Question

Calcium carbonate decomposes at high temperatures to give calcium oxide and carbon dioxide as shown below. CaCO3(s) CaO(s) + CO2(g) The KP for this reaction is 1.16 at 800°C. A 5.00 L vessel containing 10.0 g of CaCO3(s) was evacuated to remove the air, sealed, and then heated to 800°C. Ignoring the volume occupied by the solid, what will be the mass of the solid in the vessel once equilibrium is reached?

Answer 1

To calculate the mass of the solid CaCO3 in the vessel once equilibrium is reached, we need to use the equilibrium constant (K) and the given information. The mass of the solid CaCO3 can be calculated by subtracting the mass of the solid CaO from the initial mass of CaCO3. The molar concentration of CaCO3 can be calculated using the given mass and volume, and the equilibrium constant can be used to determine the concentrations of CaO and CO2.

Step-by-step explanation:

To calculate the mass of the solid CaCO3 in the vessel once equilibrium is reached, we need to use the equilibrium constant (K) and the given information. The equilibrium constant (K) is 1.16 at 800 °C. We are given a 5.00 L vessel containing 10.0 g of CaCO3(s) and the volume occupied by the solid is ignored. To calculate the mass of the solid in the vessel, we can use the equation n = m/M to calculate the number of moles of CaCO3 and then multiply it by the molar mass of CaCO3 to get the mass.

Given that the volume of the vessel is 5.00 L, we can calculate the initial concentration of CaCO3 (Molar concentration = moles/volume) as [CaCO3] = (10.0g / 100.09 g/mol) / 5.00 L = 0.02 mol/L.

Now we can use the equation Q = [CaO][CO2] / [CaCO3] = 1.16 (as it's at equilibrium) and substitute the concentrations as 0.02 mol/L and calculate the concentration of [CaO] and [CO2]. Since the volume occupied by the solid is ignored, the concentrations of CaO and CO2 will be the same. Now we can multiply the concentration of CaO by the volume to get the moles of CaO. Finally, multiply the moles of CaO by its molar mass to calculate the mass of the solid CaO.

Therefore, the mass of the solid CaCO3 in the vessel once equilibrium is reached is the initial mass of CaCO3 minus the mass of the solid CaO, which is calculated above.

Answer 2

`\boxed{\text{10.0 g}}`

Step-by-step explanation:

The balanced equation is

CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = 1.16

Calculate Kc

T= 800 °C = 1073 K; Δn= 1

`K_{\text{p}} = K_{\text{c}}{RT}^(\Delta n)\\\\1.1 = K_(c){(RT)}^(1)\\\\K_(c) = (1.1)/(RT) = (1.1)/(8.314* 1073)= (1.1)/(8921) = 1.233 * 10^(-4)`

Equilibrium concentration of CO₂

`K_(c) = [\text{CO}_(2)] = 1.233 * 10^(-4)`

Moles of CO₂ formed

`n= \text{5.00 L } * \frac{\text{1.233 $*$ 10$^(-4)$ \text{ mol} }}{\text{1 L}} = 6.155 * 10^(-4)\text{ mol}`

Moles of CaCO₃ used up

Moles of CaCO₃ used up = moles of CO₂ formed = 6.155 × 10⁻⁴ mol

Mass of CaCO₃ used up

`m = 6.155 * 10^(-4)\text{ mol } * \frac{\text{100.09 g}}{\text{1 mol}} = \text{0.0616 g}`

Moles of CaO formed

Moles of CaO formed = moles of CO₂ formed = 6.155 × 10⁻⁴ mol

Mass of CaO formed

`m = 6.155 * 10^(-4)\text{ mol } * \frac{\text{56.08 g}}{\text{1 mol}} = \text{0.0345 g}`

Mass of solid at equilibrium

m = 10.0 g – 0.0616 g + 0.0345 g = 10.0 g

`\text{The mass of solid remaining at equilibrium is } \boxed{\textbf{10.0 g}}`