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Notice that the ΔH values changed as well.ĥ⁄ 2O 2 ⇒ first & sum of second and third equation We need one H 2 on the reactant side and that's what we have.Ģ) Rewrite all three equations with changes applied:ĢCO 2(g) + H 2O(ℓ) -> C 2H 2(g) + 5⁄ 2O 2(g) ΔH° = +1299.5 kJĢC(s) + 2O 2(g) -> 2CO 2(g) ΔH° = −787 kJ H 2(g) + 1⁄ 2O 2(g) -> H 2O(ℓ) ΔH° = −285.8 kJġ) Determine what we must do to the three given equations to get our target equation:Ī) first eq: flip it so as to put C 2H 2 on the product sideī) second eq: multiply it by two to get 2CĬ) third eq: do nothing. If a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation equals the sum of the enthalpy changes of the other chemical equations.Įxample #1: Calculate the enthalpy for this reaction:ĢC(s) + H 2(g) -> C 2H 2(g) ΔH° = ? kJ The enthalpy of a given chemical reaction is constant, regardless of the reaction happening in one step or many steps. Germain Henri Hess, in 1840, discovered a very useful principle which is named for him: Using three equations and their enthalpies Hess' Law: three equations and their enthalpies - Problems 1 - 10 Hess' Law: two equations and their enthalpies Hess' Law: standard enthalpies of formation Hess' Law: three equations and their enthalpies - Problems 11 - 25 Hess' Law: four or more equations and their enthalpies Hess' Law: bond enthalpies Thermochemistry menu ChemTeam: Hess' Law - using three equations and their enthalpies