Thermochemistry (Energetics)
CollegeThe heat of reactions
Thermochemistry studies energy — usually heat — exchanged in chemical change. At constant pressure this heat is the enthalpy change, ΔH.
- Exothermic (ΔH < 0) — releases heat to surroundings; products lower in energy than reactants (combustion, neutralisation).
- Endothermic (ΔH > 0) — absorbs heat; products higher in energy (thermal decomposition, photosynthesis).
Breaking bonds costs energy (endothermic); forming bonds releases energy (exothermic). The net enthalpy is the difference:
Heat absorbed by a substance is measured by calorimetry:
Hess's law: because enthalpy is a state function, the total ΔH of a reaction is the same whatever route it takes — so enthalpy changes can be added like the steps of an algebraic path.
Exothermic (ΔH < 0)
- Releases heat to the surroundings
- Products lower in energy than reactants
- Surroundings feel warmer
- Bonds formed release more than bonds broken cost
- e.g. combustion, neutralisation
Endothermic (ΔH > 0)
- Absorbs heat from the surroundings
- Products higher in energy than reactants
- Surroundings feel colder
- Bonds broken cost more than bonds formed release
- e.g. thermal decomposition, photosynthesis
Energy is bookkeeping
A chemical reaction is an energy transaction. Reactant bonds must be broken (an energy investment) and product bonds formed (an energy payout). If the payout exceeds the investment, the surplus escapes as heat and the reaction is exothermic; if the investment exceeds the payout, the shortfall is drawn in from the surroundings and the reaction feels cold. Nothing is created or destroyed — energy is merely conserved and relocated (the first law of thermodynamics).
Hess's law is powerful because many reactions cannot be measured directly — carbon burning to only CO, for instance, is impossible to isolate. But if you know the enthalpies of steps that do connect the same start and end points, you can add them to get the answer, just as the altitude gained climbing a mountain depends only on start and finish, not the path. Enthalpy is a "state function": it cares only about the endpoints.
Worked example — enthalpy of combustion by calorimetry
Burning 0.50 g of a fuel raises the temperature of 200 g of water by 15.0 °C. Find the enthalpy released per gram. (cwater = 4.18 J g⁻¹ K⁻¹.)
- Calculate heat absorbed by the water: q = m c ΔT = 200 × 4.18 × 15.0.
- Compute: q = 12 540 J = 12.54 kJ.
- Assume this heat came from the burning fuel (ignoring losses): energy released ≈ 12.54 kJ from 0.50 g.
- Scale to per gram: 12.54 kJ ÷ 0.50 g = 25.1 kJ g⁻¹.
- Assign sign: combustion is exothermic, so ΔHcombustion ≈ −25 kJ g⁻¹. (Real bomb calorimetry captures heat lost to the apparatus, giving higher, more accurate values.)
Is bond breaking exothermic or endothermic? Explain.
Endothermic. A bond is a stable, low-energy arrangement; pulling the atoms apart requires an energy input to overcome their attraction. Conversely, forming bonds releases energy. A reaction's overall ΔH is the balance of the two.
Use bond energies to find ΔH for H₂ + Cl₂ → 2HCl. (H–H 436, Cl–Cl 242, H–Cl 431 kJ mol⁻¹.)
Bonds broken = 436 + 242 = 678 kJ. Bonds formed = 2 × 431 = 862 kJ. ΔH = 678 − 862 = −184 kJ mol⁻¹. Negative, so the reaction is exothermic.
How much heat is needed to raise 500 g of water from 20 °C to 100 °C?
q = mcΔT = 500 × 4.18 × (100 − 20) = 500 × 4.18 × 80 = 167 200 J ≈ 167 kJ.
What does it mean that enthalpy is a "state function", and why is that useful?
It means ΔH depends only on the initial and final states, not the route taken. This underpins Hess's law: enthalpy changes of individual steps can be added to find the ΔH of an overall reaction, letting us calculate values that are impossible or dangerous to measure directly.
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The key facts, visualised
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1How much heat raises 100 g of water by 20 K? (c = 4.18 J/g K)
- Use q = m c deltaT.
- q = 100 x 4.18 x 20.
- Compute: 100 x 4.18 x 20 = 8360 J.
Example 2Burning 0.50 g of fuel raises 200 g water by 10 K. Find heat released. (c = 4.18)
- q = m c deltaT for the water absorbing the heat.
- q = 200 x 4.18 x 10 = 8360 J.
- The fuel released this much heat, so it is exothermic.
Now you try
Work each one out first, then tap to reveal the worked answer.