1. IB Chemistry Power Points
Topic 05
Energetics
www.pedagogics.ca
Lesson
Hess’s Law
Bond Enthalpies

2. Some taken from
ENTHALPY
CHANGES
Great thanks to
JONATHAN HOPTON & KNOCKHARDY PUBLISHING
www.knockhardy.org.uk/sci.htm

3. HESS’S LAW
“The overall enthalpy change of a chemical process is
independent of the path taken”
The enthalpy change going from A to B can be found by
adding the values of the enthalpy changes for the reactions
A to X, X to Y and Y to B.
ΔHr = ΔH1 + ΔH2 + ΔH3

4. Consider three reactions
1. NaOH(s) + H2O  NaOH (aq) + H2O ΔH1
Dissolving solid sodium
hydroxide in water
This process produces sodium and
hydroxide ions ie. NaOH (aq)
solution.

5. 2. NaOH (aq) +HCl (aq)  NaCl (aq) + H2O ΔH2
Na+ OH-
Reacting the sodium hydroxide
solution with a hydrochloric acid
solution = neutralization

6. 3. NaOH(s) + HCl (aq)  NaCl (aq) + H2O ΔH3
Alternatively - add solid
sodium hydroxide directly to
hydrochloric acid solution.
H3O+
Cl-

7. Recap:
1. NaOH(s) + H2O  NaOH (aq) + H2O ΔH1
2. NaOH (aq) +HCl (aq)  NaCl (aq) + H2O ΔH2
3. NaOH(s) + HCl (aq)  NaCl (aq) + H2O ΔH3
Show that equation 1 plus equation 2 is the same as equation 3
What conclusion about enthalpy can be made?

8. Represented as an enthalpy level diagram
NaOH (s) + HCl
(aq)
+ H2O ΔH1 - 42 kJ/mol
NaOH (aq) +HCl (aq) - 99 kJ/mol ΔH3
- 57 kJ/mol ΔH2
NaCl (aq) + H2O

9. Represented as an enthalpy cycle
ΔH3
NaOH (s) + HCl NaCl (aq) + H2O
+ H2O + HCl
ΔH1 ΔH2
NaOH (aq)

11. Enthalpy of reaction from bond enthalpies
Theory Imagine that, during a reaction, all the bonds of reacting species are broken
and the individual atoms join up again but in the form of products. The
overall energy change will depend on the difference between the energy
required to break the bonds and that released as bonds are made.
energy released making bonds > energy used to break bonds ... EXOTHERMIC
energy used to break bonds > energy released making bonds ... ENDOTHERMIC
Step 1 Energy is put in to break bonds to form separate, gaseous atoms
Step 2 The gaseous atoms then combine to form bonds and energy is released
its value will be equal and opposite to that of breaking the bonds
Applying Hess’s Law ΔHr = Step 1 + Step 2

12. Enthalpy of reaction from bond enthalpies
Calculate the enthalpy change for the hydrogenation of ethene
DH2 1 x C=C bond @ 611 = 611 kJ
4 x C-H bonds @ 413 = 1652 kJ
1 x H-H bond @ 436 = 436 kJ
Total energy to break bonds (reactants) = 2699 kJ
DH3 1 x C-C bond @ 346 = 346 kJ
6 x C-H bonds @ 413 = 2478 kJ
Total energy to break bonds (products) = 2824 kJ
DH = bonds broken – bonds made = (2699 – 2824) = – 125 kJ