Hess's Law

AP Chemistry· AP Chemistry CED — Thermodynamics· 14 min read

1. What is Hess's Law?★★☆☆☆⏱ 3 min

Hess's law (full name: Hess's law of constant heat summation) states that the total enthalpy change for a chemical reaction is independent of the path taken between initial reactants and final products, and depends only on the enthalpy difference between reactants and products. This is a direct consequence of enthalpy being a state function, a core principle of thermodynamics.

The law allows us to calculate ΔH for reactions that cannot be measured directly in a lab, such as reactions with very high activation energy or competing side reactions, by combining ΔH values from other known reactions. On the AP Chemistry exam, Hess's law accounts for ~3-5% of your total score, appearing in both multiple-choice and free-response sections.

📘 Definition

Hess's Law of Constant Heat Summation

The total enthalpy change for a chemical reaction depends only on the difference in enthalpy between reactants and products, and is independent of the reaction path taken.

Example:

Used to calculate ΔH for methane formation, which cannot be measured directly experimentally

2. Manipulating Source Reactions to Calculate ΔH★★★☆☆⏱ 5 min

To solve a basic Hess's law problem, you start with a target reaction (whose ΔH you need to find) and a set of source reactions with known ΔH values. Three valid manipulations are allowed, each with a corresponding change to ΔH:

Reverse a reaction: Reversing a reaction flips the direction of energy flow, so the sign of ΔH is flipped.
Scale stoichiometry: Multiplying/dividing all coefficients by a constant requires scaling ΔH by the same constant, since enthalpy is an extensive property.
Add reactions: After modification, add reactions and cancel common intermediate species that do not appear in the target. Sum the modified ΔH values to get the total ΔH.

📐 Worked Example

Given the following reactions with known enthalpies:

$C (s) + O_{2} (g) \to CO_{2} (g) Δ H_{1} = - 393.5 kJ/mol$
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) Δ H_{2} = - 571.6 kJ/mol$
$CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (l) Δ H_{3} = - 890.3 kJ/mol$

Calculate ΔH for the target reaction: $C (s) + 2 H_{2} (g) \to CH_{4} (g)$

1
Match species to the target: C(s) and H₂(g) are reactants, matching reactions 1 and 2 as written. CH₄(g) is a product but is a reactant in reaction 3, so reverse reaction 3.
2
Reverse reaction 3 and flip the sign of ΔH₃:
$CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \quad -\Delta H_3 = +890.3 \text{ kJ/mol}$
3
Add all modified reactions together:
$C(s) + O_2(g) + 2H_2(g) + O_2(g) + CO_2(g) + 2H_2O(l) \rightarrow CO_2(g) + 2H_2O(l) + CH_4(g) + 2O_2(g)$
4
Cancel common species on both sides: 2 mol O₂, 1 mol CO₂, and 2 mol H₂O cancel completely, leaving the target reaction.
5
Sum the modified ΔH values:
$\Delta H = -393.5 - 571.6 + 890.3 = -74.8 \text{ kJ/mol}$

Exam tip:

When checking for cancellation, always cross off one mole of a species on the left for every one mole on the right. If you end up with a partial mole of an intermediate left over, you likely forgot to scale a source reaction to match the target stoichiometry.

3. ΔH from Standard Enthalpies of Formation★★★☆☆⏱ 4 min

A common AP exam application of Hess's law is calculating the standard enthalpy of reaction from standard enthalpies of formation. By definition, $Δ H_{f}^{\circ} = 0$ for any element in its standard state.

📘 Definition

Standard Enthalpy of Formation

$Δ H_{f}^{\circ}$

Enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states (1 atm, 25°C, most stable form).

Example:

$Δ H_{f}^{\circ} = 0$ for $O_{2} (g)$ , but not for $O (g)$ or $O_{3} (g)$

Using Hess's law, any reaction can be broken into two steps: (1) decompose all reactants into their constituent elements (reverse of formation, so ΔH is negative sum of reactant $Δ H_{f}^{\circ}$ ), (2) combine elements to form products (sum of product $Δ H_{f}^{\circ}$ ). This gives the shortcut formula:

\Delta H^\circ_{rxn} = \sum n \Delta H^\circ_f (\text{products}) - \sum m \Delta H^\circ_f (\text{reactants})

📐 Worked Example

Calculate the standard enthalpy of reaction for the combustion of propane: $C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 CO_{2} (g) + 4 H_{2} O (g)$ . Use the following data: $Δ H_{f}^{\circ} (C_{3} H_{8} (g)) = - 103.8 kJ/mol$ , $Δ H_{f}^{\circ} (CO_{2} (g)) = - 393.5 kJ/mol$ , $Δ H_{f}^{\circ} (H_{2} O (g)) = - 241.8 kJ/mol$ , $Δ H_{f}^{\circ} (O_{2} (g)) = 0 kJ/mol$ .

1
Calculate the sum of enthalpies of formation for products, multiplied by stoichiometric coefficients:
$\sum n \Delta H^\circ_f (\text{products}) = (3 \times -393.5) + (4 \times -241.8) = -1180.5 - 967.2 = -2147.7 \text{ kJ}$
2
Calculate the sum of enthalpies of formation for reactants:
$\sum m \Delta H^\circ_f (\text{reactants}) = (1 \times -103.8) + (5 \times 0) = -103.8 \text{ kJ}$
3
Subtract reactant sum from product sum per the formula:
$\Delta H^\circ_{rxn} = (-2147.7) - (-103.8) = -2043.9 \text{ kJ}$
4
Verify the result: Combustion of a hydrocarbon is exothermic, so the negative sign matches expectations.

Exam tip:

Always remember the order is products minus reactants, not the reverse. It is easy to mix up the order under test pressure, so write the formula down before plugging in any values.

4. ΔH from Average Bond Enthalpies★★★★☆⏱ 3 min

Another common application of Hess's law is estimating $Δ H_{r x n}$ from average bond enthalpies. A bond enthalpy is the energy required to break 1 mole of a specific covalent bond in the gaseous state. Breaking bonds is always endothermic (ΔH positive), and forming bonds is always exothermic (ΔH negative).

Using Hess's law, we split the reaction into two steps: break all bonds in reactants to form gaseous atoms, then form all bonds in products from the atoms. This gives the shortcut formula:

\Delta H_{rxn} = \sum (\text{Bond enthalpies of bonds broken}) - \sum (\text{Bond enthalpies of bonds formed})

📐 Worked Example

Estimate ΔH for the hydrogenation of gaseous ethene to form ethane: $C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g)$ . Use the following average bond enthalpies (kJ/mol): C=C = 614, C-H = 413, H-H = 436, C-C = 348.

1
Draw Lewis structures to count bonds broken and formed: Bonds broken (reactants): 1 C=C, 4 C-H, 1 H-H. Bonds formed (products): 1 C-C, 6 C-H.
2
Calculate total energy required to break bonds:
$(1 \times 614) + (4 \times 413) + (1 \times 436) = 614 + 1652 + 436 = 2702 \text{ kJ/mol}$
3
Calculate total bond enthalpy for bonds formed:
$(1 \times 348) + (6 \times 413) = 348 + 2478 = 2826 \text{ kJ/mol}$
4
Apply the formula to get ΔH:
$\Delta H = 2702 - 2826 = -124 \text{ kJ/mol}$
5
The negative sign confirms hydrogenation is exothermic, which matches experimental results.

Exam tip:

Always confirm all species are gaseous when using bond enthalpies. If any species is liquid or solid, you must add the enthalpy of phase change to get the correct total ΔH.

5. Common Pitfalls

Wrong move:

Forgetting to change the sign of ΔH when reversing a source reaction

Why:

Students often adjust stoichiometry correctly but forget reversing a reaction flips energy flow.

Correct move:

Immediately flip the sign of ΔH after reversing a reaction, and write it down before moving to the next step.

Wrong move:

Subtracting product enthalpies from reactant enthalpies when calculating ΔH from enthalpies of formation

Why:

The 'products minus reactants' rule is easy to flip under test pressure.

Correct move:

Write the full formula $Δ H = \sum products - \sum reactants$ at the top of your work before plugging in values.

Wrong move:

Not scaling ΔH proportionally when scaling stoichiometric coefficients of a source reaction

Why:

Students remember to change the equation but forget ΔH is an extensive property that depends on the amount of reactant.

Correct move:

Multiply ΔH by the same scaling factor immediately after adjusting the reaction coefficients.

Wrong move:

Counting extra or too few bonds in bond enthalpy calculations

Why:

Students often count all bonds instead of only those that change, or miscount C-H bonds in hydrocarbons.

Correct move:

Draw full Lewis structures for all molecules, and cross off bonds that are unchanged on both sides to simplify counting.

Wrong move:

Assuming ΔHf of any form of an element is zero

Why:

Students memorize that oxygen has ΔHf = 0, but forget this only applies to the element in its standard state.

Correct move:

Confirm any element is in its standard state (most stable form at 1 atm/25°C) before setting ΔHf to zero.

Wrong move:

Summing ΔH values before confirming the net reaction matches the target

Why:

Students rush to get a numerical answer and miss incorrect intermediate cancellation.

Correct move:

After adding all modified reactions, confirm the net reaction matches the target exactly before summing ΔH values.

6. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Core Hess's Law	Total $Δ H = \sum$ modified $Δ H$ of source reactions	Enthalpy is a state function, so path does not affect total ΔH
Reverse a reaction	$Δ H_{reverse} = - Δ H_{forward}$	Flipping reaction direction flips energy flow
Scale a reaction	$Δ H_{scaled} = c \times Δ H_{original}$	ΔH is extensive, scales with moles of reaction
ΔH from Enthalpies of Formation	$Δ H_{r x n}^{\circ} = \sum n Δ H_{f}^{\circ} (products) - \sum m Δ H_{f}^{\circ} (reactants)$	$Δ H_{f}^{\circ} = 0$ for elements in their standard states
ΔH from Bond Enthalpies	$Δ H_{r x n} = \sum (bonds broken) - \sum (bonds formed)$	Only applies to all gaseous species; results are approximate
Standard Enthalpy of Formation	Enthalpy change to form 1 mole of compound from elements in standard states	Standard state = 1 atm, 25°C, most stable form of the element
Intermediate Cancellation	All intermediates (species not in target) must cancel completely	If intermediates remain, your reaction manipulations are incorrect

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Calculate ΔH from given reaction data
2022 · FRQ
ΔH from enthalpies of formation
2021 · MCQ
Bond enthalpy ΔH estimation

Going deeper

unit overviewAP Chemistry Unit 6 Thermodynamics Overview

What's Next

Mastering Hess's law is a critical foundation for all subsequent thermodynamics topics on the AP Chemistry exam, from entropy and Gibbs free energy to thermochemical calculations in equilibrium problems. The principles you learned here—manipulating state function values based on reaction path independence—will reappear when you calculate entropy changes and Gibbs free energy of reaction, so it is important to solidify these calculation skills now. Hess's law questions are consistently high-weight on both MCQ and FRQ, so practicing the manipulation rules and avoiding common pitfalls will directly boost your exam score.