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

Answer:

$\boxed{\text{10.0 g}}$

Explanation:

The balanced equation is

CaCO₃(s) ⇌ CaO(s) + CO₂(g); Kp = 1.16

Calculate Kc

T= 800 °C = 1073 K; Δn= 1

$K_{\text{p}} = K_{\text{c}}{RT}^{\Delta n}\\\\1.1 = K_{c}{(RT)}^{1}\\\\K_{c} = \dfrac{1.1}{RT} = \dfrac{1.1}{8.314\times 1073}= \dfrac{1.1}{8921} = 1.233 \times 10^{-4}$

Equilibrium concentration of CO₂

$K_{c} = [\text{CO}_{2}] = 1.233 \times 10^{-4}$

Moles of CO₂ formed

$n= \text{5.00 L } \times \dfrac{\text{1.233 \times 10 ^{-4} \text{ mol} }}{\text{1 L}} = 6.155 \times 10^{-4}\text{ mol}$

Moles of CaCO₃ used up

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

Mass of CaCO₃ used up

$m = 6.155 \times 10^{-4}\text{ mol } \times \dfrac{\text{100.09 g}}{\text{1 mol}} = \text{0.0616 g}$

Moles of CaO formed

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

Mass of CaO formed

$m = 6.155 \times 10^{-4}\text{ mol } \times \dfrac{\text{56.08 g}}{\text{1 mol}} = \text{0.0345 g}$

Mass of solid at equilibrium

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

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