At 850°C, CaCO3 undergoes substantial decomposition to yield CaO and CO2. Assuming that the ΔH o f values of the reactant and products are the same at 850°C as they are at 25°C, calculate the enthalpy change (in kJ) if 68.10 g of CO2 is produced in one reaction.

The enthalpy if 68.10 grams of CO2 is produced is - 189.04 kJ

Explanation:

Step 1: Data given

temperature = 850 °C

Mass of 68.10 grams of CO2

ΔH°f (CaO) = - 635.6 kJ/mol

ΔH°f (CO2) = - 693.5 kJ/mol

ΔH°f (CaCO3) = -1206.9 kJ/mol

Step 2: The balanced equation

CaCO3 (s) → CaO (s) + CO2 (g)

Step 3: Calculate ΔH°reaction

ΔH°reaction = ΣΔH°f (products) - ΣΔH°f (reactants)

ΔH°reaction = (ΔH°f (CaO) + ΔH°f (CO2)) - ΔH°f (CaCO3)

ΔH°reaction = (-635.6 kJ/mol + - 693.5 kJ/mol) + 1206.9 kJ/mol

ΔH°reaction = - 122.2 kJ / mol

Step 4: Calculate moles of CO2

Moles CO2 = mass CO2 / Molar mass CO2

Moles CO2 = 68.10 grams / 44.01 g/mol

Moles CO2 = 1.547 moles

Step 5: Calculate the enthalpy change for 68.10 grams of CO2

-122.2 kJ/mol * 1.547 moles = - 189.04 kJ

The enthalpy if 68.10 grams of CO2 is produced is - 189.04 kJ