Using the standard enthalpies of formation, calculate ΔH°rxn for the reaction: CaO(s) + CO2(g) → CaCO3(s). Given: ΔH°f [CaO(s)] = -635.1 kJ/mol, ΔH°f [CO2(g)] = -393.5 kJ/mol, ΔH°f [CaCO3(s)] = -1206.9 kJ/mol.

• A. -178.3 kJ/mol
• B. -200.0 kJ/mol
• C. -250.0 kJ/mol
• D. -150.0 kJ/mol

Answer: (A)

Verified step by step guidance

1. Understand the concept of standard enthalpy of formation (ΔH°f), which is the change in enthalpy when one mole of a compound is formed from its elements in their standard states.
2. Recall the formula for calculating the standard enthalpy change of a reaction (ΔH°rxn): ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants).
3. Identify the reactants and products in the given chemical reaction: Reactants are CaO(s) and CO2(g), and the product is CaCO3(s).
4. Apply the formula: Calculate the sum of the standard enthalpies of formation for the products and subtract the sum of the standard enthalpies of formation for the reactants. Use the given values: ΔH°f [CaCO3(s)] = -1206.9 kJ/mol, ΔH°f [CaO(s)] = -635.1 kJ/mol, ΔH°f [CO2(g)] = -393.5 kJ/mol.
5. Perform the calculation: ΔH°rxn = (-1206.9 kJ/mol) - [(-635.1 kJ/mol) + (-393.5 kJ/mol)]. This will give you the standard enthalpy change for the reaction.