Unit 6 Thermochemistry

AP Chemistry Enthalpy

What enthalpy and ΔH mean, the rules for scaling and reversing thermochemical equations, how to find the heat for any amount of reactant, and how enthalpy differs from entropy, in four worked examples.

Enthalpy is the quantity behind every heat calculation in thermochemistry. The sign of ΔH tells you whether a reaction is endothermic or exothermic; its size, combined with stoichiometry, tells you how much heat a given amount of reaction absorbs or releases. This page covers what ΔH means, how a thermochemical equation works as a conversion factor, and the rules that set up Hess’s law.

What Enthalpy and ΔH Mean

Enthalpy (H) is a measure of a system’s heat content at constant pressure. Its absolute value cannot be measured, but its change can: ΔH = H_products − H_reactants, which equals the heat absorbed or released when a reaction runs at constant pressure.

ΔH < 0: exothermic. The products are lower in enthalpy, and heat flows out to the surroundings.
ΔH > 0: endothermic. The products are higher in enthalpy, and heat flows in from the surroundings.
State function: ΔH depends only on the starting and ending states, not on the path between them.

How a Thermochemical Equation Works

A thermochemical equation is a balanced equation with its ΔH attached:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = −890.3 kJ/mol

The “per mol” means per mole of reaction as written: 890.3 kJ is released when 1 mol CH₄ reacts with 2 mol O₂. That makes ΔH a conversion factor, just like a mole ratio in stoichiometry: −890.3 kJ per 1 mol CH₄, or per 2 mol O₂. States of matter matter too, because forming liquid water releases more heat than forming water vapor.

Fig. 1 · Energy released per gram of common fuels

Dividing −ΔH by molar mass turns a per-mole value into a per-gram value. Hydrogen releases far more energy per gram than any carbon fuel because its molar mass is so small, and ethanol releases less than the hydrocarbons because it is already partly oxidized.

The Rules for Reversing and Scaling ΔH

Reverse the reaction: change the sign of ΔH.
Multiply the coefficients by a factor: multiply ΔH by the same factor.
Add reactions together: add their ΔH values. This is Hess’s law.

All three rules follow from ΔH being a state function: the enthalpy difference between two states does not care which direction or how many times you travel between them.

How Enthalpy Differs From Entropy

	Enthalpy change, ΔH	Entropy change, ΔS
Measures	Heat absorbed or released at constant pressure	Change in how dispersed matter and energy are
Typical units	kJ/mol	J/(mol·K)
Favorable sign	Negative (exothermic)	Positive (more dispersal)
Covered in	This page	Entropy

Neither one alone decides whether a process happens. They combine in Gibbs free energy, ΔG = ΔH − TΔS. Watch the units: convert ΔS to kJ before combining it with ΔH.

Worked Example 1: Finding the Heat Released by Burning Methane

How much heat is released when 8.00 g of methane burns completely? CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = −890.3 kJ/mol.

Moles of CH₄: 8.00 g ÷ 16.04 g/mol = 0.499 mol.
Heat: q = 0.499 mol × (−890.3 kJ/mol) = −444 kJ, so 444 kJ is released.

The pattern is grams → moles → kJ, the same chain as any stoichiometry problem with ΔH as the last conversion factor. The negative sign of q matches an exothermic reaction.

Worked Example 2: Reversing and Scaling the Ammonia Equation

N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.2 kJ/mol. Find ΔH for (a) the decomposition of 2 mol NH₃ and (b) the formation of 1 mol NH₃.

(a) Reverse: 2NH₃ → N₂ + 3H₂, so ΔH = +92.2 kJ/mol.
(b) Halve: ½N₂ + 3/2H₂ → NH₃, so ΔH = −92.2 ÷ 2 = −46.1 kJ/mol.

Fractional coefficients are fine in a thermochemical equation, because the coefficients count moles, not molecules. Part (b) is the enthalpy of formation of ammonia, the kind of value Hess’s law combines.

Worked Example 3: Finding the Mass of Fuel for a Required Heat

What mass of propane must burn to release 1000. kJ? C₃H₈ + 5O₂ → 3CO₂ + 4H₂O, ΔH = −2219.9 kJ/mol.

Moles needed: 1000. kJ ÷ 2219.9 kJ/mol = 0.4505 mol C₃H₈.
Mass: 0.4505 mol × 44.10 g/mol = 19.9 g of propane.

Working backward from heat to mass uses the same conversion factor upside down. Keep the heat as a positive amount released and the arithmetic stays simple.

Worked Example 4: An Endothermic Process That Still Happens

Dissolving ammonium nitrate in water has ΔH = +25.7 kJ/mol, yet it dissolves readily, which is how instant cold packs work. How can an endothermic process happen on its own?

Enthalpy: positive, so heat is absorbed and the water gets colder; enthalpy alone does not favor dissolving.
Entropy: the ions spread out from an ordered crystal into solution, so ΔS is positive.
Conclusion: at room temperature, TΔS is larger than ΔH, so ΔG = ΔH − TΔS is negative and the process is thermodynamically favored.

A negative ΔH helps a process along, but it is not required. That is the core difference between enthalpy and the full test, Gibbs free energy.

Five Enthalpy Practice Questions With Answers

For 2H₂ + O₂ → 2H₂O(l), ΔH = −571.6 kJ/mol. What is ΔH for 2H₂O(l) → 2H₂ + O₂?
Show answer
+571.6 kJ/mol. Reversing the reaction changes the sign.
Using the equation in question 1, how much heat is released when 10.0 g of H₂ burns?
Show answer
10.0 g ÷ 2.016 g/mol = 4.96 mol H₂; 4.96 × (571.6 ÷ 2) = 1418 kJ released, about 1420 kJ.
How much heat is released when 50.0 g of ethanol burns? ΔH = −1366.8 kJ per mol of ethanol.
Show answer
50.0 ÷ 46.07 = 1.085 mol; 1.085 × 1366.8 = 1483 kJ released.
Is ΔH for melting ice positive or negative? Explain.
Show answer
Positive. Melting absorbs heat to overcome the hydrogen bonds holding the solid together, so it is endothermic.
A reaction has ΔH = −50 kJ/mol. If every coefficient is tripled, what is the new ΔH?
Show answer
−150 kJ/mol. Multiplying the coefficients multiplies ΔH by the same factor.

Common Enthalpy Mistakes

Forgetting to flip the sign when reversing a reaction.
Forgetting to scale ΔH with the coefficients. Doubling the equation doubles ΔH.
Using ΔH per mole of the wrong substance. Check which coefficient the ΔH is tied to.
Mixing kJ and J. ΔH is usually in kJ, ΔS in J/(mol·K).
Ignoring states. H₂O(l) and H₂O(g) give different ΔH values.
Thinking exothermic means spontaneous. Entropy matters too; use ΔG.

Thermochemical equations show up in calculation and justification questions throughout Unit 6. To see how stronger thermochemistry answers would move your overall estimate, try the AP Chem Score Calculator.

Frequently Asked Questions

What is enthalpy?

Enthalpy (H) is a measure of the heat content of a system at constant pressure. Chemists work with the change in enthalpy, ΔH, which equals the heat absorbed or released by a reaction at constant pressure: negative for exothermic reactions and positive for endothermic ones.

What are the units of enthalpy change?

For a thermochemical equation, ΔH is usually given in kJ/mol, meaning kilojoules per mole of reaction as written, that is, for the molar amounts shown by the coefficients. For a specific sample, the heat is in kJ.

What happens to ΔH when you reverse or multiply a reaction?

Reversing a reaction changes the sign of ΔH. Multiplying every coefficient by a factor multiplies ΔH by the same factor. Both follow from ΔH being a state function.

What is the difference between enthalpy and entropy?

Enthalpy change (ΔH) is the heat absorbed or released, in kJ/mol. Entropy change (ΔS) measures how the dispersal of matter and energy changes, in J/(mol·K). Both feed into Gibbs free energy, ΔG = ΔH − TΔS, which decides whether a process is thermodynamically favored.

How do you calculate the heat released by a reaction?

Find the moles of a reactant or product, then use the thermochemical equation as a conversion factor: q = n × ΔH per mole of that substance. Burning 8.00 g (0.499 mol) of methane, with ΔH = −890.3 kJ per mol CH₄, releases about 444 kJ.

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