1. List as many forms of energy and examples that you can think of:
BELL RINGER
What is energy to you?
List as many forms of energy and examples that you can think of:

3. Chapter 5 Thermochemistry
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 5 Thermochemistry

4. Energy The ability to do work or transfer heat.
Work: Energy used to cause an object that has mass to move.
Heat: Energy transferred from hotter to colder objects

5. Kinetic Energy
Energy of motion 1 2
KE =  mv2

6. Calculate kinetic energy
Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s.
 KE = 0.5*m*v2
KE = (0.5) * (625 kg) * (18.3 m/s)2
KE = 1.05 x105 Joules

7. Calculate potential energy
A cart is loaded with a brick and pulled at constant speed along an inclined plane to the height of a seat-top. If the mass of the loaded cart is 3.0 kg and the height of the seat top is 0.45 meters, then what is the potential energy of the loaded cart at the height of the seat-top?
PE = m*g*h
PE = (3 kg ) * (9.8 m/s/s) * (0.45 m)
PE = 13.2 J

8. Potential Energy
Energy an object possesses by virtue of its position or chemical composition.
Ep= mgh g = 9.8 m/s2

9. Electrostatic Potential Energy
arises from the interaction between charged particles
Eel = (kQ1Q2) / d
k= 8.99 x 10^9 J-m/C2 C=Coulomb, unit of electrical charge
Q1,Q2= electrical charges
Both + = charges repel each other Eel = +
Opposite charges = attract each other Eel = -
*The lower the energy of a system= the more stable

10. Units of Energy The SI unit of energy is the joule (J). kg m2
The calorie (cal) is the amount of energy required to raise the temperature of 1 g of water by 1 C.
1 cal = J
Calorie (Cal) is energy unit for nutrition.
1 Cal = 1000 cal = 1 kcal
1 J = 1 
kg m2 s2

12. Calculate the amount of calories and joules are in 1 serving of Doritos:

13. Calculate the amount of calories and joules are in 1 serving of Coke:
140 Calories= 140,000 calories= 585,760 J

14. System and Surroundings
The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).
The surroundings are everything else (here, the cylinder and piston).

15. Closed System- energy can be exchanged with surroundings but not matter
Isolated System- neither exchanged
Open System- matter and energy can be exchanged with surroundings

16. Work Energy used to move an object over some distance. w = F  d,
where w is work, F is the force, and d is the distance over which the force is exerted.

17. Heat Energy can also be transferred as heat.
Heat flows from warmer objects to cooler objects.

18. Transferal of Energy
The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.

19. Transferal of Energy
The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.
As the ball falls, its potential energy is converted to kinetic energy.

20. Transferal of Energy
The potential energy of this ball of clay is increased when it is moved from the ground to the top of the wall.
As the ball falls, its potential energy is converted to kinetic energy.
When it hits the ground, its kinetic energy falls to zero (since it is no longer moving); some of the energy does work on the ball, the rest is dissipated as heat.

21. 5.2 First Law of Thermodynamics
Energy is neither created nor destroyed.
In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa.
Use Fig. 5.5

22. Internal Energy
The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

23. Internal Energy
By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:
E = Efinal − Einitial
In chemical reaction initial = reactants; final = products
Use Fig. 5.5

24. Changes in Internal Energy
If E > 0, Efinal > Einitial
Therefore, the system absorbed energy from the surroundings.
This energy change is called endergonic.

25. Changes in Internal Energy
If E < 0, Efinal < Einitial
Therefore, the system released energy to the surroundings.
This energy change is called exergonic.

26. Changes in Internal Energy
When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).
That is, E = q + w.

27. E, q, w, and Their Signs

28. The internal energy for Mg(s) and Cl2(g) is greater than that of MgCl2(s). Sketch an energy diagram that represents the reaction MgCl2(s) Mg(s) + Cl2(g)
Mg(s) + Cl2(g)
MgCl2

29. Exchange of Heat between System and Surroundings
When heat is absorbed by the system from the surroundings, the process is endothermic.
(feels cold)

30. Exchange of Heat between System and Surroundings
When heat is released by the system to the surroundings, the process is exothermic.
(feels hot)

31. A system can lose energy in two ways:
  the system can release energy (q) to the surroundings
  the system can do work (w) on the surroundings
 A system does 20 J of work on the surroundings. The change in energy of the system, E, is ______ J.
 A system has a change in energy of -25 J. No work is done by or on the system, so _______ J of heat
must have been _____________________ (absorbed / released).
 A system has a change in energy of +45 J. If 30 J of work was done on the system, then _______ J of heat must have been _____________________ (absorbed / released).

32. Determine the change in energy, E, for each system:
A system gives off 25.0 kJ of heat and has 15.0 kJ of work done on it. _____
lose 25.0, gain 15.0 = kJ
A system takes in 75.0 kJ of heat and has 25.0 kJ of work done on it. _____
gains 75.0, gain 25.0 = kJ
A system does 45.0 kJ of work and loses 80.0 kJ of heat. _____
lose 45.0, lose 80.0 = kJ

33. Classify each statement as talking about an [EXO]thermic or [ENDO]thermic reaction:
_____  surroundings get hot
_____  PE diagram is uphill
_____  energy is a product
_____  H is positive
_____  reactants have more energy
_____ H is negative
_____ PE diagram is downhill
_____ surroundings get cold
_____ products have more energy
_____ energy is a reactant
EXO
ENDO

34. State Functions
Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

35. State Functions
However, we do know that the internal energy of a system is independent of the path by which the system achieved that state.
In the system below, the water could have reached room temperature from either direction.

36. State Functions Therefore, internal energy is a state function.
It depends only on the present state of the system, not on the path by which the system arrived at that state.
And so, E depends only on Einitial and Efinal.

37. Many paths, same destination

38. State Functions However, q and w are not state functions.
Whether the battery is shorted out or is discharged by running the fan, its E is the same.
But q and w are different in the two cases.

39. 5.3 Enthalpy Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g) video clip
Hydrogen gas produced expands against atmosphere which requires the system to do work

40. Work
When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

41. Work
We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston: pressure-volume work
w = −PV
P=pressure, V = Final – initial volume
negative sign necessary to conform to sign conventions from table

42. Greek word enthalpein= to warm
Enthalpy
Greek word enthalpein= to warm
If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system.
Enthalpy is the internal energy plus the product of pressure and volume:
H = E + PV

43. Enthalpy
When the system changes at constant pressure, the change in enthalpy, H, is
H = (E + PV)
This can be written
H = E + PV

44. Enthalpy
Since E = q + w and w = −PV, we can substitute these into the enthalpy expression:
H = E + PV
H = (q+w) − w
H = q
So, at constant pressure the change in enthalpy is the heat gained or lost.

45. Endothermicity and Exothermicity
A process is endothermic, then, when H is positive.

46. Endothermicity and Exothermicity
A process is endothermic when H is positive.
A process is exothermic when H is negative.

47. Enthalpy is a state function bc internal energy, pressure, and volume are all state functions.

48. Determine the sign of ΔH
Endothermic= process in which heat is absorbed = (+)
Exothermic= process in which heat is released = (-)
a. ice cube melts
+ ice cube absorbs heat from surroundings= endothermic
b. 1 g butane is combusted into CO2 and H2O
- reaction gives off heat = exothermic

49. 5.4 Enthalpies of Reaction
The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:
H = Hproducts − Hreactants

50. Enthalpies of Reaction
This quantity, H, is called the enthalpy of reaction, or the heat of reaction.
2H2(g) + O2(g)  2H2O(g) ΔH = kJ
Balloon filled with Hydrogen and oxygen gas

51. The Truth about Enthalpy
Enthalpy is an extensive property.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ΔH=-890 kJ
2CH4(g)+4O2(g)2CO2(g)+4H2O(l) ΔH=-1780 kJ
2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.
CO2(g) + 2H2O(l) CH4 + 2O2(g) ΔH=890 kJ
3. H for a reaction depends on the state of the products and the state of the reactants.
CH4(g)+2O2(g)  CO2(g) + 2H2O(g) ΔH=-802 kJ

52. Practice Hydrogen peroxide can decomposed into water and oxygen:
2H2O2(l)  2H2O(l) + O2(g) ΔH= -196kJ
Calculate the value of q when 5.00g of H2O2(l) decomposes at constant pressure.
-14.4 kJ

53. Homework
pg 203 # odd

54. Calorimetry
Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.
Calorimeter device used to measure heat flow

55. Heat Capacity and Specific Heat
The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity.
The greater the heat capacity the more heat required to increase the temperature.
We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K.

56. Heat Capacity and Specific Heat
heat transferred
mass  temperature change
s = q
m  T

57. Practice
Large beds of rock are used in some solar-heated homes to store heat. Specific heat of rocks=0.82 J/gK.
a. Calculate heat absorbed by 50.0 kg of rocks if temperature increased by 12°C.
4.9 x 10^5 J
b. What temperature change would these rocks undergo if they emitted 450 kJ of heat?
11 K decrease

58. Constant Pressure Calorimetry
By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

59. Constant Pressure Calorimetry
Because the specific heat for water is well known (4.184 J/mol-K), we can measure H for the reaction with this equation:
q = m  s  T

60. Bomb Calorimetry
Reactions can be carried out in a sealed “bomb,” such as this one, and measure the heat absorbed by the water.

61. Bomb Calorimetry
Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H.
For most reactions, the difference is very small.

62. Hess’s Law
H is well known for many reactions inconvenient to measure H for every reaction we are interested.
However, we can estimate H using H values that are published.

63. Hess’s Law
States that “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

64. Hess’s Law
REMEMBER H is a state function Therefore total enthalpy change depends only on initial state of reactants and final state of the products.

68. Enthalpies of Formation
By using the methods we have discussed, we can calculate the H change for a great many reactions from tabulate ΔH values:
enthalpies of vaporization (ΔH for converting liquid to gas)
enthalpies of fusion (ΔH for converting solid to liquid)
enthalpies of combustion (ΔH for combusting substance in O2)

69. Standard Enthalpies of Formation
Standard enthalpies of formation, Hf, are measured under standard conditions (25°C (298K) and 1.00 atm pressure).
Hf= The change in H for the reaction that forms 1 mole of compound from its elements:
elements (STP)  Compound(STP) 

70. Standard Enthalpy Formation of Ethanol
C(graphite) + 3H2(g) + ½ O2(g) C2H5OH(l) ΔH°f= kJ
kJ /mol = bc 1 mol
Use O2 (not O or O3) bc most stable form at STP
H2 is the most stable form of H
C graphite is the most stable form of carbon
The standard enthalpy formation of the most stable form of any element is zero bc there is no formation reaction needed when element already in its standard state  ΔH°f for C(graphite), H2, O2 are zero

71. The standard enthalpy formation of C2H2(g) is 226. 7 kJ/mol
The standard enthalpy formation of C2H2(g) is kJ/mol. Write the thermochemical equation associated with ΔH°f for this substance.
2C(graphite)+ H2(g) C2H2 (g) ΔH°f=226.7 kJ/mol

72. The standard enthalpy formation of CCl4(l) is -135. 4 kJ/mol
The standard enthalpy formation of CCl4(l) is kJ/mol. Write the thermochemical equation associated with ΔH°f for this substance.
C(graphite)+ 2Cl2(g)  CCl4(l) ΔH°f = kJ/mol

75. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in 3 steps:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)

76. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in 3 steps:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)

77. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in 3 steps:
C3H8 (g)  3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)

78. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
The sum of these equations is:
C3H8 (g)  3 C(graphite) + 4 H2 (g) ΔH1= -ΔH°fC3H8(g) = -( kJ)
3 C(graphite) + 3 O2 (g)  3 CO2 (g) ΔH2= 3ΔH°f CO2(g) = 3( kJ)
4 H2 (g) + 2 O2 (g)  4 H2O (l) ΔH3= 4ΔH°f H2O(l) = 4( kJ)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) ΔH

79. Calculation of H We can use Hess’s law in this way:
H = nHf(products) - mHf(reactants)
where n and m are the stoichiometric coefficients.  

80. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H = [3( kJ) + 4( kJ)] - [1( kJ) + 5(0 kJ)]= [( kJ) + ( kJ)] - [( kJ) + (0 kJ)]
= ( kJ) - ( kJ)
= kJ

81. Energy in Foods
Most of the fuel in the food we eat comes from carbohydrates and fats.

82. Fuels
The vast majority of the energy consumed in this country comes from fossil fuels.