1. Chapter 6 - Thermochemistry
DE Chemistry
Dr. Walker

2. Terms To Know Energy Law of Conservation of Energy Potential Energy
The capacity to do work or to produce heat.
Law of Conservation of Energy
Energy can be converted from one form to another but can be neither created nor destroyed.
Potential Energy
Energy due to position or composition.
Chemicals store potential energy in their bonds that can later be released at heat

3. Terms To Know Kinetic energy
energy due to the motion of the object and depends on the mass of the object m and its velocity v
Heat
- The transfer of energy between two objects due
to a temperature difference
Work
- Force acting over a distance

4. State Functions
A state function refers to a property of the system that depends only on its present state.
State functions do not depend on what has happened in the system, or what might happen in the system in the future
State functions are independent of the pathway taken to get to that state.

5. Examples of State Functions
Elevation
Denver – 5280 ft, Chicago – 674 ft
The difference is 4606 ft, no matter what route was taken between those cities.
A liter of water behind a dam has the same potential energy for work regardless of whether it flowed downhill to the dam, or was taken uphill to the dam in a bucket. The potential energy is a state function dependent only on the current position of the water, not on how the water got there.

6. Chemical Energy Exothermic reactions
Reactions that give off energy as they progress
Some of the potential energy stored in the chemical bonds is converted to thermal energy (random KE) through heat
Products are generally more stable (stronger bonds) than reactants

7. Energy Diagram of an Exothermic Reaction
CH4 + 2 O H2O + CO2

8. Chemical Energy Endothermic reactions
Reactions in which energy is absorbed from the surroundings
Energy flows into the system to increase the potential energy of the system
Products are generally less stable (weaker bonds) than the reactants

9. Exothermic Reaction – Energy Diagram
N2 + O NO

10. Thermodynamics The study of energy and its interconversions
First Law of Thermodynamics
The energy of the universe is constant

11. Internal Energy
Total sum of all potential and kinetic energies in the system.
The internal energy (E) of a system can be changed by a flow of work (w), heat (q), or both.

12. E = q + w E = change in internal energy of a system
q = heat flowing into or out of the system
-q if energy is leaving to the surroundings
+q if energy is entering from the surroundings
w = work done by, or on, the system
-w if work is done by the system on the
surroundings
+w if work is done on the system by the
surroundings

13. Example
Calculate E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.

14. Example
Calculate E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ of work is done on the system.
q = kJ
w = kJ
E = q + w
E = kJ kJ = kJ

15. Work, Pressure, and Volume
Expansion
Compression
+V (increase)
-V (decrease)
-w results
+w results
Esystem decreases
Esystem increases
Work has been done by the system on the surroundings
Work has been done on the system by the surroundings

16. PV Work Example
Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm.

17. PV Work Example
Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm.
W = -PDV
W = - (15 atm)(64 L – 46 L)
W = L.atm (this is an energy unit, just not joules)
Negative work – system is doing work on the surroundings with expansion
270 L atm

18. Example
A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate E for the process. (To convert between L atm and J, use 1 L atm = J.)

19. Example
A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes from x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate E for the process. (To convert between L atm and J, use 1 L atm = J.)
E = q + W
q = 1.3 x 108 J
w = - (1.0 atm)(4.50 x x 106 L) = 5.0 x 105 L atm
W = x 105 L atm (101.3 J/L atm) = x 107 J
E = 1.3 x 108 J + (- 5.1 x 107 J) = 8 X 107 J
The heat addition (+q) overcomes the work done on the surroundings (-PV) to have a positive energy for the process

20. Energy Change in Chemical Processes
Endothermic:
Reactions in which energy flows into the
system as the reaction proceeds.
+ qsystem - qsurroundings

21. Energy Change in Chemical Processes
Exothermic:
Reactions in which energy flows out of the system as the reaction proceeds.
- qsystem qsurroundings

22. Calorimetry
The amount of heat absorbed or released during a physical or chemical change can be measured, usually by the change in temperature of a known quantity of water in a calorimeter.

23. Bomb Calorimetry
Steel vessel capable of withstanding large pressures and exploding force (if necessary) of burning reagents.
Container has constant volume = no work done, DE = q

24. Units for Measuring Heat
The Joule is the SI system unit for measuring heat:
The calorie is the heat required to raise the temperature of 1 gram of water by 1 Celsius degree
1 BTU is the heat required to raise the temperature of 1 pound of water by 1 F

25. Specific Heat
The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.
Substance
Specific Heat (J/g·K)
Water (liquid)
4.18
Ethanol (liquid)
2.44
Water (solid)
2.06
Aluminum (solid)
0.897
Carbon (graphite,solid)
0.709
Iron (solid)
0.449
Copper (solid)
0.385
Mercury (liquid)
0.140
Lead (solid)
0.129
Gold (solid)
q = c . m . DT
q = Heat lost or gained
c = Specific Heat Capacity
T = Temperature change
m = mass

26. More on Heat and Terminology
q = c . m . DT
Remember q is heat and heat is a form of energy
heat = q
DH = enthalpy = energy (E)
q = E at constant volume because no “work” is done (w = PDV = 0)
These are all interchangeable terms used in this chapter
This applies to reaction coordinates, calorimetry, and specific heat capacity

27. Calorimetry Example
When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0 C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25 C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1 C.
Assumptions:
Calorimetry absorbs no heat (not true, we’ll see in our lab)
Specific heat of solution is 4.18 J/C g
Density of solution is 1.0 g/mL, gives us mass = 2.0 x 103 g
Calculate the enthalpy change per mole of BaSO4 formed.

28. Calorimetry Example q = c . m . DT
m = 2.0 x 103 g
c = 4.18 J/C g
DT = 28.1 – 25 = 3.1 C
q = (2.0 x 103 g)(4.18 J/C g)(3.1 C) = x 104 J
Why is the sign negative? Temp goes up = exothermic reaction = energy loss by the system

29. A Second, Harder Example
A 30.0-g sample of water at 280. K is mixed with 50.0 g of water at 330. K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. Specific heat of water = 4.18 J/g C

30. A Second, Harder Example
A 30.0-g sample of water at 280. K is mixed with 50.0 g of water at 330. K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. Specific heat of water = 4.18 J/g C
What to do:
If there is no heat loss, the energy of the solutions is equal
Set up specific heat equation for both equal sides
Tf (final temp) is the unknown, plug in everything else and solve
First solution’s temp goes up, Second solution’s temp goes down

31. A Second, Harder Example
A 30.0-g sample of water at 280. K is mixed with 50.0 g of water at 330. K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. Specific heat of water = 4.18 J/g C
q = c . m . DT (solution 1) = c . m . DT (solution 2)
(4.18 J/g C)(30.0 g)(Tf – 280 K) = (4.18 J/g C)(50.0 g)(330 – Tf)
(30.0Tf – 8400) = (16500 – 50.0Tf)
80Tf = 24900
Tf =

32. Hess’s Law
“In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”
Enthalpy is a state function
Only the overall change is important, not the pathway.

33. Hess’s Law Example
Notice that whether the process occurred in one or two steps, the overall process took the same amount of energy overall

34. Enthalpy Changes with Hess’s Law
If a reaction is reversed the sign on DH is reversed
The magnitude of DH is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of DH is multiplied by the same integer

35. Using Hess’s Law Work backward from the final reaction
Reverse reactions as needed, being sure to also reverse DH
Remember that identical substances found on both sides of the summed equation cancel each other

36. Calculate H for the combustion of methane, CH4:
Hess’s Law Example
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ

37. Calculate H for the combustion of methane, CH4:
Hess’s Law Example
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

38. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
C + O2  CO kJ
Step #2: Keep reaction #2 unchanged, because CO2 belongs on the product side

39. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
C + O2  CO kJ
2H2 + O2  2 H2O kJ
Step #3: Multiply reaction #2 by 2

40. Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
    Reaction
Ho  
C + 2H2  CH4 kJ
C + O2  CO2 kJ
H2 + ½ O2  H2O kJ
CH4  C + 2H kJ
C + O2  CO kJ
2H2 + O2  2 H2O kJ
CH4 + 2O2  CO2 + 2H2O kJ
Step #4: Sum up reaction and H

41. Standard Enthalpy (Heat) of Formation
The change in enthalpy that accompanies the formation of mole of a compound from its elements in their standard state

42. Standard State

43. Calculation of Heat of Reaction from Enthalpies of Formation
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Hrxn =  Hf(products) -   Hf(reactants)
    Substance
Hf  
CH4 kJ O2
0 kJ
CO2 kJ
H2O kJ
DHf for elements = 0

44. Calculation of Heat of Reaction from Enthalpies of Formation
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Hrxn =  Hf(products) -   Hf(reactants)
    Substance
Hf  
CH4 kJ O2
0 kJ
CO2 kJ
H2O kJ
DHf for elements = 0
Hrxn = [ kJ + 2( kJ)] – [-74.80kJ]
Hrxn = kJ

45. More Examples of Heat of Formation
Calculate the standard change in enthalpy for the thermite reaction
Hrxn =  Hf(products) -   Hf(reactants)

46. More Examples of Heat of Formation
Calculate the standard change in enthalpy for the thermite reaction
Hrxn =  Hf(products) -   Hf(reactants)
( kJ/mol) (-826 kJ/mol) = kJ/mol

48. Heat of Reaction Example
Give the heat of reaction for the combustion of gasoline, assuming it is pure octane (which it isn’t)
2 C8H18 (l) + 25 O2 (g) CO2 (g) + 18 H2O (g)

49. Heat of Reaction Example
Give the heat of reaction for the combustion of gasoline, assuming it is pure octane (which it isn’t)
2 C8H18 (l) + 25 O2 (g) CO2 (g) + 18 H2O (g)
SDHf (products) = (16 moles x -394 kJ/mol) +
(18 moles x -286 kJ/mole) = kJ
SDHf (reactants) = (2 moles x kJ/mol) +
(0) = -538 kJ
SDHf (products) - SDHf (reactants) =
kJ – (-538 kJ) = kJ
kJ/ 2 moles = kJ/mole