Multiple Choice
Using average bond energies and the standard enthalpy of formation of C(g) (718.4 kJ/mol), estimate the standard enthalpy of formation of gaseous benzene, C6H6(g). Which of the following values is closest to the correct estimate?
8. Thermochemistry
Enthalpy of Formation
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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
Verified step by step guidance1
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.
Recall the formula for calculating the standard enthalpy change of a reaction (ΔH°rxn): ΔH°rxn = ΣΔH°f(products) - ΣΔH°f(reactants).
Identify the reactants and products in the given chemical reaction: Reactants are CaO(s) and CO2(g), and the product is CaCO3(s).
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.
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.
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