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 (25ºC)
and 100kPa
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
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
