1. Chapter 15.1 notes: Energy Cycles Chapter 15.2: entropy & spontaneity
15.1: The concept of the energy change in a single step reaction being equivalent to the summation of smaller steps can be applied to changes involving ionic compounds.
15.2: a reaction is spontaneous if the overall transformation leads to an increase in total entropy (system plus surroundings). The direction of spontaneous change always increases the total entropy of the universe at the expense of energy available to do useful work. This is known as the second law of thermodynamics.

2. Important terms for these sections
Chapter 15.1
Electron affinity
Ionic lattice
Lattice enthalpy
Born-Haber cycle
Ionic model
Hydrated
Hydration enthalpiee
Chapter 15.2
Spontaneous change
Entropy
Gibbs free energy

3. Energy cycles First ionization energy Electron affinity Ionic lattice
Energy needed to form a positive ion
Electron affinity
The enthalpy change when 1 mol gaseous atoms attracts 1 mol of electrons
See table 8 of the Data Booklet
Ionic lattice
Ionic compounds arrange in a lattice
Occurs when oppositely charged gaseous ions come to together

4. Energy cycles Given NaCl formation:
Na(g)  Na+(g) + e-(g) ΔHiѳ = +496 kJ mol-1 (first ionization energy)
Cl(g) + e-(g)  Cl-(g) ΔHeѳ = kJ mol-1 (electron affinity)
Then the formation of NaCl is:
Na(g) + Cl(g)  Na+(g) + Cl-(g) ΔH = = +147 kJ mol-1
This means the reaction is endothermic… would you expect this?
In reality, the ions react readily:
Na+(g) + Cl-(g)  NaCl(s) ΔHѳ = -790 kJ mol-1
This last part explains the overall exothermic reaction!

5. Lattice enthalpy
Expresses the enthalpy change in terms of the reverse endothermic process
Relates to the formation of gaseous ions from 1 mol of solid crystal breaking into gaseous ions
NaCl(s)  Na+(g) + Cl-(g)
ΔHѳ = +790 kJ mol-1

6. Born-Haber cycle Derived and related to Hess’ Law
Enthalpy level diagram breaking down the formation of an ionic compound into a series of simpler steps

7. Lattice enthalpy

8. Lattice enthalpy Problem
Draw a Born-Haber cycle for magnesium oxide and use it to determine the second electron affinity of oxygen.
Given:
ΔHatѳ[Mg(s)] = +150 kJ mol-1
ΔHatѳ[O2(g)] = +248 kJ mol-1
First ionization energy (Mg) = +736 kJ mol-1
Second ionization energy (Mg) = kJ mol-1
First electron afiinity (O) = -142 kJ mol-1
ΔHfѳ(MgO) = -602 kJ mol-1
Δhlattѳ(MgO) = kJ mol-1

9. Lattice enthalpy Problem HINts
1st ionization Mg(g)  Mg+(g) + e-
2nd ionization Mg+(g)  Mg2+(g) + e-
1st electron affinity O(g) + e-  O-(g)
2nd electron affinity O-(g) + e-  O2-(g)
All ionization energies are endothermic
All second electron affinities are endothermic (e- repulsion)

10. ΔHatѳ[Mg(s)] = +150 kJ mol-1
ΔHatѳ[O2(g)] = +248 kJ mol-1
First ionization energy (Mg) = +736 kJ mol-1
Second ionization energy (Mg) = kJ mol-1
First electron afiinity (O) = -142 kJ mol-1
ΔHfѳ(MgO) = -602 kJ mol-1
Δhlattѳ(MgO) = kJ mol-1

11. We are skipping page 242

12. Comparison of lattice energies (size & charge)
Effect of charge:
The higher the charges on ions, the more strongly they will attract each other
Thus, greater lattice enthalpy
The force is proportional to the product of the charges so force of attraction increases along series
1+/1- < 1+/2- < 2+/2-
Effect of size:
Lattice enthalpy is larger for smaller ions
Example, MgCl2 has a larger lattice enthalpy than NaCl because Mg2+ has a higher charge than Na+ but also because the Mg2+ ion is smaller
The effect of charge causes a larger change in lattice enthalpy than size variations

13. Enthalpies of solution and hydration
ΔHsolѳ = enthalpy change when 1 mol of an ionic substance dissolves in water to give a solution of infinite dilution
Enthalpy change of hydration of an ion:
ΔHhydѳ = enthalpy change when 1 mol of gaseous ions is dissolved to form an infinitely dilute solution of 1 mol of aqueous ions at STP

14. Enthalpies of hydration
Enthalpy change of hydration of an ion:
ΔHhydѳ = enthalpy change when 1 mol of gaseous ions is dissolved to form an infinitely dilute solution of 1 mol of aqueous ions at STP
Smaller ions have more exothermic values
More electrostatic attraction when the ion is smaller
Larger charges also have more exothermic values

15. Enthalpies of solution, hydration & lattice
Enthalpy changes of solution, hydration and lattice are all related
Example: dissolving of NaCl in water is broken into 2 stages:
Lattice enthalpy – breaking apart the lattice into gaseous ions (endothermic)
Hydration of the ions – surrounding the gaseous ions by water molecules (exothermic)
Breaking apart the lattice is endothermic by this definition
Ion-dipole forces are formed between ions and water molecules so hydration enthalpies are always exothermic
Here, you are solving for
ΔHsol will be exothermic if the total enthalpy change
of hydration is more negative than the lattice enthalpy
is positive
**video from online book**

16. Enthalpies of solution, hydration & lattice
Calculation goes like this: ΔHsol(XnYm) = ΔHlatt + nΔHhyd(Xm+) + mΔHhyd(Yn-)
So try it here: (Use Data Booklet sect. 18 & 20)
Dissolve MgCl2 and find the ΔHsol

17. Chapter 15.2: Entropy and Spontaneity
Entropy, S : a measure of how the available energy is distributed among particles
AKA how disordered the system becomes  more disorder = higher entropy
Related to the way energy can be distributed among available energy states in the system (sitting vs running)
Actual values of S can be calculated unlike with enthalpy (only changes can be determined)

18. Chapter 15.2: Entropy and Spontaneity
S has units JK-1 mol-1
Note they are NOT kJ and they are in KELVIN
Changes in S:
ΔSѳ is positive if there is an increase in ‘disorder’ and negative if order is increased
For comparison of similar atoms, entropy increases solid < liquid < gas
Looking at the period 2 elements:
Li Be B C N2 O2 F2 Ne
SOLIDS GASES

19. Chapter 15.2: Entropy and Spontaneity
Predicting the sign of an entropy change
If an increase in number of moles of gas, then ΔSѳ = + (entropy increases)
If a decrease in the number of moles of gas, then ΔSѳ = - (entropy decreases)
Examples:
N2(g) + 3H2(g)  2NH3(g) decrease, ΔSѳ = negative
CaCO3(s)  CaO(s) + CO2(g) increase, ΔSѳ = positive
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) decrease, ΔSѳ = negative
C2H4(g) + H2(g)  C2H6(g) decrease, ΔSѳ = negative
F2(g) + Cl2(g)  2FCl(g) no change, ΔSѳ = approx. zero

20. Chapter 15.2: Entropy and Spontaneity
Calculating entropy changes:
Calculate the entropy change of the reaction N2(g) + 3H2(g)  2NH3(g)
Given:
N2 Sѳ = 192 kJ mol-1
H2 Sѳ = 191 kJ mol-1
NH3 Sѳ = 193 kJ mol-1
ΔSѳ = (2 x 193) – [192 + (3 x 191)]
ΔSѳ = -379 kJ mol-1
A decrease in entropy is what was predicted!

21. Chapter 15.2: Entropy and Spontaneity
Predicting Spontaneous reactions and the Entropy of the Universe!
Spontaneous reaction: one that occurs without any outside influence
Examples:
4Na(s) + O2(g)  2Na2O(s)
Left alone in a container will spontaneously react
C(diamond)  C(graphite)
Also spontaneous but takes a very, very long time to occur
IF the entropy of the Universe increases, then a reaction can be called spontaneous
Energy spreads out from low entropy to high entropy
Low entropy has a higher concentration of energy which means change can occur
High entropy has a low concentration of energy which means change cannot occur

22. Chapter 15.2: Entropy and Spontaneity
ΔSUniverse = ΔSsurroundings + ΔSsystem
When a reaction occurs, if ΔSUniverse is positive, then the reaction is spontaneous
Because the energy release of a reaction can be related to both entropy and enthalpy, we have another equation:
ΔG = ΔH – TΔS
Where ΔG = the change in Gibbs free energy in kJ mol-1
ΔGѳ = ΔHѳ – TΔSѳ
Where ΔGѳ = the standard free energy change in kJ mol-1
For a reaction to be spontaneous, ΔG must be negative

23. Chapter 15.2: Entropy and Spontaneity
ΔGѳ = ΔHѳ – TΔSѳ
For a reaction to be spontaneous, ΔG must be negative
How do the pieces of the equation affect the spontaneity of the reaction?