1. ENTHALPY AND HESS’ LAW

2. Chemical energy is stored in:
– moving electrons
– vibration of chemical bonds
– rotation and translation of
molecules
– stored nuclear energy of protons &
neutrons
– energy stored in chemical bonds
ENTHALPY AND HESS’ LAW

3. Enthalpy
enthalpy (H) – total kinetic and potential energy of
a system at a constant pressure
change in enthalpy (∆H) – change in heat of a
system
∆H = Hfinal - Hinitial
∆H = Hproducts - Hreactants
ENTHALPY AND HESS’ LAW

4. Enthalpy
For an exothermic reaction:
Hproducts < Hreactants
∴∆H is negative
Describe what has happened.
H
reaction progress
reactants
products
ENTHALPY AND HESS’ LAW

5. Enthalpy
For an endothermic reaction:
Hproducts > Hreactants
∴∆H is positive
Describe what has happened.
H
reaction progress
reactants
products
ENTHALPY AND HESS’ LAW

6. Enthalpy
How does enthalpy relate to heat with respect
to an isolated system?
∆H = q
when using a calorimeter.
What is the formula for q?
ENTHALPY AND HESS’ LAW

7. Enthalpy
What is the formula for q?
q = mcΔT
Since q = ΔH
ΔH = mcΔT
ENTHALPY AND HESS’ LAW

8. Enthalpy
How are enthalpy and heat related?
∆Hsystem = -q exothermic reaction
∆Hsystem = +q endothermic reaction
This is only true at constant pressure, which is
always the case in a calorimeter.
Standard conditions for enthalpy changes: temperature at 298K
ENTHALPY AND HESS’ LAW

9. Enthalpy
The enthalpy change for a balanced chemical
reaction is constant.
e.g. 2 H2(g) + O2(g)  2 H2O(g) ∆H° = - 483.6 kJ
∆H° – standard enthalpy of a reaction
– must be a balanced chemical
equation
Read as - 483.6 kJ per 2 mol of H2 being reacted.
ENTHALPY AND HESS’ LAW
This is called a
thermochemical equation

10. Enthalpy
∆H° values are proportional to coefficient
changes of a chemical equation.
N2O4 (g)  2 NO2 (g) ∆H° = +57.93 kJ
3 N2O4 (g)  6 NO2 (g) ∆H° = +173.79 kJ
½ N2O4 (g)  NO2 (g) ∆H° = +28.96 kJ
ENTHALPY AND HESS’ LAW

11. Enthalpy
N2O4 (g)  2 NO2 (g) ∆H° = +57.93 kJ
2 NO2 (g)  N2O4(g) ∆H° = -57.93 kJ
Reversing a chemical reaction causes a sign
change in front of the ∆H° value.
ENTHALPY AND HESS’ LAW

12. Enthalpy
∆H° values of unknown reactions can be solved
when other known reactions are given.
ENTHALPY AND HESS’ LAW

13. Example #1
C(s) + O2(g)  CO2(g) ∆H° = ?
C(s) + ½ O2(g)  CO(g) ∆H° = -110.5 kJ
CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ
ENTHALPY AND HESS’ LAW
What parts of the above reactions are the same
as the first reaction?

14. Rules for solving thermochemical
equation questions:
For any reaction that can be written in steps,
the ∆H° is the same as the sum of the values
of the ∆H° for each individual step.
Hess’s Law of Summation
ENTHALPY AND HESS’ LAW

15. ∆H° Rules
1. If all the coefficients of an eqn
are multiplied
or divided by a common factor, the ∆H°
must be changed likewise.
ENTHALPY AND HESS’ LAW
N2O4 (g)  2 NO2 (g) ∆H° = +57.93 kJ
3 N2O4 (g)  6 NO2 (g) ∆H° = +57.93kJ x 3
= +173.79 kJ
½ N2O4 (g)  NO2 (g) ∆H° = +57.93kJ ÷ 2
= +28.96 kJ

16. ∆H° Rules
2. When a reaction is reversed, the sign of ∆H°
must also be reversed.
ENTHALPY AND HESS’ LAW
N2O4 (g)  2 NO2 (g) ∆H° = +57.93 kJ
2 NO2 (g)  N2O4(g) ∆H° = -57.93 kJ

17. ∆H° Rules
3. When canceling compounds for Hess’s Law,
the state of the compounds is important.
ENTHALPY AND HESS’ LAW
H2(g) + ½ O2(g)  H2O(g) ΔH° = +57.93 kJ
K2SO4(s) + 2H2O(l)  H2SO4(aq) + 2KOH(s) ΔH° = +342.4kJ
These two CANNOT cancel each other out

18. C(s) + O2(g)  CO2(g) ∆H° = ?
C(s) + ½ O2(g)  CO(g) ∆H° = -110.5 kJ
CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ
ENTHALPY AND HESS’ LAW
Ex #1
C(s) + O2(g)  CO2(g) ∆H° = -393.5 kJ
The enthalpies are
added together
Therefore the enthalpy is -393.5 kJ

19. 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = ?
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) ∆H° = -26.7 kJ
CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ
ENTHALPY AND HESS’ LAW
Ex #2
This is not as easy to do!
STEP 1: Number each given thermochemical equation.
1
2

20. 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = ?
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) ∆H° = -26.7 kJ
CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ
ENTHALPY AND HESS’ LAW
Ex #2
1
2
STEP 2: Arrange the equations so that your desired
reactants are on the left side, and your desired
products are on the right side
2 Fe(s) + 3 CO2(g)  Fe2O3(s) + 3 CO(g) ∆H° = +26.7 kJ
CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ
1-
2

21. 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = ?
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) ∆H° = -26.7 kJ
CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ
ENTHALPY AND HESS’ LAW
Ex #2
1
2
2 Fe(s) + 3 CO2(g)  Fe2O3(s) + 3 CO(g) ∆H° = +26.7 kJ
3CO(g) + 3/2 O2(g)  3CO2(g) ∆H° = -849.0 kJ
1-
2
STEP 3: Multiply the equations by factors such that
they may match your desired equation. Remember
to multiply the enthalpy by the same factor.
3 x

22. 2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = ?
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g) ∆H° = -26.7 kJ
CO(g) + ½ O2(g)  CO2(g) ∆H° = -283.0 kJ
ENTHALPY AND HESS’ LAW
Ex #2
1
2
2 Fe(s) + 3 CO2(g)  Fe2O3(s) + 3 CO(g) ∆H° = +26.7 kJ
3CO(g) + 3/2 O2(g)  3CO2(g) ∆H° = -849.0 kJ
1-
2
STEP 4: Add the two equations and enthalpies
together. Remember to cancel repeating chemicals.
3 x
2 Fe(s) + 3/2 O2(g)  Fe2O3(s) ∆H° = -822.3 kJ

23. Enthalpy
H2O (s)  H2O (l)
What is happening?
Is a ∆H° value involved? Why or why not?
ENTHALPY AND HESS’ LAW
Heat is absorbed by the ice in order to melt.
ΔH° is positive, because heat is required

24. Enthalpy
Enthalpy is also involved in physical changes.
ENTHALPY AND HESS’ LAW