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
6.1
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. Do now!
Let, play, Mr. Bond
Before you have a cow

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

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

16. Bond energies C-H = 435 Kj O=O = 497 Kj
Total for breaking bonds = 4x x497 = 2734 KJ/mol
H-O = 464 Kj
C=O = 803 Kj
Total for making bonds = 2x x464 = 3462 KJ/mol
Total energy change = = KJ/mol

18. Hess’s Law

19. 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

20. 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

21. 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)

22. 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

23. Enthalpy of reaction from bond enthalpies
Calculate the enthalpy change for the hydrogenation of ethene
DH2 1 x C=C bond @ = kJ
4 x C-H bonds @ = kJ
1 x H-H bond @ = kJ
∑ Total energy to break bonds of reactants = kJ
DH3 1 x C-C bond @ = kJ
6 x C-H bonds @ = kJ
∑ Total energy to break bonds of products = kJ
Applying Hess’s Law DH1 = DH2 – DH3 = (2699 – 2824) = – 125 kJ

24. Another Example CH3Br + HI  CH3I + HBr DH = -31 kJ/mol Broken:
3 C-H = kJ/mol
1 C-Br = kJ/mol
1 H-I = kJ/mol
Made:
3 C-H = kJ/mol
1 C-I = kJ/mol
1 H-Br = kJ/mol
DH = -31 kJ/mol

25. “The enthalpy change is independent of the path taken”
HESS’S LAW
“The enthalpy change is independent of the path taken”
How The enthalpy change going from
A to B can be found by adding the
values of the enthalpy changes for
the reactions A to X, X to Y and Y to B.
DHr = DH DH DH3
If you go in the opposite direction of an arrow, you subtract the value of
the enthalpy change.
e.g DH2 = - DH DHr DH3

26. Standard Enthalpy Changes of Reactions
Standard Enthalpy Change of Neutralization
The standard enthalpy change of neutralization (Hneut) is the enthalpy change when one mole of water is formed from the neutralization of an acid by an alkali under standard conditions. ø
e.g. H+(aq) + OH-(aq)  H2O(l) Hneut = kJ mol-1 ø
e.g. The standard enthalpy change of neutralization between HNO3 and NaOH is kJ mol-1
e.g. The standard enthalpy change of neutralization between HCl and NaOH is kJ mol-1

27. Standard Enthalpy Changes of neutralization
-57.1
-57.2
-52.2
-68.6
NaOH
KOH
NH3
HCl HF
Hneu
Alkali
Acid ø
H+(aq) + OH-(aq)  H2O(l)

28. Standard Enthalpy Changes of Reactions
Standard Enthalpy Change of Solution
The standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is dissolved to form an infinitely dilute solution under standard conditions. ø
e.g. NaCl(s) + water  Na+(aq)+Cl-(aq) Hsoln=+ 3.9 kJ mol-1 ø
e.g. LiCl(s) + water  Li+(aq) + Cl-(aq) Hsoln= kJ mol-1 ø
Note that enthalpy changes of solution associate with physical changes.

29. Standard Enthalpy Changes of solution
-44.7
+3.9
-57.8
+20.0
NaOH
NaCl
KOH
KBr
Hsoln(kJ mol-1)
Salt ø

30. 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

31. Standard Enthalpy Changes of Reactions
What is Hf [N2(g)] ? ø
N2(g)  N2(g)
Hf [N2(g)] = 0 ø
The enthalpy change of formation of an element is always zero.

32. 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.

33. 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)

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

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

36. 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

37. 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?

38. Standard Enthalpy Changes of Reactions
Standard Enthalpy Change of Combustion
The standard enthalpy change of combustion (Hc) of a substance is the enthalpy change when one mole of the substance burns completely under standard conditions. ø
e.g. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
Hc = kJ mol mole ø
The standard enthalpy change of combustion of propane is kJ mol-1

39. 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.

40. Develop an analogy for soccer and scoring a goal.

41. Develop an analogy for soccer and scoring a goal.

42. Hess’s Law
Flip Method

43. H2O(g) + C(s) → CO(g) + H2(g)
Example 1
H2O(g) + C(s) → CO(g) + H2(g)
C(s) + ½ O2(g) → CO(g) ∆H = kJ H2(g) + ½ O2(g) → H2O(g) ∆H = kJ
_____________________________________
C(s) + H2O(g)→H2(g) + CO(g) ∆H = kJ

44. 4C(s) + 5H2(g) → C4H10(g)
Example 2
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

45. 4C(s) + 5H2(g) → C4H10(g) 4C(s) + 5H2(g) → C4H10(g) ∆H = -125.6kJ/mol
Example 2
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)

46. Hess’s Law Start Finish A State Function: Path independent.
Both lines accomplished the same result, they went from start to finish.
Net result = same.
#(min University of Berkley)

47. 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.

48. 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 and 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

49. 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?

50. 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.

51. 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

52. enthalpy is a state function and is path independent.
Summary:
enthalpy is a state function
and is path independent.

53. Hess’s Law A + B C + D E Route 1 H1 H3 Route 2 H2 H1 = H2 + H3
Hess’s Law states that the total enthalpy change accompanying a chemical reaction is independent of the route by which the chemical reaction takes place.
Why?

54. 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
But the enthalpy change of such reactions can be determined indirectly by applying Hess’s Law.

55. Enthalpy Change of Formation of CO(g)
Given: Hf [CO2(g)] = kJ mol-1; Hc [CO(g)] = kJ mol-1 ø
C(graphite) + ½O2(g)
CO(g)
Hf [CO(g)] ø
H2
+ ½O2(g)
CO2(g)
H1
+ ½O2(g)
Hf [CO(g)] + H2 = H1 ø
Hf [CO(g)] = H1 - H2 ø
= ( )
= kJ mol-1

56. Calorimetry
Fuse school ( calorimeters and heats of combustion )

57. Introduction to Thermodynamics
10.2
Thermodynamics is the study of the interconversion of heat and other kinds of energy.
In thermodynamics, there are three types
of systems:
An open system can exchange mass and
energy with the surroundings.
A closed system allows the transfer
of energy but not mass.
An isolated system does not exchange
either mass or energy with its
surroundings.

58. Experimental Determination of Enthalpy Changes by Calorimetry
Calorimeter = a container used for measuring the temperature change of solution
Determination of Enthalpy Change of Neutralization

59. Determination of Enthalpy Change of Combustion
Experimental Determination of Enthalpy Changes by Calorimetry
Determination of Enthalpy Change of Combustion

60. A simple calorimeter

61. Specific heat capacity of water = 4.18 KJ kg-1 K-1 or 4.18 J g-1 K1
Specific heat capacity is the amount of heat needed to raise the temperature of 1g of substance by 1K
Specific heat capacity of water = 4.18 KJ kg-1 K-1
or 4.18 J g-1 K1
Be careful with the units it could also be quoted as KJ g-1 K-1
Ensure you use the correct units in your calculation!
To measure the heat released in a process we arrange for the heat to be transferred to a substance (usually water) then measure the temperature rise.

62. Specific Heat and Heat Capacity
The heat associated with a temperature change may be calculated:
m is the mass.
(c) is the specific heat.
ΔT is the change in temperature (ΔT = Tfinal – Tinitial).
C is the heat capacity.
q = mcΔT
q = CΔT

63. Calculate the heat to take 255 g of water from 25.2 C to 90.5 C.
Strategy Use q = mcΔT to calculate q c = J/g∙°C, m = 255 g, ΔT = 90.5°C – 25.2°C = 65.3°C.
Solution
q = 4.184J/gC
× 255 g × 65.3°C = 6.97×104 J or 69.7 kJ
SPECIFIC HEAT PROBLEM
It takes J to heat 25 grams of copper from 25 °C to 75 °C.
What is the specific heat in Joules/g·°C?
q = CΔT
487.5 J = (25 g) c (75 °C - 25 °C) J = (25 g) c (50 °C)
Answer: The specific heat of copper is 0.39 J/g·°C.

64. Cal Problems worked 4 min https://www.youtube.com/watch?v=Hs5x0-IU2F4

65. Qsystem = - Qsurroundings
Example Problem
A 12.9 gram sample of an unknown metal at 26.5°C is placed in a Styrofoam cup containing 50.0 g of water at 88.6°C. The water cools down and the metal warms up until thermal equilibrium is achieved at 87.1°C. Determine the specific heat capacity of the unknown metal. The specific heat capacity of water is 4.18 J/g/°C.
ANSWER:
This problem is like two problems in one. At the center of the problem-solving strategy is the recognition that the quantity of heat lost by the water (Qwater) = (Qmetal). 
Part 1: Determine the Heat Lost by the Water
Given: m = 50.0 g C = 4.18 J/g/°C Tinitial = 88.6°C Tfinal = 87.1°C ΔT = -1.5°C
Solve for Qwater= m•C•ΔT = (50.0 g)•(4.18 J/g/°C)•(-1.5°C)
Qwater = J (unrounded)
(The - sign indicates that heat is lost by the water)
Part 2: Determine the value of Cmetal
Given: Qmetal = J (use a + sign since the metal is gaining heat)
Rearrange Qmetal = mmetal•Cmetal•ΔTmetal to obtain Cmetal = Qmetal / (mmetal•ΔTmetal)
Cmetal = Qmetal / (mmetal•ΔTmetal) = (313.5 J)/[(12.9 g)•(60.6°C)]
Cmetal = J/g/°C

66. Strategy Water constitutes the surroundings; the pellet is the system.
A metal pellet with a mass of g, originally at 88.4°C, is dropped into 125 g of water at 25.1°C. The final temperature is 31.3oC. Calculate the heat capacity C of the pellet.
Strategy Water constitutes the surroundings; the pellet is the system.
Use qsurr = mcΔT to determine the heat absorbed by the water;
then use q = CΔT to determine the heat capacity of the metal pellet, no mass needed as C is J/C
mwater = 125 g, cwater = J/g∙C, and ΔTwater = 6.2°C.
The heat absorbed by the water must be released by the pellet:
qwater = -qpellet ( J lost by pellet)
and ΔTpellet = -57.1°C.
From q = CΔT we have
J = Cpellet × (-57.1°C)
Thus,
Cpellet = 56.8 J/°C
4.184 J
g∙°C
× 125 g × 6.2°C = J gained by water

67. Excess powdered zinc was added to 100ml of 0
Excess powdered zinc was added to 100ml of 0.2 mol/L copper (II) sulphate solution. A temperature rise of 10oC was recorded. Find the MOLAR enthalpy change for the reaction.
H = m x c x T
H = 100g x 4.18 KJ kg-1 K-1 x 10 KJ
1000
= KJ
This is for the no of moles of CuSO4 used in the experiment
No of moles of CuSO4 = 0.2 x 100 = 0.02 moles
H = = 209 KJ mol-1
0.02
The reaction is exothermic so we need to put in a negative sign
Hr = KJ mol-1 of CuSO4

68. Combustion
To find the heat of combustion of a substance a known mass of the substance is burned, the heat released transferred to water and the enthalpy change found as before
In an experiment to find the heat of combustion of ethanol the following results were obtained
Initial mass of lamp + ethanol = g
Final mass of lamp = g
Final temperature of water = oC
Initial temperature of water = oC
Mass of the water = g
What are the products of complete combustion of ethanol?
What mass of ethanol was burnt? How many moles is this?
What quantity of heat was transferred to the water?
Find Hc of ethanol
Identify any sources of error
Is ethanol a good fuel?

69. C2H5OH + 3O2  2CO H2O
q=mc∆t
H = x KJ kg-1 K-1 x 18.6K = 23.3KJ
1000
H lost = h gained
(so 23.3 KJ gained by calorimeter water was lost in burning)
Mass of ethanol used = 0.92g
0.92g = 0.92 g = mol
46g/mol
H=mc∆t / n
Hc = = KJ/mol
0.020

70. Errors
Heat lost to surroundings (air, can thermometer)
Errors in measuring temperature change (unavoidable error in reading thermometer)
Errors in measuring masses (unavoidable error in reading balance)
The enthalpy change of combustion is high ethanol is a good fuel

71. Chemistry in Action: Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l) DH = kJ/mol
Per gram Sugar = 2999 KJ / 180 g
= KJ/gram
1 cal = J
1 Cal = 1000 cal = 4184 J =4.184KJ
If 1.0 gram sugar is = KJ Or
15.7 KJ / KJ/Cal = 3.75 Cal
So sugar burned is approx 4Cal/g
Here 26g x 4 Cal/g = 104 Cal from sugar

72. Foods and Fuels Cheetos burned v Cereal
Foods
1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal.
Energy in us comes from carbohydrates and fats TED X (3 min, Calories)
Intestines: carbohydrates converted into glucose:
C6H12O6 + 6O2  6CO2 + 6H2O, DH = kJ
Fats break down as follows:
2C57H110O O2  114CO H2O, DH = -75,520 kJ
Fats contain more energy; are not water soluble, so are good for energy storage.

73. The END