Hess’s law
• 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 = – 110.5 kJ
CO + ½O2  CO2
H = – 283.0 kJ
C + CO + O2  CO + CO2 H = – 393.5 kJ
I.e. C + O2  CO2
H = – 393.5 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” …
Steps in drawing enthalpy diagrams
1. Balance the equation(s).
2. Sketch a rough draft based on H values.
3. Draw the overall chemical reaction as an
enthalpy diagram (with the reactants on one
line, and the products on the other line).
4. Draw a reaction representing the intermediate
step by placing the relevant reactants on a line.
5. Check arrows: Start: two leading away
Finish: two pointing to finish
Intermediate: one to, one away
6. Look at equations to help complete balancing
(all levels must have the same # of all atoms).
7. Add axes and H values.
C + ½O2  CO
CO + ½O2  CO2
H = – 110.5 kJ
H = – 283.0 kJ
 CO2
H = – 393.5 kJ
C
+ O2
C + O2
Reactants
Products
Enthalpy
Intermediate
H = – 110.5 kJ
CO + ½O2
H =
– 393.5 kJ
H =
– 283.0 kJ
CO2
Note: states such as (s) and (g) have been ignored to reduce
clutter on these slides. You should include these in your work.
Practice Exercise 6 (pg. 167) with Diagram
Using example 5.6 as a model, try PE 6.
Draw the related enthalpy diagram.
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) H= –1411.1 kJ
2CO2(g) + 3H2O(l)  C2H5OH(l) + 3O2(g) H= +1367.1 kJ
Intermediate
H =
– 1411.1 kJ
Reactants
Products
Enthalpy
C2H4(g) + H2O(l)  C2H5OH(l)
H= – 44.0 kJ
C2H4(g) + H2O(l) + 3O2(g)
C2H5OH(l)+ 3O2(g) H=
H =
+1367.1 kJ
2CO2(g) + 3H2O(l)
– 44.0 kJ
5.51 (pg. 175)
GeO(s)
 Ge(s) + ½ O2(g) H= + 255 kJ
Ge(s) + O2(g)  GeO2(s)
H= – 534.7 kJ
GeO(s) + ½ O2(g) GeO2(s)
H= – 279.7 kJ
Reactants
Products
Enthalpy
Intermediate
H =
– 534.7 kJ
Ge(s) + O2(g)
H = +255 kJ
GeO(s) + ½ O2(g)
GeO2(s)
H=
– 280 kJ
5.52 (pg. 175)
NO(g)
 ½ N2(g) + ½ O2(g) H= – 90.37 kJ
½ N2(g) + O2(g)  NO2(g)
H= + 33.8 kJ
H= – 56.57 kJ
NO(g) + ½ O2(g)  NO2(g)
NO + ½ O2(g)
Intermediate
H =
– 90.37 kJ
Products
Enthalpy
Reactants
NO2(g)
H = +33.8 kJ
½ N2(g) + O2(g)
H=
– 56.6 kJ
Hess’s law: Example 5.7 (pg. 166)
We may need to manipulate equations further:
2Fe + 1.5O2  Fe2O3 H=?, given
Fe2O3 + 3CO  2Fe + 3CO2 H= – 26.74 kJ
CO + ½O2  CO2
H= –282.96 kJ
1: Align equations based on reactants/products.
2: Multiply based on final reaction.
3: Add equations.
2Fe + 3CO2  Fe2O3 + 3CO H= + 26.74 kJ
CO + ½O2  CO2
H= –282.96 kJ
3CO + 1.5O2  3CO2
H= –848.88 kJ
2Fe
+ 1.5O2  Fe2O3
H= –822.14 kJ
Don’t forget to add states. Try 5.55, 5.57, 5.58, 5.61 (pg. 175)
For more lessons, visit
www.chalkbored.com