1. Energetics

2. 5.1 Endothermic and Exothermic Reactions
Endothermic rxn  heat is taken in from the surroundings (rxn vessel gets cooler)
Exothermic rxn rxn that result in the release of heat (rxn vessel gets warmer)
Enthalpy change amount of heat energy taken in/ given out in a chemical rxn (ΔH)
Positive for endo. Rxn
Negative for exo. rxn

3. Exothermic Rxn Energy= Enthalpy Products more stable than reactants
Heat energy has been released
ΔH is negative

4. Endothermic Rxn Energy= Enthalpy Products less stable than reactants
ΔH is positive

5. Stability Exo. produces a more stable product
Ex. Cdiamond Cgraphite ΔH= -1.9 kJ mol-1
Kinetic vs. Thermodynamic stability
Graphite is more thermodynamically more stable than diamond
Diamond is kinetically stable
What does this mean?
Rxn will go very very slowly
Thermodynamic (at a lower energy level)
Kinetic (how well products will react)

6. Activation Energy Define Activation Energy
Minimum energy that colliding particles must have before collision results in a chemical rxn
This is an exo. rxn
Define Activation Energy

7. Activation energy High AE= faster or slower rxn?
Can the conditions be altered?
If yes, in what ways?
Does how endo- or exothermic a rxn is tell us how quickly the rxn will run?
Ex. Which rxn will run faster?
-52 kJ mol-1
-16 kJ mol-1
Do endo. or exo. rxns usually occur spontaneously? (under normal conditions)
1. Slower
2. Yes
3. catalyst, heat, pressure, etc
4. no, there is no connection between rxn speed and enthalpy change
5. Exo.

8. 5.2 Calculations of enthalpy changes from experimental data
Specific heat capacity (c) energy required to raise the temp. of 1 g of substance by 1 K (1°C) or, the energy to raise 1 kg of substance by 1 K.
Units-> J g-1 K-1
J g-1 °C-1
kJ kg-1 K-1
J kg-1 K-1

9. Calculating c
q= mcΔT
q= heat energy
m= mass
ΔT= change in temperature
How does heat capacity affect how easily a substance can be heated?
Can c be calculated for a substance undergoing cooling?
1. Higher the c, the more difficult it is to heat
Ex. Iron is nearly twice aluminum, thus the temp of iron will go up twice that of Al (if all others (q, m) are equal)
2. Yes

10. Measuring enthalpy change of combustion rxns
Worked ex. Page 185
Literature value for ΔH ethanol is kJ mol-1
What are some reasons the calculated value is different?
A bomb calorimeter could have been used so that the system was heavily insulated and provide a plentiful supply of oxygen
Heat loss to surroundings
could have determined c for the can to take that into account
could have insulated the can to reduce heat loss
using something to reduce convection currents around the system
incomplete combustion
evaporation of water and/ or alcohol

11. Enthalpy changes in solution
General method for measuring
Measure known amounts of reagents
Record initial temps
Mix in a polystyrene cup
Record max/ min temperatures observed
Assume that c for the final solution is the same as water

12. ΔH of Solutions: Definitions
Enthalpy change of neutralisation (ΔHn) enthalpy change when 1 mol of water molecules are formed when acid reacts with alkali under standard conditions
H+(aq) + OH-(aq)  H2O(l)
Enthalpy change of solution (ΔHsol) the enthalpy change when 1 mol of solute is dissolved in excess solvent to form a solution of ‘infinite dilution’ under standard conditions
NH4NO3(s) NH4+(aq) + NO3-(aq)
1.ΔHn always exo
2. ‘infinite dilution’ any further dilution of solution produces no further enthalpy change
3. Δhsol endo or exo

13. 5.3 Hess’s Law
The enthalpy change accompanying a chemical reaction is independent of the pathway between the initial and final states
What does this mean?
Ex. Find ΔHr for the reaction of AB
Knowns:
ΔHr= ΔH1 +ΔH2
AC =ΔH1
BC= ΔH2
What is C?
What needs to change about the BC step?
The pathway doesn’t matter, only the initial and final values of enthalpy
An intermediate
The sign must be reversed because we go from ACB to get the final enthalpy

14. Hess’s law: Definitions
State function pathway does not matter
Standard conditions pressure= 1 atm (or, 1.01E5 Pa), 298K (or, 25°C)
Standard enthalpy change (ΔHrΘ) the enthalpy change when molar amounts of reactants as shown in the stoichiometric equation react together under standard conditions to give products (Θ= under standard conditions)

15. Working out enthalpy changes
Hess’s law can be used to determine enthalpy changes of unknows from knowns
P. 194 Worked example

16. 5.4 Bond enthalpies
The enthalpy change when 1 mole of covalent bonds, in a gaseous molecule, are broken under standard conditions (aka bond energy)
Ex. Then enthalpy of H-H bond is 436 kJ mol-1
H2(g)  2H(g) ΔHΘ= +436 kJ mol-1
How many H-H bonds were broken? 1

17. Bond Enthalpy
What state must a substance be in to calculate bond enthalpy?
Consider this process:
Br2(l)  2Br(g) Br-Br= 193 kJ mol-1
What is the ΔHΘ? kJ mol-1
Why is this higher than the bond enthalpy?
The reactants are not in a gaseous state
We must also account for the energy required for vaporisation of the reactants
This process is called atomisation
Gaseous

18. Bond breaking Bond breaking is… endothermic or exothermic?
Endothermic! What does that mean about ΔH?
Positive! What will bond making be?
Exothermic with a negative ΔH

19. Average bond enthalpy
The average amount of energy required to break 1 mole of covalent bonds, in a gaseous molecule under standard conditions
These are the values used to calculate bond enthalpies

20. Using bond enthalpies to work out enthalpy changes in a rxn
Must draw out the structural formulas for rxn
Imagine the rxn happening and ALL bonds being broken
Add up the total energy for all broken bonds
Draw in all the bonds formed in products
Add up the total energy of all bonds made
Determine signs for the total enthalpy changes
Broken positive
Made negative
Add the changes to get the overall enthalpy change of the rxn

21. Example
Consider the rxn between ethene and bromine, to produce 1,2-dibromoethane,
C2H4(g) + Br2(g) C2H4Br2(g)
What bonds are broken?
What bonds are made?
Follow your steps! p

22. Using a cycle
Same concept as previous calculations, but a process is drawn out to see all the steps
P. 203

23. HL2

24. 5.5 Calculating enthalpy changes: Definitions
Standard enthalpy change of combustion (ΔHcΘ) the enthalpy change when 1 mole of a substance is completely burnt in oxygen under standard conditions.
If ΔHcΘ is always negative, what does this mean?
Standard enthalpy change of formation (ΔHfΘ) the enthalpy change when 1 mole of the substance is formed from its elements in their standard states under standard conditions
Endo and exo rxns are dependent on the type of substance
ΔHfΘ for any element in its standard state is zero
1. Always an exothermic rxn

25. Using ΔHcΘ to calculate enthalpy change
Method 1: Construct an enthalpy cycle
P. 208
Method 2: rearrange the equations to give the overall equations related to the enthalpy change
P. 209
Method 3: use an enthalpy level diagram for calculations
P. 210
Method 4: use the equation,
ΔHr = ΣΔHc (reactants)- ΣΔHc (products)

26. Using ΔHfΘ to calculate other enthalpy chages
Method 5: similar to method 1, but used for formation rather than combustion
P. 214
Method 6: refer method 2 (be sure equations are running in the correct direction)
Method 7: draw enthalpy level diagram for formation (method 3)
Method 8: use the equation,
ΔHr = ΣΔHf (products)- ΣΔHf (reactants)

27. Choosing your method
Choose a method based on the data you are given, NOT on what needs to be found
If needing the enthalpy of combustion and given the enthalpy of formation, use one of the methods 5-8
Once the basic principle of the methods are understood, there is no need to have any distinctions between them

28. 5.6 Enthalpy changes for ionic compounds
First ionisation energy
Second ionisation energy
First electron affinity enthalpy change when one electron is added to each atom in 1 mol of gaseous atoms under standard conditions (always EXOTHERMIC)
X (g) +e-  X- (g)
Second electron affinity (always ENDOTHERMIC) why?
Lattice enthalpy(ΔHΘlatt) the enthalpy change when 1 mol of an ionic compound is broken apart into iest constituent gaseous ions under standard conditions
1. It is an unfavorable process to add e- to an ion that is already negative

29. Born-Haber cycles
Enthalpy level diagram breaking down the formation of an ionic compound into a series of simpler steps
1. put the equation for the enthalpy of formation
2. add lattice enthalpy
3. convert to gaseous form (why?)
Two steps
Must convert ALL reactants to gaseous form
Connect the cycle by adding the electrons removed from one reactant to the more electronegative reactant
#3 In order for a substance to be ionised, it must be in the gaseous state

30. Draw a Born-Haber cycleP
Na and Cl example

31. Comparisons of lattice enthalpy
What is lattice enthalpy the result of?
Electrostatic attractions of + and – ions
If the attractions of great, will more or less energy need to be supplied to break the bonds?
More

32. Effect of charge and size
How does the charge of the ions effect lattice enthalpy?
The higher the ion charge, the greater the lattice enthalpy
Does NaCl or MgCl2 have great lattice enthalpy?
MgCl2
How does size effect lattice enthalpy?
The larger the ions the weaker the forces, the smaller the lattice enthalpy
Which has the larger lattice enthalpy, CsCl or NaCl?
NaCl

33. Theoretical vs. experimental
Theoretical assumes a totally ionic model
What is this?
Bonding is solely due to attractive forces between oppositely charged ions
Experimental use the Born-Haber cycle to find
These are compared to determine how ionic a particular compound is

34. How to use theoretical and experimental values
If values are exactly the same, complete ionic bonding is suggested
If values are significantly different, it is suggested that the bonding has a significant degree of covalency
Ex. Silver iodide
Theoretical value/ kJ mol-1  736
Experimental value/ kJ mol-1 876
What do the values suggest?
There is a fairly large degree of covalent character, thus a purely ionic model for the bonding is incorrect

35. Covalent character What is covalent character the result of?
Polarisation of the negative ion by the positive one
How does size of the anion effect this?
The polarisation effect is greater

36. 5.7 ENtropY A measure of randomness or disorder of a system
Especially significant in the case of endothermic processes occurring at standard conditions (ice melting at room temp, water evaporating, NaCl dissolving in water etc)
Endo rxns can only occur if there is an increase in entropy
Represented by S
Units: J K-1 mol-1

37. Standard entropy Represented by SΘ
Positive ΔSΘ indicates increased entropy
Less order
Negative ΔSΘ indicates decreased entropy
More order

38. Predicting sign of entropy change
Which state of matter has the higher entropy?
Gases
Which state has the least entropy?
Solids

39. Period 2 entropies Element Li Be B C N2 O2 F2 Ne State Solid Gas
SΘ / J K-1 mol-1 29 10 6
192
205
203
146

40. Predicting sign
Must consider whether the system’s disorder increases or decreases
Good to consider whether moles of gas have increased or decreased
What would an increase in moles of gas mean?
If moles of gas remain constant, our prediction of a change in entropy would be approximately zero
1. Increase in disorder

41. Calculating entropy change for a rxn
ΔSΘ= ΣSΘproducts -- ΣSΘreactants

42. 5.8 Spontaneity
Spontaneous reaction one that occurs without any outside influence

43. Predicting spontaneity
A reaction being spontaneous does not mean it will run quickly!
Whether a rxn is spontaneous or not under a certain set of conditions can be deduced by looking at the change in the “entropy of the Universe”
ΔSUniverse = ΔSsuroundings + Δssystem
If ΔSUniverse is positive, the entropy of the universe increases and the rxn occurs spontaneously
When heat is given out in a rxn, the entropy of the surroundings get hotter

44. Gibbs free energy Represented by ΔG ΔG = ΔH – TΔS
Also called free energy change
ΔG = ΔH – TΔS
ΔH and ΔS are referring to the system
Under standard conditions this symbol is used: ΔGΘ
For a reaction to be spontaneous, ΔG for the rxn must be NEGATIVE
Units kJ mol-1

45. Calculating ΔG Method 1 use ΔGΘ = ΔHΘ – TΔSΘ
Method 2 use ΔGΘ = ΣΔGfΘ(products) – ΣΔGfΘ(reactant)
Standard free energy of formation the free energy change for the formation of 1 mol of substance from its elements in their standard state under standard conditions
Temp must be in K
If no temp is given and it is standard conditions, assume 298K

46. Outside influence such as; temp, catalysts, etc.
Non-spontaneous rxns
If a rxn is non-spontaneous, does that mean it will never happen? NO
What will make it run?
Outside influence such as; temp, catalysts, etc.

47. Effects of temp on spontaneity
Refer to the equation:
ΔG = ΔH – TΔS
If ΔS is positive, temp must be high or low to be spontaneous?
High
ΔG must be negative to be spontaneous, thus TΔS must be higher than ΔH

48. Temp and spontaneity
If ΔS is negative --TΔS will be positive and the rxn cannot be spontaneous
Endo rxns will only occur spontaneously in entropy is increased and temp is significantly high
Exo rxns will always be spontaneous at some temp
If the rxn involves a decrease in entropy the rxn will be spontaneous at a lower temp; becomes less spontaneous as temp increases

49. Signs ΔH ΔS --TΔS ΔG Spontaneous? -- + Negative At all temps
Becomes more neg as temp increases
Becomes more spontaneous as temp increases
Becomes less neg as temp increases
Becomes less spontaneous as temp increases
Positive
Never