1. 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)

2. 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

3. 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)

4. 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

5. 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

6. 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 = kJ 2 H2(g) + O2 (g) > 2H2O (g) ΔH = -968 kJ C (s) + O2 (g) > CO2 (g) ΔH = kJ