Enthalpy changes

MDCAT Chemistry · Chemistry · Thermochemistry and energetics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

Test this topic free All MDCAT Chemistry topics

The idea

Every reaction breaks bonds, which takes in energy, and makes bonds, which gives out energy. The balance is the enthalpy change, ΔH. In an exothermic reaction heat passes to the surroundings, the temperature rises and ΔH is negative. In an endothermic reaction heat is absorbed and ΔH is positive.

To compare reactions fairly we use standard enthalpy changes, ΔH°, measured at 298 K and 1 atm for one mole of the substance named in the definition.

ΔH can be worked out from bond energies, or measured by calorimetry: the heat is passed to water or a solution and found from the temperature change. The same method gives the energy content of foods.

Rules to remember

Formation, ΔH°_f: one mole of a compound is formed from its elements in their standard states. For an element in its standard state ΔH°_f = 0.
Combustion, ΔH°_c: one mole of a substance burns completely in excess oxygen. It is always negative.
Atomisation: one mole of gaseous atoms is formed from the element. It is always positive.
Neutralisation: one mole of water is formed from an acid and a base; about −57 kJ mol⁻¹ for any strong acid with a strong base.
ΔH = (sum of bond energies of bonds broken) − (sum of bond energies of bonds formed).
Heat = mass × specific heat capacity × temperature change (q = mcΔT); for a whole calorimeter q = CΔT with C in J K⁻¹. Then ΔH = −q ÷ moles, in kJ mol⁻¹.
Food energy: fat gives about 38 kJ g⁻¹, carbohydrate and protein about 17 kJ g⁻¹ each.

Common mistake

Students subtract the wrong way round with bond energies. It is bonds broken minus bonds formed, and every bond must be counted, including the number of molecules in the equation.

Worked example

Use the bond energies N≡N 945, H–H 436 and N–H 391 kJ mol⁻¹ to find ΔH for N₂(g) + 3H₂(g) → 2NH₃(g).

Bonds broken: one N≡N and three H–H = 945 + (3 × 436) = 945 + 1308 = 2253 kJ.
Bonds formed: two NH₃ molecules contain six N–H bonds = 6 × 391 = 2346 kJ.
ΔH = broken − formed = 2253 − 2346 = −93 kJ.

Answer: ΔH = −93 kJ for the equation as written (exothermic).

Ask about this

Type what you are stuck on. The helper finds the explanation, worked examples and the notes to read, free and instant.

Keep going

← Activation energy and catalysis First law and Hess's law →Take a MDCAT Chemistry test