1. Unit 5: Thermochemistry
Enthalpy

2. H = ΣHproducts – ΣHreactants
Enthalpy
Enthalpy (H) is the heat energy absorbed or lost in a reaction (H if talking about change)
H = ΣHproducts – ΣHreactants

3. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
Enthalpy Example
Calculate ΔH for the following reaction:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
Substance
H (kJ/mol)
CH4
-74.8 O2
CO2
-393.5
H2O

4. Hess’s Law
Hess’s Law: overall enthalpy changes are the sum of all steps of the reaction process
Add enthalpies (H) from all reaction steps
Note:
If a reaction is reversed, the sign of H is also reversed
The magnitude of H depends on the quantities of reactants and products
Multiply H by the coefficient for that substance

5. Hess’s Law Example
Calculate the heat of reaction, H, for the overall reaction:
N2(g) + 2O2(g)  2NO2(g)
Given:
2NO(g)  N2(g) + O2(g) H=-180 kJ 2NO(g) + O2(g)  2NO2(g) H=-112 kJ

6. Hess’s Law Practice
Calculate the heat of reaction, H, for the overall reaction:
2 S (s) + 3 O2 (g)  2 SO3 (g)
Given:
S (s) + O2 (g) SO2 (g)  ∆H  =  -297 kJ
2 SO3 (g)2 SO2 (g) + O2 (g)  ∆H  =  198 kJ 

7. PbCl2 (s) + Cl2 (g) → PbCl4 (l)
Hess’s Law Exit Slip
Calculate ΔH for this reaction:
PbCl2 (s) + Cl2 (g) → PbCl4 (l)
Given:
Pb (s) + Cl2 (g) → PbCl2 (s) ∆ H = kJ
Pb (s) + 2 Cl2 (g) → PbCl4 (l) ∆ H = kJ