1. Topics 5 & 15 Chemical Thermodynamics
IB CHEMISTRY
Topics 5 & 15 Chemical Thermodynamics
Jeff Venables Northwestern High School

2. Standard Enthalpy of Combustion
ΔHc = The enthalpy change when one mole of the compound undergoes complete combustion in excess oxygen under standard conditions.

3. Born-Haber Cycle Application of Hess’s Law
Vaporize the metal (enthalpy of vaporization)
Na (s)  Na (g)
Break diatomic nonmetal molecules (if applicable) (bond enthalpy)
½ Cl2 (g)  Cl-
Remove electron(s) from metal (ionization energy)
Na (g)  Na+ (g) + e-
Add electron(s) to nonmetal (electron affinity)
Cl (g) + e-  Cl- (g)
Put ions together to form compound (lattice energy)
Na+ + Cl-  NaCl (s)

4. Overall Reaction:
Na (s) + ½ Cl2 (g)  NaCl (s)
This is useful because all quantities are directly measurable except lattice energy. The Born-Haber cycle can be used to calculate lattice energy from the other values.

5. Step
Energetics 1 2 3 4 5

6. Step
Energetics 1
endothermic 2 3 4
Exothermic 5
EXOTHERMIC (highly)
If the theoretical lattice energy is significantly different than the experimentally determined lattice energy for an ionic compound, this indicates that the bond has some covalent character. (15.2.4)

7. Spontaneous Processes
Thermodynamics is concerned with the question: can a reaction occur?
First Law of Thermodynamics: energy is conserved.
Any process that occurs without outside intervention is called spontaneous.
When two eggs are dropped they spontaneously break.
The reverse reaction is not spontaneous.
We can conclude that a spontaneous process has a direction.

9. Entropy
Entropy, S, is a measure of the disorder of a system.
Spontaneous reactions proceed to lower energy or higher entropy (or both).
In ice, the molecules are very well ordered because of the H-bonds.
Therefore, ice has a low entropy.
As ice melts, the intermolecular forces are broken (requires energy), but the order is interrupted (so entropy increases).
Water is more random than ice, so ice spontaneously melts at room temperature.

10. For a system, S = Sfinal - Sinitial.
If S > 0 the disorder increases, if S < 0 the order increases.

11. The Second Law of Thermodynamics
The second law of thermodynamics explains why spontaneous processes have a direction.
In any spontaneous process, the entropy of the universe increases.
Suniv = Ssys + Ssurr: the change in entropy of the universe is the sum of the change in entropy of the system and the change in entropy of the surroundings.
Entropy is not conserved: Suniv is increasing.

12. The Molecular Interpretation of Entropy
A gas is less ordered than a liquid that is less ordered than a solid.
Aqueous ions are less ordered than pure solids and liquids, but more ordered than gases
Any process that increases the number of gas molecules leads to an increase in entropy.
When NO(g) reacts with O2(g) to form NO2(g), the total number of gas molecules decreases, and the entropy decreases.

14. Examples – Determine the sign of ΔS for each of the following:
Na (s) + ½ Cl2 (g)  NaCl (s)
N2 (g) + 3 H2 (g)  2 NH3 (g)
2 H2 (g) + O2 (g)  2 H2O (l)
H2O (l)  H2O (g)
NaCl (s)  Na+ (aq) + Cl- (aq)

15. Examples – Determine the sign of ΔS for each of the following:
Na (s) + ½ Cl2 (g)  NaCl (s) negative
N2 (g) + 3 H2 (g)  2 NH3 (g) negative
2 H2 (g) + O2 (g)  2 H2O (l) negative
H2O (l)  H2O (g) positive
NaCl (s)  Na+ (aq) + Cl- (aq) positive

16. The Molecular Interpretation of Entropy
There are three atomic modes of motion:
translation (the moving of a molecule from one point in space to another),
vibration (the shortening and lengthening of bonds, including the change in bond angles),
rotation (the spinning of a molecule about some axis).

17. Energy is required to get a molecule to translate, vibrate or rotate.
The more energy stored in translation, vibration and rotation, the greater the degrees of freedom and the higher the entropy.
In a perfect crystal at 0 K there is no translation, rotation or vibration of molecules. Therefore, this is a state of perfect order (zero entropy).

18. The Molecular Interpretation of Entropy

19. Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is zero.
Entropy changes dramatically at a phase change.
As we heat a substance from absolute zero, the entropy must increase.
If there are two different solid state forms of a substance, then the entropy increases at the solid state phase change.

21. Boiling corresponds to a much greater change in entropy than melting.
Entropy will increase when
liquids or solutions are formed from solids,
gases are formed from solids or liquids,
the number of gas molecules increase,
the temperature is increased.

22. Entropy Changes in Chemical Reactions
Absolute entropy can be determined from complicated measurements.
Standard molar entropy, S: entropy of a substance in its standard state. Similar in concept to H.
Units: J mol-1 K-1. Note units of H: kJ mol-1.
Standard molar entropies of elements are not zero.
For a chemical reaction which produces n moles of products from m moles of reactants:

23. Examples – Calculate ΔS for each of the following reactions:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
N2 (g) + 3 H2 (g)  2 NH3 (g)
2 SO3 (g)  2 SO2 (g) + O2 (g)
HCl (g)  H+ (aq) + Cl- (aq)

24. Examples – Calculate ΔS for each of the following reactions:
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
-5.0 J K-1
N2 (g) + 3 H2 (g)  2 NH3 (g)
J K-1
2 SO3 (g)  2 SO2 (g) + O2 (g)
189.6 J K-1
HCl (g)  H+ (aq) + Cl- (aq)
J K-1