# Hess's Law

> AP Chemistry · Unit 6 Thermodynamics
> Source: https://www.owlsprep.com/study/ap-chemistry-u6-hess-s-law/

This subtopic covers Hess's law of constant heat summation, core reaction manipulation rules, and three common AP Chemistry applications: calculating ΔH from source reactions, standard enthalpies of formation, and average bond enthalpies.

**Prerequisites:** Enthalpy change (ΔH) definition and sign conventions for endo/exothermic reactions; Difference between state functions and path functions; Stoichiometric relationships between moles and energy change

## Learning objectives

- Define Hess's law of constant heat summation
- Manipulate reaction enthalpies to calculate ΔH for unknown reactions
- Calculate standard reaction enthalpy from standard enthalpies of formation
- Estimate reaction enthalpy from average bond enthalpies
- Avoid common exam pitfalls in Hess's law problems

## What is Hess's Law?

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.

**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

## Manipulating Source Reactions to Calculate ΔH

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:

1. **Reverse a reaction**: Reversing a reaction flips the direction of energy flow, so the sign of ΔH is flipped.
2. **Scale stoichiometry**: Multiplying/dividing all coefficients by a constant requires scaling ΔH by the same constant, since enthalpy is an extensive property.
3. **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:
1. $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_1 = -393.5 \text{ kJ/mol}$
2. $2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) \quad \Delta H_2 = -571.6 \text{ kJ/mol}$
3. $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H_3 = -890.3 \text{ kJ/mol}$

Calculate ΔH for the target reaction: $\text{C}(s) + 2\text{H}_2(g) \rightarrow \text{CH}_4(g)$

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.

## ΔH from Standard Enthalpies of Formation

A common AP exam application of Hess's law is calculating the standard enthalpy of reaction from standard enthalpies of formation. By definition, $\Delta H^\circ_f = 0$ for any element in its standard state.

**Standard Enthalpy of Formation** — 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).

*Notation:* $\Delta H^\circ_f$

*Example:* $\Delta H^\circ_f = 0$ 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 $\Delta H^\circ_f$), (2) combine elements to form products (sum of product $\Delta H^\circ_f$). 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: $\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)$. Use the following data: $\Delta H^\circ_f (\text{C}_3\text{H}_8(g)) = -103.8 \text{ kJ/mol}$, $\Delta H^\circ_f (\text{CO}_2(g)) = -393.5 \text{ kJ/mol}$, $\Delta H^\circ_f (\text{H}_2\text{O}(g)) = -241.8 \text{ kJ/mol}$, $\Delta H^\circ_f (\text{O}_2(g)) = 0 \text{ kJ/mol}$.

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.

## ΔH from Average Bond Enthalpies

Another common application of Hess's law is estimating $\Delta H_{rxn}$ 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:

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

> **note**
>
> Bond enthalpies are average values over many different compounds, so ΔH calculated from bond enthalpies is always approximate, unlike the exact value obtained from standard enthalpies of formation.

**Worked example:** Estimate ΔH for the hydrogenation of gaseous ethene to form ethane: $\text{C}_2\text{H}_4(g) + \text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g)$. 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.

## Common pitfalls

- **Wrong:** Forgetting to change the sign of ΔH when reversing a source reaction
  - Why it fails: Students often adjust stoichiometry correctly but forget reversing a reaction flips energy flow.
  - Correct: Immediately flip the sign of ΔH after reversing a reaction, and write it down before moving to the next step.
- **Wrong:** Subtracting product enthalpies from reactant enthalpies when calculating ΔH from enthalpies of formation
  - Why it fails: The 'products minus reactants' rule is easy to flip under test pressure.
  - Correct: Write the full formula $\Delta H = \sum \text{products} - \sum \text{reactants}$ at the top of your work before plugging in values.
- **Wrong:** Not scaling ΔH proportionally when scaling stoichiometric coefficients of a source reaction
  - Why it fails: Students remember to change the equation but forget ΔH is an extensive property that depends on the amount of reactant.
  - Correct: Multiply ΔH by the same scaling factor immediately after adjusting the reaction coefficients.
- **Wrong:** Counting extra or too few bonds in bond enthalpy calculations
  - Why it fails: Students often count all bonds instead of only those that change, or miscount C-H bonds in hydrocarbons.
  - Correct: Draw full Lewis structures for all molecules, and cross off bonds that are unchanged on both sides to simplify counting.
- **Wrong:** Assuming ΔHf of any form of an element is zero
  - Why it fails: Students memorize that oxygen has ΔHf = 0, but forget this only applies to the element in its standard state.
  - Correct: Confirm any element is in its standard state (most stable form at 1 atm/25°C) before setting ΔHf to zero.
- **Wrong:** Summing ΔH values before confirming the net reaction matches the target
  - Why it fails: Students rush to get a numerical answer and miss incorrect intermediate cancellation.
  - Correct: After adding all modified reactions, confirm the net reaction matches the target exactly before summing ΔH values.

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Core Hess's Law | Total $\Delta H = \sum$ modified $\Delta H$ of source reactions | Enthalpy is a state function, so path does not affect total ΔH |
| Reverse a reaction | $\Delta H_{\text{reverse}} = -\Delta H_{\text{forward}}$ | Flipping reaction direction flips energy flow |
| Scale a reaction | $\Delta H_{\text{scaled}} = c \times \Delta H_{\text{original}}$ | ΔH is extensive, scales with moles of reaction |
| ΔH from Enthalpies of Formation | $\Delta H^\circ_{rxn} = \sum n \Delta H^\circ_f (\text{products}) - \sum m \Delta H^\circ_f (\text{reactants})$ | $\Delta H^\circ_f = 0$ for elements in their standard states |
| ΔH from Bond Enthalpies | $\Delta H_{rxn} = \sum (\text{bonds broken}) - \sum (\text{bonds formed})$ | 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 |

## 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.

- [Unit 6 Thermodynamics Overview](https://www.owlsprep.com/study/ap-chemistry-u6-overview/)
- [Equilibrium Overview](https://www.owlsprep.com/study/ap-chemistry-u7-overview/)
- [Introduction to Equilibrium](https://www.owlsprep.com/study/ap-chemistry-u7-introduction-to-equilibrium/)

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