibrium, and no shift occurs.

Equilibrium Constant (Kp)

The equilibrium constant, \( K_p \), offers insight into the position of equilibrium for a chemical reaction at a specified temperature. It is defined for reactions involving gases and calculated using the equilibrium partial pressures of reactants and products. For the decomposition of calcium carbonate:\[ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \]only the gaseous \( CO_2 \) is included, so:

• \( K_p = P_{\mathrm{CO}_2} \) at equilibrium.

The value of \( K_p \) reflects the extent of the reaction:

• If \( K_p \) is large, the equilibrium position favors products, indicating a "product-dominant" reaction.
• If \( K_p \) is small, the equilibrium position favors reactants, indicating a "reactant-dominant" reaction.

For the reaction provided, \( K_p = 1.04 \) at \( 900^{\circ} \mathrm{C} \), suggesting a balanced position where neither reactants nor products are dominantly favored.

Le Chatelier's Principle

Le Chatelier's Principle is a cornerstone concept in understanding how systems at equilibrium respond to external changes. This principle states that if an equilibrium system experiences a change in concentration, temperature, or pressure, the system will adjust to minimize that change and restore a new equilibrium.For example, if in our reaction \( \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) \), extra \( CO_2 \) is added to the system, the pressure of \( CO_2 \) increases, resulting in a larger \( Q_p \). Following Le Chatelier’s Principle, the system will act to reduce the pressure by shifting the equilibrium towards the left, thereby producing more reactants (\( \mathrm{CaCO}_3 \)).Conversely, if the \( CO_2 \) pressure is reduced, \( Q_p \) becomes lower than \( K_p \), prompting a shift to the right to restore balance by producing more \( CO_2 \).It’s a dynamic process where the system "compensates" for changes, striving to maintain stability.

Gas Phase Reactions

Gas phase reactions, such as the decomposition of \( \mathrm{CaCO}_3 \) to \( \mathrm{CaO} \) and \( \mathrm{CO}_2 \), are affected by the pressures and concentrations of the gaseous components. Working with such reactions often involves using partial pressures to analyze changes at equilibrium.In these reactions, pressure plays a significant role. If pressure is increased by reducing volume, the equilibrium will shift towards the side with fewer moles of gas, according to Le Chatelier's Principle. For our specific example:- The decomposition side has only one molecule of gas (\( \mathrm{CO}_2 \)), so reducing pressure would encourage the decomposition of \( \mathrm{CaCO}_3 \), while increasing pressure tends to favor the reactants.Temperature is also a pivotal factor in gas reactions. Raising the temperature generally favors endothermic reactions (which absorb heat), whereas lowering it favors exothermic ones (which release heat).Understanding how these factors affect gas phase reactions is crucial for predicting system behavior and achieving balanced reactions.