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🔥 Unit 6 · Thermodynamics 🏠 Unit Hub 🗂 Flashcards 🗺 Cheat Sheet Essentials 🎨 Visual Review 📝 MC Practice ✍️ SAQ Practice

AP Chemistry Unit 6 SAQ Practice

Practice a College Board-style short-answer question on Thermodynamics. Write your response, then reveal the model answer to see exactly what earns each point.

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Short Answer Question · Unit 6 · Hess's Law & Enthalpy
A student wants to determine ΔH for the reaction: 2C(s) + H₂(g) → C₂H₂(g) (the formation of acetylene), but this reaction cannot be measured directly in the lab. The student is given the following two combustion reactions and their enthalpy changes:
ReactionΔH (kJ/mol)
(1) C(s) + O₂(g) → CO₂(g)−393.5
(2) 2H₂(g) + O₂(g) → 2H₂O(l)−571.6
(3) 2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(l)−2598.8
A
Explain why Hess's law allows the student to determine ΔH for the formation of acetylene using these three combustion reactions, even though the target reaction was never measured directly.

✓ Model answer (earns the point)

Because enthalpy is a state function, ΔH depends only on the initial reactants and final products, not on the specific reaction pathway used to get there. This means the target reaction's ΔH can be found by algebraically combining (adding, reversing, and/or scaling) other reactions whose ΔH values are known, as long as the combined reactions sum to the same overall equation as the target reaction.

Why it scores: Names enthalpy as a state function specifically AND explains why that property allows path-independent calculation via combination of known reactions.
B
Describe what must be done to reaction (3) in order to use it in this Hess's law calculation, and explain how this affects its ΔH value.

✓ Model answer (earns the point)

Reaction (3) must be reversed (since C₂H₂ is a reactant in reaction 3, but the target reaction needs C₂H₂ as a product) and then divided by 2 (since the target reaction forms only 1 mol of C₂H₂, while reaction 3 involves 2 mol). Reversing flips the sign of ΔH to +2598.8 kJ, and dividing by 2 halves it to +1299.4 kJ.

Why it scores: Correctly identifies BOTH operations needed (reverse AND divide by 2) AND correctly applies both the sign flip and the scaling to the ΔH value, showing the resulting number.
C
Using reactions (1), (2), and the modified version of reaction (3) from Part B, calculate ΔH for the formation of acetylene: 2C(s) + H₂(g) → C₂H₂(g).

✓ Model answer (earns the point)

The target reaction requires 2× reaction (1) [2C(s) + 2O₂(g) → 2CO₂(g), ΔH = 2 × (−393.5) = −787.0 kJ], ½× reaction (2) [H₂(g) + ½O₂(g) → H₂O(l), ΔH = ½ × (−571.6) = −285.8 kJ], and the reversed, halved reaction (3) [2CO₂(g) + H₂O(l) → C₂H₂(g) + 2.5O₂(g), ΔH = +1299.4 kJ]. Adding these three ΔH values: −787.0 + (−285.8) + 1299.4 = +226.6 kJ. So ΔH for the formation of acetylene is approximately +226.6 kJ/mol.

Why it scores: Correctly scales each reaction (and its ΔH) to match the target equation's coefficients, sums all three ΔH values with correct signs, AND arrives at a final answer with the correct sign and units.

How to score points on SAQs