1. Hess's Law

2. Hess’s Law
Germain Henri Hess

3. A. Hess’s Law
Hess’s law states that if you can add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction
2S(s) + 3O2(g)  2SO3(g) H = ?

4. A. Hess’s Law

5. Hess’s law Examples
Hess’s Law states that the heat of a whole reaction is equivalent to the sum of it’s steps.
For example: C + O2  CO2 (pg. 165)
The book tells us that this can occur as 2 steps
C + ½O2  CO H = – kJ
CO + ½O2  CO2 H = – kJ
C + CO + O2  CO + CO2 H = – kJ
I.e. C + O2  CO2 H = – kJ
Hess’s law allows us to add equations.
We add all reactants, products, & H values.
We can also show how these steps add together via an “enthalpy diagram” …

6. C + ½ O2  CO H = – 110.5 kJ C + O2  CO2 H = – 393.5 kJ C + O2
CO + ½ O2  CO2 H = – kJ
C + O2  CO2 H = – kJ
C + O2
Reactants
H = – kJ CO
+ ½ O2
Intermediate
Enthalpy
H =
– kJ
H =
– kJ
Products
CO2
Note: states such as (s) and (g) have been ignored to reduce space on these slides.

7. Practice Exercise 6 with Diagram
Using example as a model, Draw the related enthalpy diagram.
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H= – kJ
2CO2(g) + 3H2O(l)  C2H5OH(l) + 3O2(g) H= kJ
C2H4(g) + H2O(l)  C2H5OH(l)
H= – kJ
C2H4(g) + H2O(l)
+ 3O2(g)
Reactants
Products
C2H5OH(l)
+ 3O2(g)
H=
– 44.0 kJ
Enthalpy
– kJ
H =
H = kJ
2CO2(g) + 3H2O(l)
Intermediate

8. Practice Exercise with Diagram
NO(g)  ½ N2(g) + ½ O2(g) H= – kJ
½ N2(g) + O2(g)  NO2(g) H= kJ
H= – kJ
NO(g) + ½ O2(g)  NO2(g)
NO + ½ O2(g)
Reactants
H=
– 56.6 kJ
NO2(g)
Enthalpy
– kJ
H =
Products
H = kJ
½ N2(g) + O2(g)
Intermediate

9. THE END