Enthalpy of Reactions -We can describe the energy absorbed as heat at constant pressure by the change in enthalpy (ΔH) -the enthalpy of a reaction is the amount of energy transferred as heat during a reaction; it doesn’t take into account work, change in temperature, or the nature of the substance the heat is transferring to (ie. mass or specific heat)

Enthalpy of Reactions -a thermochemical equation will tell you how much energy is released or absorbed during a reaction -heat absorbed (endothermic) will be presented as a reactant or as positive -heat released (exothermic) will be presented as a product or as negative

ΔH = (sum of ΔH products) – (sum ofΔH reactants) Enthalpy of Reactions -total enthalpy is often calculated as: ΔH = (sum of ΔH products) – (sum ofΔH reactants) note: moles are taken into consideration as part of calculations (ex: (1 mol × ΔH)

Enthalpy of Reactions -Heats of formation for products or reactants in their elemental form are considered to be 0 kJ/mol -We can use the heats of formation to calculate the enthalpy of a reaction without having to do Hess’s Law

Enthalpy of Reactions -we can calculate the amount of enthalpy for a reaction that we don’t know by using the enthalpy of reactions that we do know To Solve: 1. You will be given several equations and you want to arrange them in a way that produces the equation for the reaction you want 2. Equations/reactions can be reversed, which means the ΔH is also reversed 3. Multiply/divide coefficients in the equation so that compounds are the same as the desired thermochemical equation 4. Add it all up

C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O(g) ΔH = ? Enthalpy Practice Calculate the enthalpy of reaction for the following reaction: C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O(g) ΔH = ? 3C (s)+ 4 H2 (g) -------> C3H8 (g) ΔH = +103.8 kJ 2 H2(g) + O2 (g) -------> 2H2O (g) ΔH = -968 kJ C (s) + O2 (g) --------> CO2 (g) ΔH = -1180.5 kJ