1. CHAPTER 6
All Bold Numbered Problems

2. Chapter 6 Outline Energy -vs- Heat Specific Heat
First Law of Thermodynamics
q (Heat)
Hess’s Law

3. THERMOCHEMISTRY or Thermodynamics3

4. Energy & Chemistry
Burning sugar (sugar reacts with KClO3, a strong oxidizing agent)
Burning peanuts supply sufficient energy to boil a cup of water.

5. Energy and Chemistry
2 H2(g) + O2(g) --> 2 H2O(g) + heat and light
This can be set up to provide ELECTRIC ENERGY in a fuel cell.
Oxidation: H2 ---> 4 H e-
Reduction: e- + O H2O ---> 4 OH-

6. 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 types of energy
light
electrical
kinetic and potential

7. Kinetic and Potential Energy
Potential energy: energy stored in chemical due to its structure.
The energy of a motionless body due to its position.

8. Kinetic and Potential Energy
Kinetic energy: energy of motion.
Translational
Rotational
Vibrational (IR)

9. Thermodynamics
Thermodynamics is the science of heat (energy) transfer.
Heat energy is associated with molecular motions.

10. Energy and Chemistry All of thermodynamics depends on the law of
THE CONSERVATION OF ENERGY.
The total energy of a system is constant.

11. 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”)
The S.I sytem uses the unit called the
JOULE
1 cal = Joules
James Joule

12. Specific Heat Capacity
Thermochemistry is the science of heat (energy) flow. A difference in temperature leads to energy transfer.
The heat “lost” or “gained” is related to
a) sample mass
b) change in T
c) specific heat capacity
Specific heat capacity =
heat lost or gained by substance (J)
(mass,
g)(T change, K)

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

14. What drives chemical reactions?
CHEMICAL REACTIVITY
What drives chemical reactions?
How do they occur?
The first is answered by THERMODYNAMICS and the second by KINETICS.
Have already seen a number of “driving forces” for reactions that are PRODUCT-FAVORED.
• formation of a precipitate
• gas formation
• H2O formation (acid-base reaction)
• electron transfer in a battery

15. CHEMICAL REACTIVITY
But ENERGY TRANSFER also allows us to predict reactivity.
In general, reactions that transfer energy to their surroundings are product-favored.
So, let us consider heat transfer in chemical processes.

16. Heat Energy Transfer in Chemical Processes
CO2 (s, -78 oC) ---> CO2 (g, -78 oC)

17. FIRST LAW OF THERMODYNAMICS
heat energy transferred (to the system)
DE = q + w
work done
(by the
surroundings)
energy
change
Energy is conserved!

18. DH = qp (for chemist in a lab which means only T changes)
The First Law of Thermodynamics is the law of conservation of energy.
DE = q + w
DE is the change in Energy, q is heat, w is work
Enthalpy (H), defined H = E + PV.
DH = DE + PDV at constant pressure (only V and T change).
If w = - PDV (because V does not change in a typical lab experiment, only T changes) and DH = q + w + PDV
DH = qp (for chemist in a lab which means only T changes)

19. ENTHALPY Most chemical reactions occur at constant P. qp = DH
DH = change in heat content of the system
DH = Hfinal - Hinitial
How do we measure q in the lab?

20. Specific Heat Capacity
Substance Spec. Heat (J/g•K)
H2O 4.18 Al
glass 0.84
If 25.0 g of Al cool from 310 oC to 37 oC, how many joules of heat energy are lost by the Al?
Specific heat capacity =
heat lost or gained by substance (J) or
(mass,
g)(T change, K)
C = Specific Heat, units of J/(gK)

21. 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?
heat gain/lost = q = (C)(mass)(DT)
where DT = Tfinal - Tinitial
q = (0.902 J/g•K)(25.0 g)( )K
q = J

22. Heat Energy Transfer in Chemical Processes
Heat flows FROM the SYSTEM into the SURROUNDINGS is an EXOTHERMIC process.
q is “-”
The reaction
feels hot.
Energy given
Off!

23. Heat Energy Transfer in Chemical Processes
Heat flows INTO the SYSTEM from the SURROUNDINGS is an ENDOTHERMIC process.
q is “+”
The reaction
feels cold.
Energy Added!

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?
q = J
q = kJ
Notice that the negative sign on q signals heat “lost by” or transferred out of Al. Is this Exo or Endothermic?

25. Specific Heat Capacity
A g piece of metal is heated to C. It is placed into 300. mL of water at C and the final temperature is C. Calculate the specific heat of the metal.
qwater qmetal = 0
qwater = -qmetal
(300. g)(4.18 J/gK)(9.0oC) = -(500.0 g)(c)(-44.0oC)
c = 0.51 J/gK

26. Specific Heat Capacity
To 150 g of water at 25oC is added 45 g of Al at 115oC, what will be the final temperature? Al

27. Heat Transfer and Changes of State
Changes of state involve energy
Ice -----> Water 333 J/g (heat of fusion)
+ energy
Exo or Endo? Sign of q?

28. Heat Transfer and Changes of State
Liquid ---> Vapor
Requires energy (heat).
This is the reason
a) you cool down after swimming
b) you use water to put out a fire.
+ energy

29. Heat and Changes of State29
Heat and Changes of State
What quantity of heat is required to change 10.0 g of ice at oC to steam at oC? Endo or Exo? Sign of q?
Heat of fusion of ice = 333 J/g
Heat of vaporization = 2260 J/g
+333 J/g
+2260 J/g

30. Heating/Cooling Curve for Water5 4 3 2 1

31. Heat and Changes of State
1. To heat ice
q = (10.0 g)(2.09 J/g•K)(50.0K) = 1050J
2. To melt ice
q = (10.0 g)(333 J/g) = 3330J
3. To heat water
q = (10.0 g)(4.18 J/g•K)(100.K) = 4180J
4. To evaporate water
q = (10.0 g)(2260 J/g) = 22600J
5. To heat steam
q = (10.0 g)(2.03 J/g•K)(100.0K) = 2030J
Total heat energy = J = kJ

32. Heat and Changes of State
Calculate the amount of heat energy necessary to change 25.0 g of copper solid at 925o C to liquid at 1083o C. Melting point = 1083o C, csolid = J/g.K, Heat of fusion = 205 J/g.
q1 = (25.0 g)(.382 J/gK)(1083oC-925oC)= 1510 J
q2 = (25.0 g)(205 J/g) = 5120 J
qT = q1 + q2 = 6630 J

33. Endo- and Exothermic qsystem > 0 qsystem < 0 EXOTHERMIC
T(system) goes up
T(system) goes down
ENDOTHERMIC
Heat goes in the system
EXOTHERMIC
Heat leaves
the system

34. USING ENTHALPY Consider the decomposition of water
H2O(g) kJ ---> H2(g) + 1/2 O2(g)
Endo or Exo?
Endothermic reaction
Heat is a “reactant”
DH = kJ

35. USING ENTHALPY Making H2 from H2O involves two steps.
Each step requires energy.
Liquid H2O
H2 + O2 gas
H2O vapor

36. USING ENTHALPY Example of HESS’S LAW
Making H2 from H2O involves two steps.
H2O(liq) + 44 kJ ---> H2O(g)
H2O(g) kJ ---> H2(g) + 1/2 O2(g)
H2O(liq) 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 other rxns.

37. USING ENTHALPY DHnet = DH1 + DH2 = -395.7 kJ
Calc. DH for: S(s) + 3/2 O2(g) --> SO3(g)
Given:
S(s) + O2(g) --> SO2(g) DH1 = kJ
SO2(g) + 1/2 O2(g) --> SO3(g) DH2 = kJ
The two equations add up to give the desired equation, so -
DHnet = DH1 + DH2 = kJ

38.  DH along one path =  DH along another path ENERGY S solid +O 2
direct path DH = 1 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

39. DH along one path = DH along another path
This equation is valid because DH is a STATE FUNCTION
State functions depend only on the state of the system and not how it got there.
Examples: V, T, P, energy — and your bank account!
Unlike V, T, and P, one cannot measure absolute H. We can only measure DH.

41. Standard Enthalpy Values
Most DH values are labeled ΔHo
Measured under standard conditions
P = 1 atmosphere or 1 bar, (105Pascals)
Concentration = 1 mol/L
T = 25 oC
with all species in standard states
e.g., C = graphite and O2 = gas, etc.

42. Standard Enthalpy Values
NIST (National Institute for Standards and Technology) gives values of
DHof = standard molar enthalpy of formation
This is the enthalpy change when 1 mole of compound is formed from elements under standard conditions.
See Table 6.2 and Appendix L

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

44. Using Standard Enthalpy Values
Use DH°f’s to calculate enthalpy change for:
H2O(g) + C(graphite) --> H2(g) + CO(g)
(product is called “water gas”)

45. Using Standard Enthalpy Values
H2O(g) + C(graphite) --> H2(g) + CO(g)
From reference books we find:
H2(g) + 1/2 O2(g) --> H2O(g)
DH°f of H2O vapor = kJ/mol
C(s) + 1/2 O2(g) --> CO(g)
DH° f of CO = kJ/mol

46. Using Standard Enthalpy Values
H2O(g) --> H2(g) + 1/2 O2(g) DHo = kJ
C(s) + 1/2 O2(g) --> CO(g) DHo = kJ
H2O(g) + C(graphite) --> H2(g) + CO(g)
DHonet = kJ
To convert 1 mol of water to 1 mol each of H2 and CO requires kJ of energy.
The “water gas” reaction is ENDOthermic.

47. Using Standard Enthalpy Values
Calculate D H of reaction?
In general, when ALL enthalpies of formation are known,
DHorxn =
 DHof (products) -  DHof (reactants)

48. 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)
DHorxn =  DHof (prod) -  DHof (react)
Use Appendix L

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

50. Using Standard Enthalpy Values
OF2(g) + H2O(g) --> O2(g) + 2 HF(g) DHorxn = kJ
DHof (H2O) = kJ and DHof (HF) = kJ
Calculate DHof (OF2)
DHorxn = {DHof (O2) DHof (HF) }
- { DHof (OF2) + DHof (H2O) }
kJ = {( 0 ) ( kJ) }
- {DHof (OF2) + ( kJ) }
DHof (OF2) = kJ /mole

51. Measuring Heats of Reaction with a BOMB CALORIMETRY
Coffee Cup
Calorimeter

52. Measuring Heats of Reaction CALORIMETRY
Calculate heat of combustion of octane.
C8H /2 O2 --> 8 CO H2O
• Burn 1.00 g of octane
Temp rises from to oC
Calorimeter contains g water
Heat capacity of bomb = 427 J/K

53. CALORIMETRY Step 1 Calc. heat transferred from reaction to water.
q = (4.18 J/g•K)(1300. g)(8.20 K) = 44,600 J
Step 2 Calc. heat transferred from reaction to bomb.
q = (bomb heat capacity)(DT)
= (427 J/K)(8.20 K) = J
Step 3 Total heat evolved =
44,600 J J = 48,100 J
Heat of combustion of 1.00 g of octane = kJ

54. Measuring Heats of Reaction CALORIMETRY
Calculate heat of combustion of ethanol.
C2H5OH O2 --> 2 CO H2O
• Burn g of ethanol
Temp rises from oC to oC
Calorimeter contains 210. g water
Heat capacity of bomb = kJ/K

55. CALORIMETRY Step 1 Calc. heat transferred from reaction to water.
q = (4.18 J/g•K)(210. g)(10.00 K) = kJ
Step 2 Calc. heat transferred from reaction to bomb.
q = (bomb heat capacity)(DT)
= (0.950 kJ/K)(10.00 K) = kJ
Step 3 Total heat evolved =
8.80 kJ kJ = kJ
( kJ/0.888g)(46.0 g/mole) = kJ/mole

56. Sample Problem
A g sample of C2H4 is burned in a calorimeter with a heat capacity of 0.47 kJ/oC and containing 2,000. g of water. The temperature increased from 25.00oC to 27.14oC. Calculate the heat of combustion of C2H4 in kJ/mole.

57. CALORIMETRY Step 1 Calc. heat transferred from reaction to water.
q = (4.18 J/g•K)(2000. g)(2.14 K) = 17,900 J
Step 2 Calc. heat transferred from reaction to bomb.
q = (bomb heat capacity)(DT)
= (0.47 kJ/K)(2.14 K) = 1.0 kJ
Step 3 Total heat evolved =
17.9 kJ kJ = kJ
(-18.9 kJ/0.105g)(28.0 g/mole) = kJ/mole

58. Practice Problems 1. How many kcal is 75.0 kJ?
2. Convert 88.6 kcal/mole to kJ/mole.
3. Calculate the kinetic energy of a lb car traveling at 50. miles per hour.
4. Is the kinetic energy of a 500. g object moving at 50. km/hr greater, less than, or the same as that of a 1.00 kg object moving at 25 km/hr?
5. How much heat is required to raise the temperature of 204 grams of lead from 22.8oC to 64.9oC? (C= cal/gK)

59. Practice Problems
6. If the specific heat of silver is cal/gK, how much heat would it take to raise the temperature of 1 kg of silver from -50.oC to 150.oC?
7. To what temperature will one pound of nickel be raised, beginning at 25.0oC, if 750 calories are absorbed by it? (c = cal/gK)
8. A 50. g chunk of unknown metal X is heated to 98.5oC and then dropped into 450 g of water initially at 25.00oC. The water temperature is observed to rise to 26.47oC. Calculate the specific heat of X.

60. Practice Problems
9. 2 H2O (l) --> 2 H2 (g) + O2 (g) DH = kcal
How many mLs of water will be decomposed to hydrogen and oxygen by 85 kcal?
10. CaCO3 (s) --> CaO (s) + CO2 (g) DH = 42.5 kcal
How much heat is required to react 186 g CaCO3?
11. C6H12O6 (s) + 6 O2 (g) --> 6 H2O (g) + 6 CO2 (g)
DH = -673 kcal. How much heat is evolved in the oxidation of 240 g of C6H12O6?
12. MgCO3(s)+2 HCl(g)->MgCl2(s)+CO2(g) + H2O(g) DH = -5.6 kcal. How many grams of MgCl2 will be produced if 80.0 kilocalories are released?

61. Practice Problems 13. 3 Cu2S(s) + 16 HNO3(l) --> 6 Cu(NO3)2(s) +
3 S(s) + 4 NO(g) + 8 H2O(l) DH = -543 kcal/mole
How much energy will be released by the reaction of 325 grams of Cu2S with HNO3.
14. 2 B5H9 (l) + 12 O2 (g) --> 9 H2O (l) + 5 B2O3 (s)
DH = kcal/mole. How much B5H9 must be burned to produce 4500 kcal of energy?
15. Upon dissolving 8.63 g of ammonium dihydrogen phosphate in 100. mL of water, it was found that DT= - 2.7oC. Calculate the value of the molar heat of reaction.

62. Practice Problems
16. A 0.25 mole sample of ammonium nitrate was dissolved in 500. mL of water at 21.0oC. As the solute dissolved the temperature of the solution dropped to a minimum value of 18.0oC. Calculate the heat of reaction in kcal/mole.
17. Given the following data: DH(kcal)
MnO2 + CO --> MnO + CO
Mn3O4 + CO --> 3 MnO + CO
3 Mn2O3 + CO --> 2 Mn3O4 + CO
Find DH in kcal for the following reaction:
2 MnO2 + CO --> Mn2O3 + CO2

63. Practice Problems 18. Given the following two reactions: DH(kcal)
2 Na(s) + 2HCl(g) --> 2 NaCl(s) + H2(g)
H2(g) + Cl2(g) --> 2 HCl(g)
What is DHfo,in kcal/mole for NaCl(s)?
19. Calculate the DHfo of glycerol, (CH2OH)2CHOH, knowing that its heat of combustion is kJ/mole.
20. Find the DH for the reaction:
2 SO2(g) H2O(g) + O2(g) --> 2 H2SO4 (l)

64. Practice Problems
21. Determine the enthalpy of formation in kcal/mole for chloroethane, C2H5Cl, from the following data:
DH(kcal)
H /2 O2 --> H2O
C + O2 --> CO
C2H5Cl --> C2H4 + HCl
2 CO H2O --> C2H O
H2 + Cl2 --> 2 HCl

65. Practice Problems
22. The combustion of g of ethene (C2H4) causes a temperature rise of 2.00oC in a bomb calorimeter that has a specific heat capacity of 600. cal/oC and containing 500. g of water. What is the heat of combustion in kJ/mole for ethene?

66. Practice Problems
23. A bomb calorimeter has a heat capacity of J/oC and containing 200. g of water. A temperature rise of 3.577oC is observed when the calorimeter is used in the combustion of a g sample of powdered tantalum in excess oxygen to product Ta2O5. Find DHfo of Ta2O5.
24. How many calories would be required to change the temperature of 750 g of water from 15.0 oC to 90.0 oC?
25. How many joules of heat would be liberated if the temperature of 300. g of iron were changed from 75 oC to 17 oC?

67. Practice Problems
26. The heat of formation of water is kcal/mole. How many joules would be liberated by burning 12 g of hydrogen?
27. 4 Fe + 3 O2 --> 2 Fe2O kcal
4 Al + 3 O2 --> 2 Al2O kcal
Calculate DH for the reaction
Fe2O3 + 2 Al --> Al2O3 + 2 Fe

68. Practice Problems
28. A g sample of C was placed in a calorimeter with a specific heat capacity of cal/oC and containing 1,000. G of water and burned. The temperature rose from oC to oC. Calculate the heat of combustion in kcal/mole?

69. Practice Problems Answers
kcal kJ/mole
3. 2 x 105 J 4. Greater
cal ,000 cal
7. 41oC J/gK
9. 11 mL kcal
x 102 kcal g
Kcal g
15. 15,000 J/mole kcal/mole
kJ kcal/mole

70. Practice Problems Answers
kJ kJ
kJ kJ/mole
kJ/mole cal
J x 106 J
kcal kcal/mole

71. Specific Heat Capacity
How much heat is required to raise the temperature of 57 g of water from 25.6oC to 66.5oC?
q = (57 g)(4.18 J/g K)(40.9oC)
q = 9700 J

72. Energy and Reactions Endothermic Exothermic C+D W+X H DH H DH A+B Y+Z
A + B + energy -->C + D
W + X -->Y + Z + energy

73. Energy and Reactions
1. The reaction of 14 g of hydrogen will produce how much energy?
3 H N2 --> 2 NH kJ
14 g ? kJ
14 g H2 mole H kJ
2.0 g H mole H2
= 210 kJ

74. Energy and Reactions
2. If 126 kJ are produced how many grams of ammonia are produced?
3 H N2 --> 2 NH kJ
? g kJ
126 kJ 2 mole NH g NH3
92 kJ mole NH3
= 47g
NH3

75. Hess’s Law N2 (g) + O2 (g) --> 2 NO (g) DH = 180.8 kJ
Find DH for the following equation:
2 NO (g) + O2 (g) --> 2 NO2 (g)
2 NO (g) --> N2 (g) + O2 (g) DH = kJ
N2 (g) O2 (g) --> 2 NO2 (g) DH = kJ 1
2 NO (g) + O2 (g) --> 2 NO2 (g)
DH = kJ

76. Hess’s Law 2 C2H4 + 2 H2 + 7 O2 --> 4 CO2 + 6 H2O DH = -3394 kJ
2 C2H O2 --> 4 CO H2O DH = kJ
Find DH for the following equation:
C2H4 + H2 --> C2H6
2 C2H4 + 2 H2 + 7 O2 --> 4 CO2 + 6 H2O DH = kJ
4 CO H2O --> 2 C2H O DH = kJ
/ kJ
/ kJ
C2H4 + H2 --> C2H6
DH = -137 kJ

77. Hess’s Law 1/2 H2 + N2 + 5/2 O2 --> HNO3 + NO2 DH = -101 kJ
2 NH + H2 --> N2H DH = -567 kJ
2 N2O5 --> 2 N O DH = 23 kJ
Find DH for the following equation:
N2H N2O5 --> 2 HNO NO NH
N2H4 --> 2 NH + H DH = 567 kJ
2 N2O5 --> 2 N O2 DH = 23 kJ kJ
N2H N2O5 --> 2 HNO NO NH
DH = 388 kJ

78. Using Standard Enthalpy Values
Calculate the DH for the following reaction:
Al2O3 (s) CO (g) --> 3 CO2 (g) Al (s)
DHorxn = { 3 DHof (CO2) DHof (Al) }
- { DHof (Al2O3) DHof (CO) }
= { 3 ( kJ) ( 0 )}
- { kJ + 3 ( kJ) }
DHorxn = kJ

79. Using Standard Enthalpy Values
C2H O2 --> 2 CO H2O (g) DH = kJ
2 C H2 --> C2H DH = kJ
H /2 O2 --> H2O (g) DH = kJ
Calculate DHf for CO2.
2 C H2 --> C2H DH = kJ
H2O (g) --> H /2 O DH = kJ
1/ / kJ 1
/ kJ f f
C + O2 --> CO2
DHf = kJ

80. Using Standard Enthalpy Values
C2H O2 --> 2 CO H2O (g) DH = kJ
2 C H2 --> C2H DH = kJ
H /2 O2 --> H2O (g) DH = kJ
Calculate DHf for CO2. f f
DHorxn = { 2 DHof (CO2) DHof (H2O) }
- {DHof (C2H4) DHof (O2) }
kJ = { 2 DHof (CO2) + 2 ( kJ )}
- { 52.3 kJ + 3 ( 0 ) }
DHof (CO2) = kJ