Chemistry · WCH12 (Unit 2)
Enthalpy changes and Hess's law
An enthalpy change, symbol ΔH, is the heat energy change at constant pressure. Hess's law lets you find ΔH values you cannot measure directly.
Signs
Exothermic reactions release heat to the surroundings: ΔH is negative. Endothermic reactions take in heat: ΔH is positive. Standard conditions are 100 kPa and a stated temperature, usually 298 K.
Hess's law
The enthalpy change of a reaction depends only on the start and end states, not the route taken. So you can add ΔH values around a cycle.
Using formation data
ΔH(reaction) = sum of ΔfH(products) - sum of ΔfH(reactants). The standard enthalpy of formation of an element in its standard state is zero.
Worked example
Find ΔH for C(s) + O2(g) -> CO2(g) via CO, given C + 1/2 O2 -> CO, ΔH = -111 kJ/mol, and CO + 1/2 O2 -> CO2, ΔH = -283 kJ/mol.
- Both steps take C to CO2.
- Add them: -111 + (-283)
Answer: ΔH = -394 kJ/mol
Check yourself
- An exothermic reaction has a ΔH that is
- positive
- zero
- always greater than 100 kJ/mol
- negative
Show answer
negative. Heat leaves the system, so ΔH < 0.
- Route 1 has steps of -50 kJ/mol and -30 kJ/mol. What is ΔH for the direct route?
Show answer
-80 kJ/mol. By Hess's law the direct route equals the sum: -80 kJ/mol.
- What is the standard enthalpy of formation of O2(g)?
- -286 kJ/mol
- +498 kJ/mol
- It depends on pressure
- 0 kJ/mol
Show answer
0 kJ/mol. Elements in their standard states have ΔfH = 0.
- ΔfH: CO2 = -394, H2O(l) = -286, CH4 = -75 kJ/mol. Find ΔcH for CH4 + 2O2 -> CO2 + 2H2O.
Show answer
-891 kJ/mol. Products: -394 + 2(-286) = -966. Reactants: -75 (O2 is 0). ΔH = -966 - (-75) = -891 kJ/mol.
- Which pressure is used for standard enthalpy changes?
- 1 kPa
- 50 kPa
- 100 kPa
- 1000 kPa
Show answer
100 kPa. Standard conditions use 100 kPa.
Sources
Last updated 7 November 2026. Spotted a mistake? Email alevel@lifereadysociety.com.