1. THERMOCHEMISTRY Thermodynamics
To play the movies and simulations included, view the presentation in Slide Show Mode.

2. ENTHALPY CHANGES be able to define enthalpy and enthalpy change (ΔH)
be able to define standard enthalpy
chemical reactions involve the making and breaking of bonds
be able to determine bond enthalpies from data
use mean bond enthalpies to estimate ΔH for reactions
be able to calculate enthalpy change from experimental data
use Hess’s law to calculate enthalpy changes in reactions

3. Internal Energy and Enthalpy
e.g Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
Heat change at constant pressure =
Change in internal energy +
Work done on the surroundings
(Heat change at constant volume)
Enthalpy change

4. THERMODYNAMICS
First Law Energy can be neither created nor destroyed but
It can be converted from one form to another
Second
Law Energy flows from hot to cold
Third Law A perfectly ordered crystal has zero entropy.
Chemical reactions either absorb or give off energy
Exothermic Energy is given out
Endothermic Energy is absorbed
Every Day Examples
Exothermic combustion of fuels
respiration (oxidation of carbohydrates)
Endothermic photosynthesis
thermal decomposition of calcium carbonate

5. Exothermic and Endothermic Reactions
An exothermic reaction is a reaction that releases heat energy to the surroundings. (H = -ve)

6. Exothermic and Endothermic Reactions
An endothermic reaction is a reaction that absorbs heat energy from the surroundings. (H = +ve)

7. Bond breaking - endothermic
Energy is always required to be inputted to break a bond. Bond breaking is always endothermic.

8. Bond making - exothermic
Energy is always released when a bond is formed. Bond making is always exothermic.

9. Enthalpy IN to Breaking Bonds
e.g. CH4 + 2O2 CO2 + 2H2O
In an exothermic reaction, the energy required in breaking the bonds in the reactants is less than the energy released in forming the bonds in the products (products contain stronger bonds).

10. Energy out, to make…. Bonds
In an endothermic reaction, the energy in breaking the bonds is more than the energy out making the bonds

11. Single bond < Double bond < Triple bond
The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy.
H2 (g)
H (g) +
DH0 = kJ
Cl2 (g)
Cl (g) +
DH0 = kJ
HCl (g)
H (g) +
Cl (g)
DH0 = kJ
O2 (g)
O (g) +
DH0 = kJ
N2 (g)
N (g) +
DH0 = kJ O
Bond Energies
Single bond < Double bond < Triple bond N
Notes • strength of bonds also depends on environment; MEAN values quoted
• making bonds is exothermic as it is the opposite of breaking a bond

12. Calculation of enthalpy of reaction from bond energies
Only the net number and types of bonds broken (red bonds above) and formed (blue bonds above) are included in the calculation. 12

13. Combustion of methane
CH4(g) + 2O2(g) 2H2O(l) + CO2(g) + ENERGY

14. Burning Methane CH4 + 2O2 2H2O + CO2 Methane Oxygen Water
Carbon dioxide

15. Bond energies C-H = 413KJ O=O = 498 KJ
Total for breaking bonds = 4x x497 = 2658 KJ/mol
H-O = 464 Kj
C=O = 805 KJ
Total for making bonds = 2x x464 = 3466 KJ/mol
Total energy change = = KJ/mol

16. Calculating ΔH CH4(g) + 2O2(g) 2H2O(l) + CO2(g)
Bond
Bond energy KJ/mol
H-H
436
Cl-Cl
242
H-Cl
431
C-H
413
C-C
347
C-O
335
O=O
498
Calculating ΔH
CH4(g) + 2O2(g) 2H2O(l) + CO2(g)
Bonds broken = 4 x (C-H) + 2 x (O=O)
= 4 x x 498
= = 2658 KJ/mol

17. ΔH CH4(g) + 2O2(g) 2H2O(l) + CO2(g) Bonds made = 4 x (O-H) + 2 x (C=O)
Bond energy KJ/mol
H-H
436
Cl-Cl
242
H-Cl
431
C-H
413
C-C
347
C-O
335
O=O
498 ΔH
CH4(g) + 2O2(g) 2H2O(l) + CO2(g)
Bonds broken = 4 x (C-H) + 2 x (O=O)
(Previous page) = 4 x x 498
= = 2658 KJ/mol
Bonds made = 4 x (O-H) + 2 x (C=O)
= 4 x x -805
= = KJ/mol

18. ΔH CH4(g) + 2O2(g) 2H2O(l) + CO2(g)
Bond
Bond energy KJ/mol
H-H
436
Cl-Cl
242
H-Cl
431
C-H
413
C-C
347
C-O
335
O=O
498
CH4(g) + 2O2(g) 2H2O(l) + CO2(g)
Bonds broken = 4 x (C-H) + 2 x (O=O)
= 4 x x 498
= = 2658 KJ/mol
Bonds made = 4 x (O-H) + 2 x (C=O)
= 4 x x -805
= = KJ/mol
Overall Energy change = = -808 KJ/mol
(Exothermic)

19. Standard Enthalpy Changes
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) H = -808 kJ mol-1
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = -890 kJ mol-1
As enthalpy changes depend on temperature and pressure, chemists find it convenient to report enthalpy changes based on an internationally agreed set standard conditions:
1. elements or compounds in their normal physical states; 2. a pressure of 1 atm ( KPa); and 3. a temperature of 250C (298 K)
Enthalpy change under standard conditions denoted by symbol: H ø

20. Standard Enthalpy Changes of Reactions
Standard Enthalpy Change of Formation
The standard enthalpy change of formation (Hf) is the enthalpy change of the reaction when one mole of the compound in its standard state is formed from its constituent elements under standard conditions. ø
e.g. 2Na(s) + Cl2(g)  2NaCl(s) H = -822 kJ mol-1 ø
Standard enthalpy change of formation of NaCl is -411 kJ mol-1.
Na(s) + ½Cl2(g)  NaCl(s) Hf = -411 kJ mol mole ø

21. STANDARD ENTHALPY OF FORMATION
Definition The enthalpy change when ONE MOLE of a compound is formed in its
standard state from its elements in their standard states.
Symbol DH°f This little f is VERY important
Example(s) C(graphite) + O2(g) ———> CO2(g)
H2(g) ½O2(g) ———> H2O(l)
2C(graphite) + ½O2(g) H2(g) ———> C2H5OH(l)
Notes Only ONE MOLE of product on the RHS of the equation
Elements In their standard states have zero enthalpy of formation.

22. Enthalpies of Formation CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Using Enthalpies of Formation, Calculate Enthalpies of Reaction
For a reaction (Method #2)
Note: n & m are stoichiometric coefficients.
Calculate heat of reaction for the combustion of methane gas giving carbon dioxide and water.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)

23. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
∆Hof (kJ/mol): Try this

24. One more time!!!!
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) + ???

25. Hess’s Law Second Method
reactants
products
Hreaction
elements
 Hf [reactants]
Methane and oxygen destroyed ø
Hf [products]
CO2 and water formed ø
Hreaction =  Hf [products] -  Hf [reactants] ø

26. Standard Enthalpy Changes of Reactions
Standard Enthalpy Change of Combustion
e.g. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l) H1 = kJ
2C3H8(g) + 10O2(g) 6CO2(g) + 8H2O(l) H2 = ?
H2 = kJ
 It is more convenient to report enthalpy changes per mole of the main reactant reacted/product formed.
ANY IDEAS WHAT IT WOULD BE FOR THE FORMATION OF g OF WATER?

27. 3rd Method For Heats of Rx
Really this is NOT another law, it is still Hess’s Law BUT It looks different, yet we get to the same place , the same answer.
There is an old Chinese proverb which says: There are many ways to the top of a mountain, but the view from the top is always the same.

28. Hess’s Law
Flip Method

29. 4C(s) + 5H2(g) → C4H10(g)
Example 1
Use these equations to calculate the molar enthalpy change to make butane gas. C4H10(g) + 6½O2(g) → 4CO2(g) + 5H2O(g) ∆H= kJ/mol C(s) + O2(g) → CO2(g) ∆H= kJ/mol H2(g) + ½O2(g) → H2O(g) ∆H= kJ/mol

30. 4C(s) + 5H2(g) → C4H10(g) 4C(s) + 5H2(g) → C4H10(g) ∆H = -125.6kJ/mol
Example 1
4C(s) + 5H2(g) → C4H10(g)
5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= kJ/mol
4C(s) + 4O2(g) → 4CO2(g) ∆H= 4(-393.5kJ/mol)
5H2(g) + 2½O2(g) → 5H2O(g) ∆H= 5(-241.8kJ/mol)
_____________________________________________________
∆H = kJ/mol
4C(s) + 5H2(g) → C4H10(g)

31. Determine the heat of reaction for the reaction:
4NH3(g) O2(g)  4NO(g) + 6H2O(g)
Using the following sets of reactions:
N2(g) + O2(g)  2NO(g) H = kJ
N2(g) + 3H2(g)  2NH3(g) H = kJ
2H2(g) + O2(g)  2H2O(g) H = kJ
Hint: The three reactions must be algebraically manipulated to sum up to the desired reaction.
and.. the H values must be treated accordingly.

32. 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
Goal:
4NH3(g) O2(g)  4NO(g) + 6H2O(g)
Using the following sets of reactions:
N2(g) + O2(g)  2NO(g) H = kJ
N2(g) + 3H2(g)  2NH3(g) H = kJ
2H2(g) O2(g)  2H2O(g) H = kJ
4NH3  2N H2 H = kJ
NH3:
Reverse & x 2
Found in more than one place, SKIP IT (its hard).
O2 :
NO: x2
2N O2  4NO H = kJ
H2O: x3
6H O2  6H2O H = kJ

33. 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
Goal:
4NH3(g) O2(g)  4NO(g) + 6H2O(g)
4NH3  2N H2 H = kJ
NH3:
Reverse and x2
Found in more than one place, SKIP IT.
O2 :
NO: x2
2N O2  4NO H = kJ
H2O: x3
6H O2  6H2O H = kJ
Cancel terms and take sum.
+ 5O2
+ 6H2O
H = kJ
4NH3 
4NO
Is the reaction endothermic or exothermic?

34. Determine the heat of reaction for the reaction:
C2H4(g) + H2(g)  C2H6(g)
Use the following reactions:
C2H4(g) O2(g)  2CO2(g) H2O(l) H = kJ
C2H6(g) + 7/2O2(g)  2CO2(g) H2O(l) H = kJ
H2(g) /2O2(g)  H2O(l) H = -286 kJ
Consult your neighbor if necessary.

35. Determine the heat of reaction for the reaction:
Goal: C2H4(g) + H2(g)  C2H6(g) H = ?
Use the following reactions:
C2H4(g) O2(g)  2CO2(g) H2O(l) H = kJ
C2H6(g) + 7/2O2(g)  2CO2(g) H2O(l) H = kJ
H2(g) /2O2(g)  H2O(l) H = -286 kJ
C2H4(g) :use 1 as is C2H4(g) O2(g)  2CO2(g) H2O(l) H = kJ
H2(g) :# 3 as is H2(g) /2O2(g)  H2O(l) H = -286 kJ
C2H6(g) : rev # CO2(g) H2O(l)  C2H6(g) + 7/2O2(g) H = kJ
C2H4(g) + H2(g)  C2H6(g) H = -137 kJ

36. Importance of Mathematical Hess’s Law
The enthalpy change of some chemical reactions cannot be determined directly because:
the reactions cannot be performed in the laboratory
the reaction rates are too slow
the reactions may involve the formation of side products