6.6 Enthalpy of Reaction

Enthalpy of Reaction

Pick a reaction and scrub moles to watch q = n × ΔH scale on a shared heat bar. Compare exo vs. endo magnitudes across combustion, thermite, and photosynthesis.

q = n × ΔH6 ReactionsHeat qExo & Endo
Topic 6.6

Introduction to Enthalpy of Reaction

Calculate the heat q absorbed or released by a system undergoing a chemical reaction in relationship to the amount of the reacting substance in moles and the molar enthalpy of reaction.

The enthalpy change of a reaction gives the amount of heat energy released (for negative values) or absorbed (for positive values) by a chemical reaction at constant pressure.

EK 6.6.A.2 traces the energy flow. When the products of a reaction are at a different temperature than their surroundings, they exchange energy with the surroundings to reach thermal equilibrium. Thermal energy is transferred to the surroundings as reactants convert to products in an exothermic reaction, and from the surroundings in an endothermic reaction.

EK 6.6.A.3 supplies the particulate explanation: the chemical potential energy of the products differs from that of the reactants because of the breaking and forming of bonds. That energy difference results in a change in the kinetic energy of the particles, which manifests as a temperature change. This is the bridge between the bond-level story (6.7) and the thermometer.

Enthalpy is an extensive property. ΔH is quoted per mole for the reaction as written. Double the coefficients and ΔH doubles:

2 H₂(g) + O₂(g) → 2 H₂O(l) ΔH = −572 kJ
H₂(g) + ½ O₂(g) → H₂O(l) ΔH = −286 kJ

The scaling calculation: q = n × ΔHrxn, where n is the moles of the substance the ΔH is referenced to. Always check which substance the given ΔH corresponds to.

Thermochemical equations may include ΔH after the equation, or embed the energy as a term: an exothermic reaction can be written with energy as a product.

Key points

  • ΔH is defined at constant pressure and is quoted per the equation as written.
  • Scaling the equation scales ΔH; reversing it flips the sign.
  • q = n × ΔHrxn, with n referenced to the correct substance.
  • Bond energy differences between products and reactants become kinetic energy, which is what the thermometer measures.

Equations

  • not on the sheet
    • moles of the reference substance

Common mistakes

  • Ignoring the coefficient the ΔH refers to. "−890 kJ" for methane combustion means per mole of CH₄, not per mole of O₂.
  • Forgetting to use the limiting reactant. Excess reactant does not produce more heat.
  • Mixing kJ and J. q = mcΔT gives joules; ΔH is normally kJ/mol.
  • Keeping the sign of ΔH when reversing the equation.

Worked example

For CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l), ΔH = −890. kJ. (a) How much energy is released when 25.0 g of CH₄ burns completely? (b) How much when 25.0 g of CH₄ reacts with 25.0 g of O₂?

(a) Excess oxygen.
n(CH₄) = 25.0 ÷ 16.04 = 1.559 mol
q = (1.559 mol)(−890. kJ/mol) = −1390 kJ (1.39 × 10³ kJ released)

(b) Limited oxygen — find the limiting reactant first.
n(CH₄) = 1.559 mol → ÷1 = 1.559
n(O₂) = 25.0 ÷ 32.00 = 0.781 mol → ÷2 = 0.391

O₂ is limiting. Moles of reaction that can occur = 0.391.
q = (0.391)(−890.) = −348 kJ

Adding more methane would release no additional energy — the oxygen is what runs out.

Full notes for topic 6.6 →