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AP Chemistry Practice Quiz: Enthalpy of Formation

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

Given the following standard enthalpies of formation (ΔHf°), calculate the standard enthalpy change (ΔH°reaction) for the combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). ΔHf° [CH₄(g)] = -74.8 kJ/mol ΔHf° [CO₂(g)] = -393.5 kJ/mol ΔHf° [H₂O(g)] = -241.8 kJ/mol

All Questions (7)

Given the following standard enthalpies of formation (ΔHf°), calculate the standard enthalpy change (ΔH°reaction) for the combustion of methane: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). ΔHf° [CH₄(g)] = -74.8 kJ/mol ΔHf° [CO₂(g)] = -393.5 kJ/mol ΔHf° [H₂O(g)] = -241.8 kJ/mol

A) -802.3 kJ/mol

B) +802.3 kJ/mol

C) -560.5 kJ/mol

D) +560.5 kJ/mol

Correct Answer: A

The standard enthalpy of reaction is calculated using the formula ΔH°reaction = ΣΔHf°products − ΣΔHf°reactants. Remember that the standard enthalpy of formation for an element in its standard state, like O₂(g), is 0 kJ/mol. ΔH°reaction = [ΔHf°(CO₂) + 2×ΔHf°(H₂O)] - [ΔHf°(CH₄) + 2×ΔHf°(O₂)] ΔH°reaction = [(-393.5 kJ/mol) + 2(-241.8 kJ/mol)] - [(-74.8 kJ/mol) + 2(0 kJ/mol)] ΔH°reaction = [-393.5 - 483.6] - [-74.8] ΔH°reaction = [-877.1] - [-74.8] = -802.3 kJ/mol.

The combustion of liquid ethanol, C₂H₅OH(l), is a highly exothermic reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). The standard enthalpy change for this reaction is -1367 kJ/mol. Using the provided standard enthalpies of formation, calculate the standard enthalpy of formation for liquid ethanol, ΔHf°[C₂H₅OH(l)]. ΔHf° [CO₂(g)] = -393.5 kJ/mol ΔHf° [H₂O(l)] = -285.8 kJ/mol

A) +277.6 kJ/mol

B) -277.6 kJ/mol

C) -687.7 kJ/mol

D) +1367 kJ/mol

Correct Answer: B

Use the formula ΔH°reaction = ΣΔHf°products − ΣΔHf°reactants and solve for the unknown ΔHf° of the reactant. Let x = ΔHf°[C₂H₅OH(l)]. ΔH°reaction = [2×ΔHf°(CO₂) + 3×ΔHf°(H₂O)] - [ΔHf°(C₂H₅OH) + 3×ΔHf°(O₂)] -1367 kJ/mol = [2(-393.5) + 3(-285.8)] - [x + 3(0)] -1367 = [-787.0 - 857.4] - x -1367 = -1644.4 - x x = -1644.4 + 1367 = -277.4 kJ/mol. The closest answer is -277.6 kJ/mol, accounting for minor rounding differences in standard values.

Which statement correctly describes a reaction with a negative standard enthalpy of reaction (ΔH°reaction < 0) based on the principles of standard enthalpies of formation?

A) The sum of the standard enthalpies of formation of the products is greater than that of the reactants, and the reaction is endothermic.

B) The sum of the standard enthalpies of formation of the products is less than that of the reactants, and the reaction is exothermic.

C) The sum of the standard enthalpies of formation of the products is greater than that of the reactants, and the reaction is exothermic.

D) The sum of the standard enthalpies of formation of the products is less than that of the reactants, and the reaction is endothermic.

Correct Answer: B

The formula is ΔH°reaction = ΣΔHf°products − ΣΔHf°reactants. If ΔH°reaction is negative, the reaction is exothermic. For the result to be negative, the value of ΣΔHf°products must be less than (more negative or less positive than) the value of ΣΔHf°reactants. This indicates that the products are in a lower enthalpy state (more stable) than the reactants.

Calculate the standard enthalpy change for the synthesis of ammonia from its elements, as shown in the reaction below. Note the states of matter. N₂(g) + 3H₂(g) → 2NH₃(g) ΔHf° [NH₃(g)] = -46.1 kJ/mol

A) -46.1 kJ/mol

B) +46.1 kJ/mol

C) -92.2 kJ/mol

D) +92.2 kJ/mol

Correct Answer: C

The standard enthalpy of reaction is ΔH°reaction = ΣΔHf°products − ΣΔHf°reactants. The standard enthalpies of formation for elements in their standard states, N₂(g) and H₂(g), are zero. ΔH°reaction = [2×ΔHf°(NH₃)] - [1×ΔHf°(N₂) + 3×ΔHf°(H₂)] ΔH°reaction = [2(-46.1 kJ/mol)] - [1(0) + 3(0)] ΔH°reaction = -92.2 kJ/mol. This value represents the enthalpy change for the formation of 2 moles of NH₃.

The standard enthalpy of formation (ΔHf°) is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. Which of the following reactions corresponds to the standard enthalpy of formation of liquid water, H₂O(l)?

A) 2H₂(g) + O₂(g) → 2H₂O(l)

B) H₂(g) + ½O₂(g) → H₂O(l)

C) H₂O(g) → H₂O(l)

D) H⁺(aq) + OH⁻(aq) → H₂O(l)

Correct Answer: B

The definition of standard enthalpy of formation requires two conditions: 1) one mole of the product is formed, and 2) the reactants are elements in their standard states. Option B is the only one that meets both criteria: it forms exactly one mole of H₂O(l) from the elements hydrogen (H₂) and oxygen (O₂) in their gaseous standard states.

The thermite reaction is highly exothermic: 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(s). Calculate the amount of heat released when 10.0 g of aluminum (Al) reacts completely with excess iron(III) oxide (Fe₂O₃). (Molar mass of Al = 27.0 g/mol) ΔHf° [Fe₂O₃(s)] = -824.2 kJ/mol ΔHf° [Al₂O₃(s)] = -1675.7 kJ/mol

A) -851.5 kJ

B) -425.8 kJ

C) -315.4 kJ

D) -157.7 kJ

Correct Answer: D

First, calculate the standard enthalpy change for the reaction as written (for 2 moles of Al). ΔH°reaction = ΣΔHf°products − ΣΔHf°reactants. ΔH°reaction = [ΔHf°(Al₂O₃) + 2×ΔHf°(Fe)] - [2×ΔHf°(Al) + ΔHf°(Fe₂O₃)] ΔH°reaction = [(-1675.7) + 2(0)] - [2(0) + (-824.2)] = -1675.7 + 824.2 = -851.5 kJ. This is the heat released per 2 moles of Al. Next, convert the mass of Al to moles: 10.0 g Al / 27.0 g/mol = 0.370 mol Al. Finally, use stoichiometry to find the heat released for this amount: 0.370 mol Al × (-851.5 kJ / 2 mol Al) = -157.7 kJ.

Calculate the standard enthalpy change (ΔH°reaction) for the reaction: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g). ΔHf° [NH₃(g)] = -46.1 kJ/mol ΔHf° [NO(g)] = +90.3 kJ/mol ΔHf° [H₂O(g)] = -241.8 kJ/mol

A) +905.2 kJ/mol

B) -197.6 kJ/mol

C) -905.2 kJ/mol

D) -105.4 kJ/mol

Correct Answer: C

Apply the formula ΔH°reaction = ΣΔHf°products − ΣΔHf°reactants, being careful to multiply each enthalpy of formation by its stoichiometric coefficient. Products: [4×ΔHf°(NO) + 6×ΔHf°(H₂O)] = [4(+90.3) + 6(-241.8)] = [361.2 - 1450.8] = -1089.6 kJ Reactants: [4×ΔHf°(NH₃) + 5×ΔHf°(O₂)] = [4(-46.1) + 5(0)] = -184.4 kJ ΔH°reaction = (-1089.6 kJ) - (-184.4 kJ) = -1089.6 + 184.4 = -905.2 kJ/mol.