Remember that chemical reactions in the laboratories do not normally involve gaseous atoms and ions - a Born-haber cycle is a theoretical construct that applys Hess'law to to an ionic lattice investigating the stability and bonding of compounds of metals and non-metals.

2. 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 kJC + 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” …

3. 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
1.Look at equations to help complete balancing

4. C + O2 → CO2 ∆H° = – 393.5 kJ
Reactants
Intermediate
Products
C + O2
CO2
CO
Enthalpy
Note: states such as (s) and (g) have been ignored to reduce
clutter on these slides. You should include these in your work.
∆H° = – 110.5 kJ
∆H° =
– 283.0 kJ
∆H° =
– 393.5 kJ
+ ½O2
C + ½O2 → CO ∆H° = – 110.5 kJ
CO + ½O2 → CO2 ∆H° = – 283.0 kJ

5. 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
C2H4(g) + H2O(l) → C2H5OH(l)
Reactants
Products
C2H4(g) + H2O(l)
C2H5OH(l)
2CO2(g) + 3H2O(l)
Enthalpy
∆H°=
–1411.1kJ
∆H° =
+1367.1 kJ
∆H°=
–44.0 kJ
+ 3O2(g)
+ 3O2(g)
∆H°= – 44.0 kJ

6. 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)
Reactants
Products
Intermediate
GeO(s) + ½ O2(g)
GeO2(s)
Ge(s) + O2(g)Enthalpy
∆H° = +255 kJ
∆H°=
–534.7kJ
∆H°=
–280 kJ
∆H°= – 279.7 kJ

7. 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
NO(g) + ½ O2(g) → NO2(g)
Reactants
Products
Intermediate
NO + ½ O2(g)
NO2(g)
½ N2(g) + O2(g)
Enthalpy
∆H°=
–90.37kJ
∆H° = +33.8 kJ
∆H°=
–56.6 kJ
∆H°= – 56.57 kJ

8. 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 + 1.5O2 → Fe2O3
3CO + 1.5O2 → 3CO2 ∆H°= –848.88 kJ
2Fe + 3CO2 → Fe2O3 + 3CO ∆H°= + 26.74 kJ
CO + ½O2 → CO2 ∆H°= –282.96 kJ
Don’t forget to add states. Try 5.55, 5.57, 5.58, 5.61 (pg. 175)
∆H°= –822.14 kJ
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