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How does Hess's law let us combine known reactions to find the enthalpy of a reaction we cannot measure directly?

Topic 6.9 Hess's Law: use Hess's law to determine the enthalpy of a reaction by combining the enthalpies of a series of reactions that add to the target, reversing and scaling as needed.

A focused answer to AP Chemistry Topic 6.9, covering Hess's law, the additivity of enthalpy as a state function, and how to reverse, scale and add reactions to find an unknown enthalpy of reaction, with full worked examples.

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  1. What this topic is asking
  2. Hess's law and state functions
  3. The two allowed operations
  4. The procedure
  5. Try this

What this topic is asking

The College Board (Topic 6.9) wants you to use Hess's law to find the enthalpy of a reaction by combining a series of known reactions that add up to the target, reversing and scaling them as needed. This is the most procedural of the thermochemistry topics: a careful exercise in manipulating equations and their enthalpies.

Hess's law and state functions

The consequence is that you can compute the enthalpy of a hard-to-measure reaction by routing through reactions you do know. The energy of the overall change is fixed by the start and end points, so any valid set of steps that connects them gives the right answer.

The two allowed operations

These two operations are all you need. Reverse a reaction to move a species from reactant to product (or vice versa); scale it so the coefficients match the target. After manipulating, add the reactions: species that appear on opposite sides in equal amounts cancel, and what remains should be exactly the target.

The procedure

Work backwards from the target. For each species in the target, find a known reaction containing it and decide whether to reverse and/or scale that reaction so the species ends up on the correct side with the right coefficient. Once all the known reactions are manipulated, add them and check that the intermediates cancel to leave the target. Finally, add the manipulated enthalpies (with their reversed signs and scaling) to get ΔH\Delta H of the target. A common shortcut is that if you have all the relevant formation reactions, this reduces to the products-minus-reactants formula of Topic 6.8.

Try this

Q1. A reaction with ΔH=−150 kJ\Delta H = -150\ \text{kJ} is reversed and halved in a Hess's law sum. State its contribution to the total. [2 points]

  • Cue. Reverse: +150+150; halve: +75 kJ+75\ \text{kJ}.

Q2. Explain why Hess's law lets us find the enthalpy of a reaction that is hard to measure directly. [2 points]

  • Cue. Enthalpy is a state function, so combining measurable reactions that sum to the target gives the same enthalpy change as the direct reaction.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). Determine ΔH\Delta H for C(s)+12O2(g)→CO(g)\text{C}(s) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g) using: Reaction 1: C(s)+O2(g)→CO2(g)\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g), ΔH1=−394 kJ\Delta H_1 = -394\ \text{kJ}; Reaction 2: CO(g)+12O2(g)→CO2(g)\text{CO}(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g), ΔH2=−283 kJ\Delta H_2 = -283\ \text{kJ}. (a) State which reaction must be reversed and why. (b) Write the manipulated reactions with their enthalpies. (c) Add them to obtain the target and calculate ΔH\Delta H. (d) Justify why Hess's law is valid.
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A 4-point quantitative FRQ on Hess's law.

(a) Reverse (1 point): reaction 2 must be reversed so that CO appears as a product (it is a product in the target but a reactant in reaction 2).
(b) Manipulated reactions (1 point): keep reaction 1 as is (ΔH=−394\Delta H = -394); reverse reaction 2 to CO2(g)→CO(g)+12O2(g)\text{CO}_2(g) \rightarrow \text{CO}(g) + \tfrac{1}{2}\text{O}_2(g), ΔH=+283 kJ\Delta H = +283\ \text{kJ}.
(c) Add and calculate (1 point): adding, CO2\text{CO}_2 cancels and 12O2\tfrac{1}{2}\text{O}_2 partly cancels, leaving C(s)+12O2(g)→CO(g)\text{C}(s) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g); ΔH=−394+283=−111 kJ\Delta H = -394 + 283 = -111\ \text{kJ}.
(d) Justify (1 point): enthalpy is a state function, so the total enthalpy change depends only on the initial and final states; any combination of steps that sums to the target gives the same ΔH\Delta H.

Markers reward identifying the reversal, the manipulated enthalpies, the summed target with its ΔH\Delta H, and the state-function justification.

AP 2021 (style)1 marksSection I (multiple choice). When a reaction in a Hess's law sum is reversed and doubled, its enthalpy change is (A) unchanged (B) reversed in sign only (C) doubled only (D) reversed in sign and doubled. Justify your choice.
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A 1-point conceptual MCQ. The answer is (D).

Reversing a reaction reverses the sign of ΔH\Delta H; scaling it by a factor multiplies ΔH\Delta H by that factor. Doing both reverses the sign and doubles the magnitude. The trap is applying only one of the two operations.

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