1. THERMOCHEMISTRY

2. 3 MAIN PROCESSES THAT WILL ALTER THE ENERGY OF A CHEMICAL SYSTEM
Heating and cooling
Phase changes
Chemical reactions

3. Definitions #1 Energy: The capacity to do work or produce heat
Potential Energy: Energy due to position or composition
Kinetic Energy: Energy due to the motion of the object

4. Definitions #2
Law of Conservation of Energy: Energy can neither be created nor destroyed, but can be converted between forms
The First Law of Thermodynamics: The total energy content of the universe is constant

5. Depend ONLY on the present state of the system
State Properties (State Functions)
The state of a system is described by its composition,
temperature and pressure.
Depend ONLY on the present state of the system
ENERGY IS A STATE FUNCTION
A person standing at the top of Mt. Everest has the same potential energy whether they got there by hiking up, or by falling down from a plane!
WORK IS NOT A STATE FUNCTION
WHY NOT???

6. SYSTEM and SURROUNDINGS
OPEN VS CLOSED SYSTEMS
THERMODYNAMIC FUNCTIONS = 3 PARTS

7. 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

8. E = q + w
Try This!
Calculate ΔE 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.
E = q + w
ΔE = 15.6 kJ kJ = 17.0 kJ
The system has gained 17.0 kJ of energy!

9. Work, Pressure, and Volume
Gas
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

10. Work, Pressure, and Volume
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 = -15 atm x 18 L = -270 L x atm
(gas is expanding , therefore it is doing
work on its surroundings!)
Energy is flowing out of the gas, so W is a negative quantity

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

12. Endothermic Reactions

13. Exothermic Reactions

14. TIME TO REVIEW CHAPTER 5 EXAM
THE FOLLOWING INDIVIDUALS WILL MOVE TO THE CENTER OF
THEIR RESPECTIVE WORKSTATIONS:
JEREMY
RANJEEKA
KUSH
SINDY
PRIYANKA
SHYAMA
PLEASE REMAIN AT YOUR WORKSTATION DURING THIS CHAPTER 6
MULTIPLE CHOICE REVIEW SESSION.

15. 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
1 calorie is the heat required to raise the temperature of 1 gram of water by 1 C
1 BTU is the heat required to raise the temperature of 1 pound of water by 1 F

16. 4.18 joule = 1 calorie The Joule
The unit of heat used in modern thermochemistry is the Joule
4.18 joule = 1 calorie

17. A Cheaper Calorimeter

18. A Bomb Calorimeter

19. Heat Capacity C heat absorbed C = Increase in temperature
Not enough!…………..When a substance is heated, the energy
required will depend on ???
AMOUNT OF SUBSTANCE
Specific Heat Capacity – J/oC x g or J/K x g
Molar Heat Capacity – J/oC x mol or J/K x mol

20. Calculations Involving Specific HeatOR
c = Specific Heat Capacity
q = Heat lost or gained
T = Temperature change
m = mass of the solution

21. ΔH
What is this?????
ENTHALPY

22. Let’s Test Our “Calorimetry “Acumen”
Suppose we mix 50.0 mL of 1.0 M HCl at 25oC with 50.0 mL of 1.0 M NaOH also at 25oC in a calorimeter. After the reactants are mixed by stirring, the temperature is observed to increase to 31.9oC. (Assume that the solution can be treated as if it were pure water with a density of 1.0 g/mL). How much energy is released?

23. Answer is………….???
2.9 x 103 J

24. 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.”

25. Hess’s Law
Δ H

26. Hints for Using Hess’s Law
work backward from the required reaction,
using the reactants and products to decide
how to manipulate the other given
reactions at your disposal.
Reverse any reactions as needed to give
the required reactants and products.
Multiply reactions to give the correct
numbers of reactants and products.

27. 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
Step #1: CH4 must appear on the reactant side, so we reverse reaction #1 and change the sign on H.

28. Hess’s Law Example Problem
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

29. Hess’s Law Example Problem
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 #3 by 2

30. Hess’s Law Example Problem
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

31. Two forms of carbon are graphite and diamond. Using the
enthalpies of combustion for graphite (-394 kJ/mol) and
diamond (-396 kJ/mol), calculate H for the conversion of
graphite to diamond:
Cgraphite(s) Cdiamond(s)
The combustion reactions are:
Cgraphite(s) + O2(g) CO2(g) H = -394 kJ
Cdiamond(s) + O2(g) CO2(g) H = -396 kJ
Cgraphite(s) + O2(g) CO2(g) H = -394 kJ
CO2(g) Cdiamond(s) + O2(g) ) H = -(-396 kJ)
Cgraphite(s) Cdiamond(s) ) H = 2 kJ

32. 6.4 Standard Enthalpies of Formation
The standard enthalpy of formation (ΔHof) of a compound
is defined as the change in enthalpy that accompanies the
formation of one mole of a compound from its elements
with all substances in their standard state
The above degree symbol indicates that the corresponding
process has been carried out under standard conditions.
The standard state for a substance is a precisely defined
reference state.
Conventional Definitions of Standard States for a compound or element
are located on pg. 246 in your textbook.

33. Key Concepts When Doing Enthalpy Calculations:
When a reaction is reversed, the magnitude of ΔH remains
the same, but its sign changes
When the balanced equation for a reaction is multiplied
by an integer, the value of ΔH for that reaction must be
multiplied by the same integer.
Elements in their standard states are not included in the
ΔH reaction calculations. That is ΔH of for an element in its
standard state is zero.

34. ΔH o reaction = Σnp ΔH of(products) - Σnr ΔH of(reactants)
The change in enthalpy for a given reaction can be
calculated from the enthalpies of formation of the reactants
and products:
ΔH o reaction = Σnp ΔH of(products) - Σnr ΔH of(reactants)

35. Calculation of Heat of Reaction
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 O
Hrxn = [ kJ + 2( kJ)] – [-74.80kJ]
Hrxn = kJ

36. Specific Heat Capacity
The amount of heat required to raise the temperature of one gram of substance by one degree Celsius.
Substance
Specific Heat (J/g·C)
Water (liquid)
4.18
Ethanol (liquid)
2.44
Water (solid)
2.06
Water (vapor)
1.87
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)