Question

The amount of carbon dioxide in a gaseous mixture of CO₂ and CO can be determined by passing the gas into an aqueous solution that contains an excess of Ba(OH)₂. The CO₂ reacts, yielding a precipitate of BaCO₃, but the CO does not react. This method was used to analyze the equiLiBrium composition of the gas obtained when 1.77 g of CO₂ reacted with 2.0 g of graphite in a 1.000 L container at 1100 K. The analysis yielded 3.41 g of BaCO₃. Part A: Use these data to calculate Kp at 1100 K for the reaction CO₂(g) + C(s) ⇌ 2CO(g)?

Answer 1

To calculate Kp at 1100 K for the reaction CO₂(g) + C(s) ⇌ 2CO(g), the partial pressures of CO and CO₂ were determined using the moles of reactants and products, resulting in Kp = 0.0835 atm².

Step-by-step explanation:

To determine Kp, we first need to find the moles of CO₂ and CO present initially and at equilibrium. Given that 1.77 g of CO₂ reacted and 3.41 g of BaCO₃ formed, we can use stoichiometry to find the moles of CO₂.

The molar mass of CO₂ = 44 g/mol. So, 1.77 g of CO₂ is 1.77/44 = 0.0402 mol of CO₂. As the reaction is 1:1 for CO₂ and CO, this is also the amount of CO formed.

The initial moles of CO is zero since only CO₂ and graphite were initially present. At equilibrium, CO₂ consumed 0.0402 mol, and the amount of CO formed is also 0.0402 mol. Therefore, the total moles of gases at equilibrium = 2 * 0.0402 = 0.0804 mol.

Given that the total pressure of the system is due to CO and CO₂, the partial pressure of each gas can be assumed to be the same. Thus, P_CO = P_CO₂ = Total pressure/total moles = 1 atm (from ideal gas law)/0.0804 mol = 12.43 atm for each gas.

Finally, using the expression for Kp, which is the partial pressure of products raised to their coefficients divided by the partial pressure of reactants raised to their coefficients, we get Kp = (P_CO)²/(P_CO₂) = (12.43)²/(12.43) = 0.0835 atm².

This calculation determines the equilibrium constant Kp for the given reaction at 1100 K using the partial pressures of CO and CO₂.

Answer 2

The calculation involves converting the given masses of reactants and products to moles, determining the moles of gases at equilibrium, calculating their partial pressures, and then determining the equilibrium constant Kp for the reaction between CO₂ and graphite producing CO.

Step-by-step explanation:

The question asks to calculate the equilibrium constant (Kp) for the reactions, CO₂(g) + C(s) ⇌ 2CO(g), using the given data from the reaction where CO₂ and graphite produce CO and BaCO₃ as a result of passing the gaseous mixture into an aqueous solution of Ba(OH)₂.

To do this, we need to use the relationship between the moles of BaCO₃ formed and the moles of CO₂ reacted, which are stoichiometrically equivalent in a 1:1 ratio, due to the reaction Ba(OH)₂ + CO₂ ⇌ BaCO₃ + H₂O.

Initially, we can convert grams of CO₂ and BaCO₃ to moles using their molar masses, which are approximately 44.01 g/mol for CO₂ and 197.34 g/mol for BaCO₃.

From the initial moles of CO₂ and the moles of BaCO₃ formed, we can determine the moles of CO₂ remaining, and therefore, the moles of CO produced.

Since the volume of the container is given as 1.000 L, we can then calculate the partial pressures of CO₂ and CO at equilibrium.

The equilibrium constant Kp is then calculated using the relationship Kp = (PCO)^2 / PCO2, where P represents the partial pressures of the respective gases.

It is important to remember that, for a pure solid like graphite, its activity is taken as one and not included in the expression for Kp.