AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/6/6-6)
Unit 6 · Topic 6.6
6.6 Introduction to Enthalpy of Reaction
The enthalpy of reaction, ΔH, is the heat a reaction releases (negative) or absorbs (positive) at constant pressure for the amounts in its balanced equation. Because heat scales with amount, you can use ΔH like a stoichiometry ratio to find the heat for any quantity that reacts.
Key terms
- enthalpy of reaction (ΔH)
- constant pressure
- kJ/mol
- thermochemical equation
What ΔH tells you
For a reaction run at constant pressure, such as one in an open beaker, the heat exchanged with the surroundings equals the enthalpy change, ΔH. A negative ΔH means the reaction releases heat; a positive ΔH means it absorbs heat. The ° in ΔH° means the reactants and products are in their standard states.
A thermochemical equation is a balanced equation with its ΔH: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH° = −890 kJ/mol. The 'per mole' here means per mole of reaction as written: when 1 mol CH₄ reacts with 2 mol O₂ to make 1 mol CO₂ and 2 mol H₂O, 890 kJ is released.
Phases matter, so always include them. If the water is formed as a gas, ΔH° is about −802 kJ/mol instead, because 88 kJ less energy is released when 2 mol of H₂O stay as vapor instead of condensing.
Where the heat comes from
Breaking and forming bonds changes how much chemical potential energy is stored in the particles, so the products and reactants sit at different potential energies. In an exothermic reaction, the lost potential energy first appears as faster-moving particles, so the reaction mixture gets hotter. The hot products then pass energy to the cooler surroundings until they reach thermal equilibrium.
In an endothermic reaction, the products gain potential energy, the particles slow down and the mixture gets colder. Energy then flows in from the surroundings.
Scaling ΔH to the amount that reacts
Heat is proportional to how much reacts. If you double the amounts, you double the heat. Treat ΔH as a conversion factor tied to the coefficients.
For 2H₂(g) + O₂(g) → 2H₂O(l), ΔH° = −571.6 kJ/mol. This is per 2 mol H₂, per 1 mol O₂, and per 2 mol H₂O. So burning 1 mol of H₂ releases 571.6 ÷ 2 = 285.8 kJ.
General path: grams of a substance → moles of that substance → moles of reaction (divide by its coefficient) → heat (multiply by ΔH).
If the reactants are not mixed in the right ratio, the limiting reactant decides how much reaction happens, and so how much heat is released.
What you won't be tested on
College chemistry courses distinguish enthalpy from internal energy. You won't need that distinction: AP treats reactions as happening at constant pressure, where ΔH is the heat of reaction.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Heat from burning a mass of fuel
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH° = −890. kJ/mol. How much heat is released when 8.00 g of CH₄ (molar mass 16.04 g/mol) burns completely?
Show the solutionHide the solution
- Step 1: Moles of CH₄: 8.00 g ÷ 16.04 g/mol = 0.4988 mol.
- Step 2: The coefficient of CH₄ is 1, so 0.4988 mol CH₄ is 0.4988 mol of reaction.
- Step 3: q = (0.4988 mol)(−890. kJ/mol) = −444 kJ. The minus sign means heat is released.
Answer: 444 kJ released
- Example 2Calculator allowed
Trap: the coefficient isn't 1
2H₂(g) + O₂(g) → 2H₂O(l), ΔH° = −571.6 kJ/mol. How much heat is released when 1.00 g of H₂ (molar mass 2.016 g/mol) burns?
Show the solutionHide the solution
- Step 1: Moles of H₂: 1.00 g ÷ 2.016 g/mol = 0.496 mol.
- Step 2: The trap is multiplying 0.496 mol by 571.6 kJ. But 571.6 kJ is released for every 2 mol of H₂, so first convert to moles of reaction: 0.496 mol H₂ × (1 mol reaction / 2 mol H₂) = 0.248 mol reaction.
- Step 3: q = (0.248 mol)(−571.6 kJ/mol) = −142 kJ.
Answer: About 142 kJ released (not 284 kJ)
- Example 3Calculator allowed
Working backward to a mass
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), ΔH° = −2220 kJ/mol. What mass of propane (molar mass 44.10 g/mol) must burn to release 1000. kJ of heat?
Show the solutionHide the solution
- Step 1: Moles of reaction needed: 1000. kJ ÷ 2220 kJ/mol = 0.450 mol. The coefficient of propane is 1, so this is also 0.450 mol C₃H₈.
- Step 2: Mass: 0.450 mol × 44.10 g/mol = 19.9 g.
Answer: About 19.9 g of propane
Common mistakes
- Ignoring coefficients: ΔH belongs to the whole balanced equation, not to one mole of every substance in it.
- Leaving off the sign or the word 'released' or 'absorbed'. 'q = 444 kJ' alone suggests heat was absorbed.
- Using the ΔH for liquid water when the equation shows water vapor, or the other way around.
- Using the reactant in excess instead of the limiting reactant to calculate heat.
On the exam
- Expect problems that convert between grams, moles and kJ using a given ΔH. Show the mole ratio step clearly; it is where most points are lost.
- You may be asked to explain a temperature change in terms of potential energy changing to kinetic energy, then flowing as heat to the surroundings.
Connected topics
Videos
Check yourself
4 questions on 6.6 Introduction to Enthalpy of Reaction. Pick an answer to see if you got it, and why.
For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), ΔH° = −92 kJ/mol. How much heat is released when 8.5 g of NH₃ (molar mass 17.03 g/mol) is produced by this reaction?
Propane burns according to C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), ΔH° = −2,220 kJ/mol. What mass of propane (molar mass 44.10 g/mol) must burn to release 555 kJ of heat?
For 2H₂(g) + O₂(g) → 2H₂O(l), ΔH° = −572 kJ/mol. A mixture of 4.0 g of H₂ (molar mass 2.02 g/mol) and 16.0 g of O₂ (molar mass 32.00 g/mol) reacts until one reactant runs out. How much heat is released?
When 0.0500 mol of a salt dissolves in water in a coffee-cup calorimeter, the temperature of the solution drops from 25.0 °C to 21.2 °C. Assume the solution has a mass of 100.0 g and a specific heat of 4.18 J/(g·°C), and that no heat is lost. What is the enthalpy of solution of the salt?
0 of 4 answered