Hess's law — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Definition of Hess's law of constant heat summation, manipulation of reaction enthalpies, calculation of reaction ΔH via rearranging given reactions, and application to enthalpies of formation and bond enthalpies for AP exam questions.

You should already know: Enthalpy change (ΔH) definition and sign conventions for endo/exothermic reactions, stoichiometric relations between moles and energy change, the difference between state functions and path functions.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. 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 difference in enthalpy between reactants and products. This is a direct consequence of enthalpy being a state function, a core property of thermodynamics. On the AP Chemistry Course and Exam Description (CED), Hess's law is a core skill in Unit 6 (Thermodynamics), accounting for approximately 15-20% of the unit's exam weight, which translates to roughly 3-5% of the total AP Chemistry exam score. Hess's law questions appear regularly on both multiple-choice (MCQ) and free-response (FRQ) sections: MCQ typically asks for calculation of ΔH from given reaction data, while FRQ often requires step-by-step manipulation of reactions and connection of ΔH results to bonding or energy topics. 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 for other known reactions.

2. Manipulating Source Reactions to Calculate ΔH

The most fundamental problem-solving approach for Hess's law starts with a target reaction (whose ΔH we need to find) and a set of source reactions with known ΔH values. Three valid manipulations can be applied to source reactions, 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 (exothermic becomes endothermic and vice versa).
2. Scale stoichiometry: Multiplying or dividing all coefficients by a constant requires scaling ΔH by the same constant, because enthalpy is an extensive property proportional to the amount of substance reacting.
3. Add reactions: After modification, add the reactions together and cancel common species (intermediates that do not appear in the target) from both sides. The total ΔH is the sum of the modified ΔH values.

Worked Example

Given the following reactions with known enthalpies:

1. $\text{C}(s)+{\text{O}}_2(g)\to {\text{CO}}_2(g)\Delta H_1=-393.5\text{ kJ/mol}$
2. $2{\text{H}}_2(g)+{\text{O}}_2(g)\to 2{\text{H}}_2\text{O}(l)\Delta H_2=-571.6\text{ kJ/mol}$
3. ${\text{CH}}_4(g)+2{\text{O}}_2(g)\to {\text{CO}}_2(g)+2{\text{H}}_2\text{O}(l)\Delta H_3=-890.3\text{ kJ/mol}$

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

1. Match species to the target: C(s) is a reactant (matches reaction 1 as written), H₂(g) is a reactant (matches reaction 2 as written), CH₄(g) is a product (it is a reactant in reaction 3, so reverse reaction 3).
2. Reverse reaction 3 and flip the sign of ΔH₃: ${\text{CO}}_2(g)+2{\text{H}}_2\text{O}(l)\to {\text{CH}}_4(g)+2{\text{O}}_2(g)-\Delta H_3=+890.3\text{ kJ/mol}$
3. Add all modified reactions together: $\text{C}(s)+{\text{O}}_2(g)+2{\text{H}}_2(g)+{\text{O}}_2(g)+{\text{CO}}_2(g)+2{\text{H}}_2\text{O}(l)\to {\text{CO}}_2(g)+2{\text{H}}_2\text{O}(l)+{\text{CH}}_4(g)+2{\text{O}}_2(g)$
4. Cancel common species: 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. Hess's Law and Standard Enthalpy of Reaction from Enthalpies of Formation

A common AP exam application of Hess's law is calculating the standard enthalpy of reaction ($\Delta H_{rxn}^{\circ }$) from standard enthalpies of formation ($\Delta H_f^{\circ }$). By definition, $\Delta H_f^{\circ }$ is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states (most stable form at 1 atm and 25°C), and $\Delta H_f^{\circ }=0$ for any element in its standard state. 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 the negative sum of ΔHf of reactants), (2) combine elements to form products (sum of ΔHf of products). Adding these steps gives the total ΔHrxn, leading to the formula: $$\Delta H_{rxn}^{\circ }=\sum n\Delta H_f^{\circ }(\text{products})-\sum m\Delta H_f^{\circ }(\text{reactants})$$ where $n$ and $m$ are the stoichiometric coefficients of products and reactants, respectively. This formula is just a shortcut for the full Hess's law rearrangement process.

Worked Example

Calculate the standard enthalpy of reaction for the combustion of propane: ${\text{C}}_3{\text{H}}_8(g)+5{\text{O}}_2(g)\to 3{\text{CO}}_2(g)+4{\text{H}}_2\text{O}(g)$. Use the following data: $\Delta H_f^{\circ }({\text{C}}_3{\text{H}}_8(g))=-103.8\text{ kJ/mol}$, $\Delta H_f^{\circ }({\text{CO}}_2(g))=-393.5\text{ kJ/mol}$, $\Delta H_f^{\circ }({\text{H}}_2\text{O}(g))=-241.8\text{ kJ/mol}$, $\Delta H_f^{\circ }({\text{O}}_2(g))=0\text{ kJ/mol}$.

1. Calculate the sum of enthalpies of formation for products, multiplied by their stoichiometric coefficients: $\sum n\Delta H_f^{\circ }(\text{products})=(3\times -393.5)+(4\times -241.8)=-1180.5-967.2=-2147.7\text{ kJ}$
2. Calculate the sum of enthalpies of formation for reactants: $\sum m\Delta H_f^{\circ }(\text{reactants})=(1\times -103.8)+(5\times 0)=-103.8\text{ kJ}$
3. Subtract reactant sum from product sum per the formula: $\Delta H_{rxn}^{\circ }=(-2147.7)-(-103.8)=-2043.9\text{ kJ}$
4. Verify the sign: 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. Hess's Law and Enthalpy from Average Bond Enthalpies

Another common application of Hess's law is estimating ΔHrxn 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: (1) break all bonds in reactants to form gaseous atoms (ΔH equals the sum of bond enthalpies of broken bonds), (2) form all bonds in products from the atoms (ΔH equals the negative sum of bond enthalpies of formed bonds). Adding these steps gives the formula: $$\Delta H_{rxn}=\sum (\text{Bond enthalpies of bonds broken})-\sum (\text{Bond enthalpies of bonds formed})$$ This is another Hess's law shortcut. A key 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: ${\text{C}}_2{\text{H}}_4(g)+{\text{H}}_2(g)\to {\text{C}}_2{\text{H}}_6(g)$. 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 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: $\Delta H=2702-2826=-124\text{ kJ/mol}$ 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 (and how to avoid them)

• 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: When calculating ΔH from enthalpies of formation, subtracting product enthalpies from reactant enthalpies instead of the reverse. Why: The "products minus reactants" rule is easy to flip under test pressure. Correct move: Write the full formula $\Delta H=\sum \text{products}-\sum \text{reactants}$ 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. 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 O₂(g), not O(g) or O₃(g). 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. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Given the following two reactions with known enthalpy changes:

1. $2\text{NO}(g)+{\text{O}}_2(g)\to 2{\text{NO}}_2(g)\Delta H_1=-116\text{ kJ}$
2. $2{\text{N}}_2\text{O}(g)+3{\text{O}}_2(g)\to 4{\text{NO}}_2(g)\Delta H_2=-128\text{ kJ}$

What is $\Delta H$ for the target reaction below? ${\text{N}}_2\text{O}(g)+\frac{1}{2}{\text{O}}_2(g)\to 2\text{NO}(g)$ A) $-52\text{ kJ}$ B) $+131\text{ kJ}$ C) $+52\text{ kJ}$ D) $-131\text{ kJ}$

Worked Solution: To get 2 moles of NO as a product, reverse reaction 1 and flip its sign: $2{\text{NO}}_2(g)\to 2\text{NO}(g)+{\text{O}}_2(g)\Delta H=+116\text{ kJ}$. Multiply this reversed reaction by 2 to get 4 moles of NO₂: $4{\text{NO}}_2(g)\to 4\text{NO}(g)+2{\text{O}}_2(g)\Delta H=2(116)=+232\text{ kJ}$. Add this to the original reaction 2, then cancel common species to get $2{\text{N}}_2\text{O}(g)+{\text{O}}_2(g)\to 4\text{NO}(g)$, which is twice the target. Sum ΔH: $232-128=+104\text{ kJ}$ for twice the target, so divide by 2 to get ΔH = +52 kJ. The correct answer is C.

Question 2 (Free Response)

Liquid methanol (${\text{CH}}_3\text{OH}(l)$) is a common alternative fuel for internal combustion engines. (a) Write the balanced chemical equation for the combustion of 1 mole of liquid methanol to form gaseous carbon dioxide and liquid water. (b) Use Hess's law and the given enthalpy of formation data to calculate the standard enthalpy of combustion of methanol. Data: $\Delta H_f^{\circ }({\text{CH}}_3\text{OH}(l))=-238.6\text{ kJ/mol}$, $\Delta H_f^{\circ }({\text{CO}}_2(g))=-393.5\text{ kJ/mol}$, $\Delta H_f^{\circ }({\text{H}}_2\text{O}(l))=-285.8\text{ kJ/mol}$, $\Delta H_f^{\circ }({\text{O}}_2(g))=0\text{ kJ/mol}$. (c) Explain why the enthalpy of combustion calculated in (b) will differ from an estimate made using average bond enthalpies.

Worked Solution: (a) Balanced equation for 1 mole of methanol: ${\text{CH}}_3\text{OH}(l)+\frac{3}{2}{\text{O}}_2(g)\to {\text{CO}}_2(g)+2{\text{H}}_2\text{O}(l)$ (b) Apply the Hess's law formula for ΔHrxn from enthalpies of formation: $$\Delta H_{comb}^{\circ }=\sum n\Delta H_f^{\circ }(\text{products})-\sum m\Delta H_f^{\circ }(\text{reactants})$$ $\sum \text{products}=(1\times -393.5)+(2\times -285.8)=-965.1\text{ kJ}$ $\sum \text{reactants}=(1\times -238.6)+(\frac{3}{2}\times 0)=-238.6\text{ kJ}$ $\Delta H_{comb}^{\circ }=-965.1-(-238.6)=-726.5\text{ kJ/mol}$ (c) Two sources of difference: First, bond enthalpies are average values over many compounds, not exact values for the bonds in methanol specifically. Second, bond enthalpies only apply to gaseous species, while methanol and water are liquid in this reaction, so enthalpies of vaporization for these species are not accounted for in a bond enthalpy estimate.

Question 3 (Application / Real-World Style)

Iron metal is produced industrially in a blast furnace via reduction of iron(III) oxide with carbon (coke). Use the following known reactions to calculate the enthalpy change for the reduction of 1 mole of ${\text{Fe}}_2{\text{O}}_3(s)$, and interpret the sign of your result:

1. $2\text{C}(s)+{\text{O}}_2(g)\to 2\text{CO}(g)\Delta H_1=-221\text{ kJ}$
2. $2\text{CO}(g)+{\text{O}}_2(g)\to 2{\text{CO}}_2(g)\Delta H_2=-566\text{ kJ}$
3. $4\text{Fe}(s)+3{\text{O}}_2(g)\to 2{\text{Fe}}_2{\text{O}}_3(s)\Delta H_3=-1648\text{ kJ}$

Target reaction (overall for 1 mole ${\text{Fe}}_2{\text{O}}_3$): ${\text{Fe}}_2{\text{O}}_3(s)+3\text{C}(s)\to 2\text{Fe}(s)+3{\text{CO}}_2(g)$

Worked Solution:

1. Reverse reaction 3 and divide by 2 to get 1 mole of ${\text{Fe}}_2{\text{O}}_3$ as a reactant: ${\text{Fe}}_2{\text{O}}_3(s)\to 2\text{Fe}(s)+\frac{3}{2}{\text{O}}_2(g)\Delta H=+824\text{ kJ}$
2. Multiply reaction 1 by $\frac{3}{2}$ to get 3 moles of C as a reactant: $3\text{C}(s)+\frac{3}{2}{\text{O}}_2(g)\to 3\text{CO}(g)\Delta H=-331.5\text{ kJ}$
3. Multiply reaction 2 by $\frac{3}{2}$ to get 3 moles of CO as a reactant: $3\text{CO}(g)+\frac{3}{2}{\text{O}}_2(g)\to 3{\text{CO}}_2(g)\Delta H=-849\text{ kJ}$
4. Add all modified reactions, cancel intermediates, and sum ΔH: $\Delta H=824-331.5-849=-356.5{\text{ kJ per mole of Fe}}_2{\text{O}}_3$

Interpretation: The negative ΔH means the overall reaction is exothermic, so the reaction releases heat once initiated, reducing the amount of external energy that must be added to the blast furnace to maintain process temperatures.

7. Quick Reference Cheatsheet

8. What's Next

Mastering Hess's law is a critical prerequisite for all remaining topics in Unit 6 Thermodynamics, as it establishes the additivity rule for state functions that applies to entropy and Gibbs free energy as well. Next, you will use the core concepts of Hess's law to calculate entropy changes and Gibbs free energy of reaction, where the same "products minus reactants" rule applies. Without a solid grasp of Hess's law and ΔH manipulation, you will struggle to solve problems involving reaction spontaneity, a heavily tested topic on the AP Chemistry exam. Hess's law also connects to later topics in electrochemistry, where you will relate cell potential to Gibbs free energy changes for redox reactions.

• Standard enthalpy of formation
• Bond enthalpy and energy changes
• Gibbs free energy and spontaneity
• Thermodynamics of electrochemical cells