1. Thermodynamics
Honors Unit 5

2. Energy: Basic Principles
Thermodynamics – the study of energy changes
Energy – the ability to do work or produce heat
Note: Work is force acting over a distance

3. Energy: Basic Principles
Kinetic Energy – energy of motion
KE =
Potential Energy –
energy due to position
or composition

4. Law of Conservation of Energy
A.k.a. first Law of Thermodynamics
Energy can be converted from one form to another but can’t be created or destroyed
This means the total energy of the universe is CONSTANT!

5. Heat vs. Temperature
Temperature – measure of the random motion of a substance
Temperature is proportional to kinetic energy (it is a measure of the average kinetic energy in a substance)
Heat (q) – flow of energy due to a temperature difference

6. Important Aspects of Thermal Energy & Temperature
Heat is NOT the same as temperature!
The more kinetic energy a substance has, the greater the temperature of its atoms and molecules.
The total thermal energy in an object is the sum of its individual energies of all the molecules.
For any given substance, its thermal energy depends not only on its composition but also on the amount of substance

7. System vs. Surroundings
A system is the part of the universe we are studying.
The surroundings
are everything else
outside
of the system.

8. Direction of Heat Flow
Heat transfer occurs when two objects are at two different temperatures.
Eventually the two objects reach the same temperature
At this point, we say that the system has reached equilibrium.

9. Thermal Equilibrium
Heat transfer always occurs with heat flowing from the HOT object to the COLD object.

10. Thermal Equilibrium
The quantity of heat lost by the hotter object and the quantity of heat gained by the cooler object are EQUAL.

11. Exothermic vs. Endothermic
Exothermic process  heat is transferred from the system to the surroundings
Heat is lost from the system (temperature in system decreases)
Endothermic process  heat transferred from the surroundings to the system
Heat is added to the system (temperature in system increases)

12. Exothermic Process

13. Endothermic Process

14. One kilojoule (kJ) = 1000 joules (J)
Units of Energy
Joule (J) is the SI unit of energy & heat
One kilojoule (kJ) = 1000 joules (J)
calorie (cal) = heat required to raise the temperature of 1.00 g of water by 1 °C
1 calorie = J

15. Units of Energy
Food is measured in Calories (also known as kilocalories) instead of calories
1 Cal = 1 kcal = 1000 calories

16. Units of Energy
3800 cal = __________ Cal = _________ J

17. Units of Energy
The label on a cereal box indicates that 1 serving provides 250 Cal. What is the energy in kJ?

18. Heat Transfer Direction and sign of heat flow – MEMORIZE!
ENDOTHERMIC: heat is added to the system & the temperature increases (+q)
EXOTHERMIC: heat is lost from the system (added to the surroundings) & the temperature in the system decreases (-q)

19. Heat Capacity
The quantity of heat required to raise an object’s temperature by 1 °C (or by 1 Kelvin)
Heat capacity is an extensive property.
Which will take more heat to raise the temperature by 1 °C?

20. Specific Heat (Specific Heat Capacity)
Specific Heat (C) - The quantity of heat required to raise the temperature of one gram of a substance by 1 °C
Intensive property
Units:
J/(g°C) or J/(g°K)
cal/(g°C) or cal/(g°K)

21. Examples of Specific Heat
At the beach, which gets hotter, the sand or the water?
Higher specific heat means the substance takes longer to heat up & cool down!

22. Examples of Specific Heat
Specific heat (C)= the heat required to raise the temperature of 1 gram of a substance by 1 °C
Cwater = J/(g°C)
Csand = J/(g°C)

23. Calculating Changes in Thermal E
q = m x C x Δt
q = mCΔt
q = heat (cal or J)
m = mass (g)
C = specific heat capacity, J/(g°C)
Δt = change in temperature, Tfinal – Tinitial
(°C or K)
***All units must match up!!!***

24. Example
q = mCΔt
How much heat in J is given off by a 75.0 g sample of pure aluminum when it cools from 84.0°C to 46.7°C? The specific heat of aluminum is J/(g°C).

25. Example
q = mCΔt
What is the specific heat of benzene if 3450 J of heat are added to a g sample of benzene and its temperature increases from 22.5 °C to 35.8 °C?

26. Example
q = mCΔt
A 50.0 g sample of water gives off kJ as it is cooled. If the initial temperature of the water was 85.0 °C, what was the final temperature of the water? The specific heat of water is 4.18 J/(g°C).

27. Calorimetry Calorimetry: measurement of quantities of heat
A calorimeter is the device in which heat is measured.

28. Calorimetry Assumptions: Heat lost = -heat gained by the system
In a simple calorimeter, no heat is lost to the surroundings

29. Coffee Cup Calorimetry
***Use styrofoam instead of a beaker to keep heat in
Steps:
Add hot solid metal to cool water
Water will heat up (T rises) as metal cools
Eventually, water & metal are at same T.
qmetal + qwater = 0
qmetal = -qwater
Heat lost by metal = -heat gained by the water

30. Calorimetry mmCmΔTm = -[mH2OCH2OΔTH2O] qmetal = -qwater
Heat lost by metal = -heat gained by water
Since q = mCΔT,
mmCmΔTm = -[mH2OCH2OΔTH2O]

31. Sample Problem
A g piece of lead was heated in water to 94.1 °C. It was removed from the water and placed into 100. mL of water in a Styrofoam cup. The initial temperature of the water was 18.7 °C and the final temperature of the lead and water was 26.1 °C. What is the specific heat of lead according to this data?

32. Bomb Calorimeter Constant volume “bomb” calorimeter
Burn sample in O2
Some heat from reaction warms water
qwater = mCH2OΔT
Some heat from reaction warms bomb
qbomb = CbombΔT
qrxn + qH2O + qbomb = 0

33. Energy & Changes of State
All changes of state involve energy changes (more in Unit 9)
Note that fusion = melting

34. State Functions
A property where the change from initial to final state does not depend on the path taken
Ex.) The change in elevation from the top to bottom of a ski slope is independent of the path taken to go down from the slope

35. Enthalpy Changes for Chemical Rxns.
Heat of reaction
The heat absorbed or given off when a chemical reaction occurs at constant T (temp.) and P (pressure)

36. Enthalpy Enthalpy (H) The heat content of a reaction (chemical energy)
ΔH = change in enthalpy
The amount of energy absorbed by or lost from a system as heat during a chemical process at constant P
ΔH = ΔHfinal - ΔHinitial

37. Properties of Enthalpy
Enthalpy is an extensive property
It does depend on quantity
Enthalpy is a state function
Depends only on the final & initial values
Every reaction has a unique enthalpy value since ΔH = Hproducts - Hreactants

38. Representation of Enthalpy as a Graph

39. Two Ways to Designate Thermochemical Equations
Endothermic:
H2 (g) + I2 (s)  2 HI (g) ΔH = 53.0 kJ
H2 (g) + I2 (s) kJ  2 HI (g)

40. Two Ways to Designate Thermochemical Equations
Exothermic:
½ CH4 (g) + O2 (g)  ½ CO2 (g) + H2O (l) ΔH = kJ
½ CH4(g) + O2(g)  ½ CO2(g) + H2O(l) kJ

41. Two Ways to Designate Thermochemical Equations
Note the meaning of the sign in ΔH in the equations above!!
Endothermic: ΔH = +
Exothermic: ΔH = -

42. Two Ways to Designate Thermochemical Equations
Note the important of designating the physical state or phase of matter. Why??
Because this will change the heat of reaction! (ΔH)

43. Thermochemical Equations
What do the coefficients stand for? How can they differ from the ones we have used before?
Coefficients = the number of moles (as before)
BUT
We can use fractional coefficients now!

44. Thermochemical Equations
What is the standard state? How do we designate conditions of temperature and pressure that are not at standard state?
Standard state = 1 atm pressure & 25 °C
ΔH° = ΔH at standard state
Must show conditions over arrow if not at standard state!

45. Thermochemical Equations
How can we find the enthalpy of reaction when we reverse it?
Reverse the reaction, reverse the sign of ΔH!
Example:
CO (g) + ½ O2 (g)  CO2 (g) ΔH = -283 kJ
CO2 (g)  CO (g) + ½ O2 (g) ΔH = +283 kJ

46. Example Given Rxn. #1, find the ΔH for Rxns. 2 & 3 Reaction #1
2 SO2 (g) + O2 (g)  2 SO3 (g) ΔH = kJ
Reaction #2
SO2 (g) + ½ O2 (g)  SO3 (g) ΔH =
Reaction #3
4 SO3 (g)  4 SO2 (g) + 2 O2(g) ΔH =

47. ΔH as a Stoichiometric Quantity
Given the reaction below, how much heat is produced when 15.0 g of NO2 are produced?
2 NO (g) + O2 (g)  2 NO2 (g) ΔH = kJ

48. ΔH as a Stoichiometric Quantity
Given: ΔH = -283 kJ
CO (g) + ½ O2 (g)  CO2 (g)
(a) Calculate the enthalpy of the above reaction when 3.00 g of product are formed

49. ΔH as a Stoichiometric Quantity
Given: ΔH = -283 kJ
CO (g) + ½ O2 (g)  CO2 (g)
(b) If only 10.0 grams of oxygen and an unlimited supply of CO are available to run this reaction, how much heat will be given off?

50. ΔH as a Stoichiometric Quantity
Given: ΔH = -283 kJ
CO (g) + ½ O2 (g)  CO2 (g)
(c) How many grams of carbon monoxide are necessary (assuming oxygen is unlimited) to produce 500 kJ of energy in this reaction?

51. ΔH as a Stoichiometric Quantity
Given: ΔH = -283 kJ
CO (g) + ½ O2 (g)  CO2 (g)
(d) Calculate the heat of decomposition of two moles of carbon dioxide.

52. Hess’s Law
The heat of a reaction (ΔH) is constant, whether the reaction is carried out directly in one step or indirectly through a number of steps.
The heat of a reaction (ΔH) can be determined as the sum of heats of reaction of several steps.

53. (Exothermic Rxn  ΔH = -241.8 kJ)
Hess’s Law: Example
Consider the formation of water:
H2(g) + ½ O2(g)  H2O(g) kJ
(Exothermic Rxn  ΔH = kJ)

54. Hess’s Law

55. Hess’s Law

56. Since ΔH is a state function!!
Hess’s Law
Σ ΔH along one path =
Σ ΔH along another path
Since ΔH is a state function!!

57. Hess’s Law Given: C(s) + O2(g)  CO2(g) ΔH = -393.5 kJ
2 CO(g) + O2(g)  2 CO2(g) ΔH = kJ
Determine the heat of reaction for:
C(s) + ½ O2(g)  CO(g)

58. Hess’s Law Given: Determine the heat of reaction for:
C(s) + O2(g)  CO2(g) ΔH = kJ
C2H4(g)+3 O2(g)  2 CO2(g)+2 H2O(l) ΔH= kJ
H2(g) + ½ O2(g)  H2O(l) ΔH = kJ
Determine the heat of reaction for:
2 C(s) + 2 H2(g)  C2H4(g)

59. Standard Enthalpies of Formation
NIST (National Institute for Standards and Technology gives values for
ΔHf° = standard molar heat of formation
Definition:
The heat content or enthalpy change when one mole of a compound is formed at 1.0 atm pressure and 25 °C from its elements under the same conditions.

60. Examples of Formation Equations
H2(g) + ½ O2(g)  H2O(g)
ΔHf°(H2O, g) = kj/mol
C(s) + ½ O2(g)  CO(g)
ΔHf°(CO, g) = -111 kj/mol
***Elements/reactants  1 mol of compound
Notice units are per mole

61. Standard Enthalpy of Formation Values
Can look up values of in reference book or textbook
By definition, ΔHf° = 0 for elements in their standard states
Example: Cl2 (g)
H2 (g)
Ca (s)

62. ΣΔH° = ΣΔHf°(products) - ΣΔHf°(reactants)
Summation Equation
In general, when all enthalpies of formation are known:
ΣΔH° = ΣΔHf°(products) - ΣΔHf°(reactants)
Must multiply all Hf values by coefficient from balanced equation!!!

63. Summation Equation Example
Use the summation equation to determine the enthalpy of the following reaction:
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
ΣΔH° = ΣΔHf°(products) - ΣΔHf°(reactants)