Chemistry 30 · Thermochemistry
Energy in reactions
Every reaction either releases energy or takes it in, and thermochemistry is the bookkeeping: what the sign of ΔH means, how a calorimeter measures it, and how Hess's law lets you find a value you could never measure directly.
- 1. Enthalpy and its sign
- 2. Potential energy diagrams
- 3. Calorimetry
- 4. Writing thermochemical equations
- 5. Hess's law
Enthalpy and its sign
The idea: ΔH is the energy difference between products and reactants. If the products hold less, the surplus left the system and ΔH is negative.
| Exothermic | Endothermic | |
|---|---|---|
| Energy | Released to the surroundings | Absorbed from the surroundings |
| ΔH | Negative | Positive |
| Products | Lower in enthalpy than reactants | Higher in enthalpy |
| Feels | Warm — the surroundings gain the energy | Cold — the surroundings lose it |
| Examples | Combustion, neutralization, condensation | Melting, evaporation, photosynthesis, most dissolving of ionic salts |
The trap is that the system and the surroundings always feel opposite. A cold pack goes cold because the reaction inside is absorbing energy from your hand — endothermic, positive ΔH, even though the thing you touch is cold.
Where the energy actually comes from
Breaking bonds always costs energy; forming bonds always releases it. A reaction is exothermic when the bonds in the products release more than the bonds in the reactants cost. Nothing is created — energy is moved between chemical potential energy and the surroundings.
Potential energy diagrams
The idea: the hump is the activation energy, the height difference between the two flat ends is ΔH, and a catalyst shortens the hump without moving the ends.
- Activation energy (Ea) is measured from the reactants up to the peak — the energy needed to reach the activated complex, whether or not the overall reaction releases energy.
- ΔH is products minus reactants, so it is the difference between the two flat ends.
- A catalyst lowers the peak by offering a different pathway. The ends do not move, so ΔH is unchanged. Catalysts speed reactions up; they do not make them more exothermic.
- The reverse reaction has the same peak measured from the other side, so its activation energy is Ea + |ΔH| for an exothermic forward reaction.
Calorimetry
The idea: you cannot measure enthalpy directly, so you measure what the reaction does to the temperature of something else — usually water — and work backwards with q = mcΔT.
| Symbol | Means | Units |
|---|---|---|
| q | Heat gained or lost | J or kJ |
| m | Mass of the substance being warmed | g |
| c | Specific heat capacity (4.19 for water) | J/(g·°C) |
| ΔT | Final temperature minus initial | °C |
Worked: finding a molar enthalpy
0.0250 mol of a substance reacts in a calorimeter and warms 50.0 g of water by 8.00 °C.
- Heat absorbed by the water: q = (50.0)(4.19)(8.00) = 1676 J = 1.68 kJ.
- Which way did it go? The water warmed, so the reaction released the energy: q(reaction) = −1.68 kJ.
- Per mole: ΔH = −1.68 kJ ÷ 0.0250 mol = −67.0 kJ/mol.
Worked: heat lost equals heat gained
25.0 g of metal at 100.0 °C is dropped into 100.0 g of water at 20.0 °C, and the mixture settles at 22.0 °C.
Water gains (100.0)(4.19)(2.0) = 838 J, so the metal lost 838 J while cooling by 78.0 °C:
c = 838 ÷ (25.0 × 78.0) = 0.43 J/(g·°C)
Using 100.0 °C as the metal's ΔT instead of 78.0 °C is the usual error — it cooled to the final temperature, not to zero.
Watch out: m and c belong to whatever is being warmed — the water — not to the reacting chemical, unless the question says the solution is the calorimeter's contents. And calorimetry always assumes no heat escapes, which is why real values come out slightly low.
Writing thermochemical equations
The idea: there are three ways to show the same energy change, and questions switch between them on purpose.
| Form | Example |
|---|---|
| ΔH written beside the equation | 2H₂(g) + O₂(g) → 2H₂O(g) ΔH = −484 kJ |
| Energy as a term in the equation | 2H₂(g) + O₂(g) → 2H₂O(g) + 484 kJ |
| Molar enthalpy, per mole of one substance | ΔHᶜ = −242 kJ/mol of H₂ |
- Energy on the product side means exothermic; on the reactant side means endothermic.
- Doubling the equation doubles ΔH. Enthalpy is an extensive quantity: it depends on how much reacted.
- Reversing the equation flips the sign.
- Molar enthalpy needs a named substance, because −484 kJ per equation is −242 kJ per mole of H₂ but −484 kJ per mole of O₂.
Hess's law
The idea: enthalpy change depends only on where you start and finish, not the route. So known equations can be added, reversed and scaled until they total the one you want.
- Write the target equation and mark where each substance needs to end up.
- Take each given equation in turn: reverse it if the substance is on the wrong side, and multiply it if the coefficient is wrong.
- Apply the same operations to ΔH: reversing flips the sign, multiplying multiplies.
- Add everything up, cancelling anything that appears on both sides, and add the adjusted ΔH values.
Worked: the enthalpy of formation of methane
Target: C(s) + 2H₂(g) → CH₄(g)
- (1) C(s) + O₂(g) → CO₂(g) ΔH = −393.5 kJ — use as written, carbon is already on the left
- (2) H₂(g) + ½O₂(g) → H₂O(l) ΔH = −285.8 kJ — × 2, the target needs 2H₂: −571.6 kJ
- (3) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ΔH = −890.3 kJ — reverse, CH₄ must be a product: +890.3 kJ
Adding: −393.5 + (−571.6) + 890.3 = −74.8 kJ. The CO₂ and H₂O cancel, leaving exactly the target equation.
Watch out: the check that catches nearly every mistake is cancelling. Write out the combined equation and make sure everything not in the target has disappeared — if something is left over, an equation needs reversing or rescaling.