1. THERMOCHEMISTRY OR THERMODYNAMICS
Chapter 6

2. Energy and Chemistry
ENERGY is the capacity to do work or transfer heat.
HEAT is the form of energy that flows between 2 samples because of their difference in temperature.
Other forms of energy —
light electrical nuclear
kinetic potential

3. Temperature v. Heat
Temperature reflects random motions of particles, therefore related to kinetic energy of the system.
Heat involves a transfer of energy between 2 objects due to a temperature difference

4. Law of Conservation of Energy
Energy can be converted from one form to another but can neither be created nor destroyed.
(Euniverse is constant)

5. Kinetic and Potential Energy
Potential energy — energy a motionless body has by virtue of its position.

6. Kinetic and Potential Energy
Kinetic energy — energy of motion.

7. Units of Energy
1 calorie = heat required to raise temp. of 1.00 g of H2O by 1.0 oC.
1000 cal = 1 kilocalorie = 1 kcal
1 kcal = 1 Calorie (a food “calorie”)
But we use the unit called the JOULE
1 cal = joules
James Joule

8. Extensive & Intensive Properties
Extensive properties depends directly on the amount of substance present.
mass
volume
heat
heat capacity (C)
Intensive properties is not related to the amount of substance.
temperature
concentration
pressure
specific heat (s)

9. System and Surroundings
System: That on which we focus attention
Surroundings: Everything else in the universe
Universe = System + Surroundings

10. Exo and Endothermic Heat exchange accompanies chemical reactions.
Exothermic: Heat flows out of the system (to the surroundings).
Endothermic: Heat flows into the system (from the surroundings).

11. Endo- and Exothermic ENDOTHERMIC EXOTHERMIC heat heat
Surroundings
Surroundings
System
System
heat
heat
E(system) goes down
E(system) goes up
ENDOTHERMIC
EXOTHERMIC

12. Enthalpy DH = Hfinal - Hinitial
If Hfinal > Hinitial then DH is positive
Process is ENDOTHERMIC
If Hfinal < Hinitial then DH is negative
Process is EXOTHERMIC

13. Upon adding potassium hydroxide to water the following reaction takes place NaOH(S) NaOH(aq) for this reaction at constant pressure, ∆H= -43 kj/mol 1 .Is the reaction exo- or endothermic 2. Does the water get warmer? 3. What is the enthalpy change for the solution if 14 g of NaOH is added?

14. Exercise Consider the combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
ΔH = –2221 kJ
Assume that all of the heat comes from the combustion of propane. Calculate ΔH in which 5.00 g of propane is burned in excess oxygen at constant pressure.
–252 kJ
(5.00 g C3H8)(1 mol / g C3H8)(-2221 kJ / mol C3H8)
ΔH = -252 kJ

15. Consider the reaction
H2(g) + O2(g) ® H2O(l) DH° = –286 kJ
Which of the following is true?
a) The reaction is exothermic.
b) The reaction is endothermic.
c) The enthalpy of the products is less than that of the reactants.
d) Heat is absorbed by the system.
e)         Both A and C are true. AC

16. Consider the reaction:
When a 24.8-g sample of ethyl alcohol (molar mass = g/mol) is burned, how much energy is released as heat? c

17. The total volume of hydrogen gas needed to fill the Hindenburg was 2
The total volume of hydrogen gas needed to fill the Hindenburg was 2.01x108 L at 1.00 atm and 24.7°C.
How much energy was evolved when it burned?
2H2(g) + O2(g) ® 2H2O(l) DH° = –286 kJ

18. 2.35 ´ 109 kJ
7.37 ´ 102 kJ

19. q = msDt Dt = “change” in temperature (Co) q = heat (J) m = mass (g)
s = specific heat (j/gCo)
Dt = “change” in temperature (Co)
Simple Calorimeter

20. Some Heat Exchange Terms
specific heat capacity (s)
heat capacity per gram = J/°C g or J/K g
molar heat capacity (s)
heat capacity per mole = J/°C mol or J/K mol

21. Heat Capacity

22. Specific Heat Capacity
Substance Spec. Heat (J/g•K)
H2O Al
glass 0.84
Aluminum

23. Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?
Spec of Al=0.902
where DT = Tfinal - Tinitial
heat gain/lost = q = m s DT

24. Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?
where DT = Tfinal - Tinitial
q = (0.902 J/g•K)(25.0 g)( )
q = J
heat gain/lost = q = m s DT

25. Specific Heat Capacity
If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?
q = J
Notice that the negative sign on q signals heat “lost by” or transferred out of Al.bv 233hhbn

26. Copper has a specific hear of .382j/goC. If 2.51 g of cooper absorbs 2.75 j of heat , what is the change in temp ?

27. Cooper has a specific heat of. 382j/g/ oC
Cooper has a specific heat of .382j/g/ oC. the temperature of an unknown mass of cooper increases by 4.50 oC when it absorbs 3.97J of heat. What is the mass of the copper?

28. Heating curves

29. In bomb calorimetry, volume is constant.
REMEMBER!!!
In regular calorimetry pressure is constant, but the volume will change.
In bomb calorimetry, volume is constant.

30. Calorimetry Constant volume calorimeter is called a bomb calorimeter.
Material is put in a container with pure oxygen. Wires are used to start the combustion. The container is put into a container of water.
The heat capacity of the calorimeter is known and tested.

31. Bomb Calorimeter thermometer stirrer full of water ignition wire
Steel bomb
sample

32. Suppose we wish to measure the energy of combustion of octane (C8H18), a component of gasoline. A g sample of octane is placed in a bomb calorimeter known to have a heat capacity of 11.3 kJ/ºC.
This means that 11.3 kJ of energy is required to raise the temperature of the water and other parts of the calorimeter by 1ºC. The octane is ignited in the presence of excess oxygen, and the temperature increase of the calorimeter is 2.25ºC.

33. The amount of energy released is calculated as follows:
Energy released by the reaction
= ΔT x heat capacity of calorimeter

34. A bomb calorimeter has a heat capacity of 9. 47 kJ/K. When a 2
A bomb calorimeter has a heat capacity of 9.47 kJ/K. When a 2.01-g sample of (C3H6) was burned in this calorimeter, the temperature increased by 4.26 K. Calculate the energy of combustion for the sample.

35. A bomb calorimeter has a heat capacity of 9. 47 kJ/K. When a 2
A bomb calorimeter has a heat capacity of 9.47 kJ/K. When a 2.01-g sample of (C3H6) was burned in this calorimeter, the temperature increased by 4.26 K. Calculate the energy of combustion for the sample.

36. A g sample of quinine (C6H4O2) is burned in a bomb calorimeter whose total heat capacity is 7.854kj/ºC. the temperature of the calorimeter plus the contents increased from 23.44ºC to 30.57ºC.
What is the heat of combustion?
per gram of quinine?
Per mole of Quinine?

37. .5865g sample of lactic acid HC3H5O3 is burned in a calorimeter whose heat capacity is 4.812kj/ºC. The temperature increases from 23.10ºC to 24.95ºC . Calculate the heat of combustion of lactic acid per gram ?

38. The heat of combustion of pentane, is -131. 64 kJ/g. Combustion of 4
The heat of combustion of pentane, is kJ/g. Combustion of 4.50 g of pentane causes a temperature rise of 2.00°C in a certain bomb calorimeter. What is the heat capacity of this bomb calorimeter?

39. The combustion of g benzoic acid increase the Temperature of a bomb calorimeter by 2.54°C. The energy released by the combustion is 26.42kj/g.Calculate the heat capacity of the bomb Calorimeter .

40. Standard States Compound For a gas, pressure is exactly 1 atmosphere.
For a solution, concentration is exactly 1 molar.
Pure substance (liquid or solid), it is the pure liquid or solid.
Element
The form [N2(g), K(s)] in which it exists at atm and 25°C.

41. Hess’s Law Reactants  Products
The change in enthalpy is the same whether the reaction takes place in one step or a series of steps.

42. Using Enthalpy Consider the decomposition of water
H2O(g) kJ ---> H2(g) + 1/2 O2(g)
Endothermic reaction — heat is a “reactant”
DH = kJ

43. Using Enthalpy Making H2 from H2O involves two steps.
H2O(l) + 44 kJ ---> H2O(g)
H2O(g) kJ ---> H2(g) + 1/2 O2(g)
H2O(l) kJ --> H2(g) + 1/2 O2(g)
Example of HESS’S LAW—
If a rxn. is the sum of 2 or more others, the net DH is the sum of the DH’s of the all rxns.

44. Calculations via Hess’s Law
1. If a reaction is reversed, H is also reversed.
N2(g) + O2(g)  2NO(g) H = 180 kJ
2NO(g)  N2(g) + O2(g) H = 180 kJ
2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer.
6NO(g)  3N2(g) + 3O2(g) H = 540 kJ

45. Using Enthalpy Calc. DH for S(s) + 3/2 O2(g) --> SO3(g)
S(s) + O2(g) --> SO2(g) kJ
SO3(g) --> SO2(g) + 1/2 O2(g) kJ
_______________________________________
S(s) + 3/2 O2(g) --> SO3(g) kJ

46.  DH along one path =  DH along another path energy S solid
direct path DH = 1 +O 2 kJ
+ 3/2 O 2
DH = SO
gas 2 kJ
+ 1/2 O 2 SO
gas 3 DH
= kJ 2
 DH along one path =
 DH along another path

47. Determine the heat of reaction for the decomposition of one mole of benzene to acetylene
C6H6(l) ® 3C2H2(g)
given the following thermo chemical equations:
2C6H6(l) + 15O2(g) ® 12CO2(g) + 6H2O(g) DH = kJ
2C2H2(g) + 5O2(g) ® 4CO2(g) + 2H2O(g) DH = kJ

48. See smart Hess Law

49. Using Standard Enthalpy Values
H2O(g) + C(graphite) --> H2(g) + CO(g)
(product is called “water gas”)

50. This equation is valid because DH is a STATE FUNCTION
DH along one path =
DH along another path
This equation is valid because DH is a STATE FUNCTION
These depend only on the state of the system and not how it got there.

51. Change in Enthalpy
Can be calculated from enthalpies of formation of reactants and products.
Hrxn° = npHf(products)  nrHf(reactants)
  H is an extensive property--kJ/mol
For the reaction: 2H2 (g) + O2 (g) ---> 2H2O(g)

52. Standard Enthalpy Values
NIST (Nat’l Institute for Standards and Technology) gives values of
DHof = standard molar enthalpy of formation
This is the enthalpy change when 1 mol of compound is formed from elements under standard conditions. DHof is always stated in terms of moles of product formed.

53. DHof, standard molar enthalpy of formation
H2(g) + 1/2 O2(g) --> H2O(g)
DHof = kJ/mol
By definition, DHof = 0 for elements in their standard states.

54. Using Standard Enthalpy Values
Use DH°f’s to calculate enthalpy change DH for
H2O(g) + C(graphite) --> H2(g) + CO(g)
From reference books we find
DH°f of H2O vapor = kJ/mol
H2(g) + 1/2 O2(g) --> H2O(g)
DH° f of CO = kJ/mol
C(s) + 1/2 O2(g) --> CO(g)

55. Using Standard Enthalpy Values
Calculate the heat of combustion of methanol, i.e., DHorxn for
CH3OH(g)+ 3/2 O2(g) -->CO2(g) +2 H2O(g)
DHof (-201.5kj) (-393.5kj) (-241.8kj)
DHorxn =  DHof (prod) -  DHof (react)

56. Using Standard Enthalpy Values
CH3OH(g) + 3/2 O2(g) --> CO2(g) + 2 H2O(g)
DHorxn =  DHof (prod) -  DHof (react)
DHorxn = ( kJ) + 2 ( kJ)
- {0 + ( kJ)}
DHorxn = kJ per mol of methanol

57. h 2Al(s) + Fe2O3(s) Al2O3(s)+Fe(s) Al2O3 DHof = 646kj/mol
Fe2O3(s) DHof =-826kj/mol

58. 2KIO3 + 12HCl 2ICl +KCL+6H20+4Cl2 KIO3= -501 6H20= -286 HCl= -92

59. The standard enthalpy of combustion of ethane gas C2H4 is –1411
The standard enthalpy of combustion of ethane gas C2H4 is – kj/mol a 298k
Given the following enthalpy of formation calculate the enthalpy of formation of ethane gas
CO2DHof =-393.5kj/mol
H20DHof =-285.8kj/mol

60. Pathway for the Combustion of Methane

61. Schematic diagram of the energy changes for the
combustion of methane.

62. Greenhouse Effect
-- a warming effect exerted by the earth’s atmosphere due to
thermal energy retained by absorption of infrared radiation.
Greenhouse Gases:
CO2 H2O
CH4 N2O