1. Chapter 6: Enthalpy Changes for Chemical Reactions
Chemistry Lecture 13
Chapter 6: Enthalpy Changes for Chemical Reactions
Chapter Highlights
energy transfer, specific heat and heat transfer
heat of fusion & heat vapourization
exothermic & endothermic reactions
first law of thermodynamics
enthalpy changes
calorimetry
Hess’s Law

2. Thermodynamics Thermodynamics: energy and its transformations.
Chemistry Lecture 13
Thermodynamics
Thermodynamics: energy and its transformations.
Thermochemistry: energy changes and chemical reactions
Force: "push" or "pull" exerted on an object.
Work: energy required to overcome a force:
Work = force x distance.
Heat: energy transferred from one object to another because of a temperature difference.

3. Kinetic and Potential Energy EK = mv2 (m = mass, v = velocity)
Chemistry Lecture 13
Kinetic and Potential Energy
Kinetic energy (EK): energy of an object due to its motion.
EK = mv2 (m = mass, v = velocity)
Potential energy (EP): energy stored by an object
due to its relative position.

4. Energy can be expressed in a wide variety of units.
Chemistry Lecture 13
Energy
Energy can be expressed in a wide variety of units.
The joule (J) is the metric unit of energy and is the energy possessed by a 2 kg object moving at a velocity of 1 m/s.
1 J (joule) =1 kg-m2/s2.
The calorie (cal) is the energy required to raise the temperature of 1 g of water by 1°C.
1 cal = J
1 Cal = 1 kcal = 1000 cal = kJ

5. 2 H2(g) + O2(g) 2 H2O(l) + energy
Chemistry Lecture 13
Energy Transfer
The environment of a chemical reaction is separated into system and surroundings.
System: reactants, products, solvents, etc. in the reaction.
Surroundings: the universe, including the vessel.
2 H2(g) O2(g) H2O(l) energy
Processes that lower a system's internal energy are spontaneous. Processes that increase a system's internal energy are nonspontaneous.

6. Energy Transfer & Specific Heat heat transferred = q = C x DT
Chemistry Lecture 13
Energy Transfer & Specific Heat
Heat capacity of an object, C, is the amount of heat energy required to raise its temperature by 1°C,
The specific heat of a substance, C, is the amount of heat required to raise the temperature of a 1 g sample of the substance by 1°C,
heat transferred = q = C x DT
C =

7. Energy Transfer & Specific Heat
Chemistry Lecture 13
Energy Transfer & Specific Heat
Example 6.2:
A lake has a surface area of 2.6 x 106 m2 and an average depth of 10 m. What quantity of heat (kJ) must be transferred to the lake to raise the temperature by 1 oC? (Assume a density of 1.0 g/cm3 the lake water)

8. q = m x C x DT = (2.6 x 1013 g)(4.184 J/g-K)(1.0 K)
Chemistry Lecture 13
Answer:
Determine the mass of water and calculate the energy required using the concept of specific heat.
Volume(H2O) = (2.6 x 106 m2)(10 m) = 2.6 x 107 m3 =
Mass(H2O) = (2.6 x 1013 cm3)(1.0 g/cm3) =
Since:
q = m x C x DT = (2.6 x 1013 g)(4.184 J/g-K)(1.0 K) =
2.6 x 1013 cm3
2.6 x 1013 g
C =
1.1 x 1011 kJ

9. Energy and Changes of State
Chemistry Lecture 13
Energy and Changes of State
500 kJ
Ice, 2.0 kg, 0oC
Iron, 1.0 kg, 0oC
0oC
1100oC
Heat transfer: heat is transferred to a substance with no structural change
Phase changes or phase transitions: a substance's structure is altered as in melting, freezing, vaporizing, and condensing.

10. Energy and Changes of State
Chemistry Lecture 13
Energy and Changes of State
Heat of fusion: the enthalpy change associated with melting a substance, DHfus (kJ/mol).
Heat of vapourization; the enthalpy associated with vapourizing a substance, DHvap (kJ/mol).
Heating a substance is endothermic. BUT... during a phase transition the temperature remains constant

11. Energy and Changes of State
Chemistry Lecture 13
Energy and Changes of State
The energy required (q) for the five possible processes is determined by the amount of sample.
For warming:
q = C x mass x DT,
(specific heat, C, is different for each physical state)
For the phase transitions: q = (# of moles) x DH.

12. Energy and Changes of State
Chemistry Lecture 13
Energy and Changes of State
Question (similar to example 6.4):
Calculate the enthalpy change upon converting 1.00 mol of ice at -25 oC to water vapour (steam) at 125 oC under a constant pressure of 1 atm. The specific heat of ice, water and steam are 2.09 J/g-K, 4.18 J/g-K and 1.84 J/g-K, respectively. Also for water; DHfus = 6.01 kJ/mol and DHvap = kJ/mol.

13. Energy and Changes of State
Chemistry Lecture 13
Energy and Changes of State
Answer:
Divide the total change into calculable segments and calculate the enthalpy change for each segment. Sum to get the total enthalpy change (Hess's Law).
Segment 1: Heat ice from -25 oC to 0 oC.
Segment 2: Convert ice to water at 0 oC.
Segment 3: Heat water from 0 oC to 100 oC. Segment 4: Convert water to steam at 100 oC. Segment 5: Heat steam from 100 oC to 125 oC.

14. Energy and Changes of State
Chemistry Lecture 13
Energy and Changes of State
3: Water
2: Melting
1: Ice
5: Steam
4: Boiling
Heat absorbed
Heat liberated
Temperature oC
Heat added (kJ)

15. Energy and Changes of State
Chemistry Lecture 13
Energy and Changes of State
Segment 1: Heating the ice from -25 oC to 0 oC.
DH1 = (1.00 mol)(18.0 g/mol)(2.09 J/g-K)(25 K) =
Segment 2: Convert ice to water at 0 oC.
DH2 = (1.00 mol)(6.01 kJ/mol) =
Segment 3: Heating the water from 0 oC to 100 oC.
DH3 = (1.00 mol)(18.0 g/mol)(4.18 J/g-K)(100 K) =
940 J
6.01 kJ
7520 J

16. Energy and Changes of State
Chemistry Lecture 13
Energy and Changes of State
Segment 4: Convert water to steam at 100oC.
DH4 = (1.00 mol)(40.67 kJ/mol) =
Segment 5: Heating the steam from 100oC to 125oC.
DH5 = (1.00 mol)(18.0 g/mol)(1.84 J/g-K)(25 K) =
Total Enthalpy Change:
DH = DH1 + DH2 + DH3 + DH4+ DH5
= (0.94 kJ) + (6.01 kJ) + (7.52 kJ) + (40.67 kJ) + (0.83 kJ) =
40.67 kJ
830 J
55.97 kJ

17. The First Law of Thermodynamics
Chemistry Lecture 13
The First Law of Thermodynamics
Energy is neither created nor destroyed; only exchanged between system and surroundings.
Internal energy is the total energy of a system. Only changes in internal energy, DE, can be measured.
DE = Efinal - Einitial
(+ve = gain by system & -ve = loss to surroundings)
DE = q w
(heat added is positive & heat withdrawn is negative)
(work done is positive & work done by is negative)
internal
energy
work
done
heat
added

18. The First Law of Thermodynamics
Chemistry Lecture 13
The First Law of Thermodynamics
Energy transferred
from surroundings
to system
Energy transferred
from system
to surroundings

19. Heat Transfer Question:
Chemistry Lecture 13
Heat Transfer
Question:
The H2(g) and O2(g) in a cylinder are ignited and the system loses 550 J to its surroundings. The expanding gas does 240 J of work on its surroundings. What is the change in internal energy of the system?

20. Heat Transfer Answer: Energy flows from the system, so q = -550 J.
Chemistry Lecture 13
Heat Transfer
Answer:
Energy flows from the system, so q = -550 J.
Work is done by the system, so w = -240 J
Therefore:
DE = q + w = (-550 J) + (-240 J) =
-790 J

21. Term Test #1 Friday October 12th, 2001 6:30 P.M.
Chemistry Lecture 13
Term Test #1
Friday October 12th, 2001
6:30 P.M.

22. A to H Education Gym I to Z Rm 104 Odette Bldg
Chemistry Lecture 13
A to H Education Gym
I to Z Rm 104 Odette Bldg

23. Contents: SIX “Problems”!!
Chemistry Lecture 13
Duration: 75 minutes
Contents: SIX “Problems”!!
Covers Material From Chapters 4 - 6
A Periodic Table & ALL Required Constants will be Supplied

24. Chapter 6: Enthalpy Changes for Chemical Reactions
Chemistry Lecture 14
Chapter 6: Enthalpy Changes for Chemical Reactions
Chapter Highlights
energy transfer, specific heat and heat transfer
heat of fusion & heat vapourization
exothermic & endothermic reactions
first law of thermodynamics
enthalpy changes
calorimetry
Hess’s Law

25. The First Law of Thermodynamics
Chemistry Lecture 14
The First Law of Thermodynamics
Energy is neither created nor destroyed; only exchanged between system and surroundings.
Internal energy is the total energy of a system. Only changes in internal energy, DE, can be measured.
DE = Efinal - Einitial
(+ve = gain by system & -ve = loss to surroundings)
DE = q w
(heat added is positive & heat withdrawn is negative)
(work done is positive & work done by is negative)
internal
energy
work
done
heat
added

26. The First Law of Thermodynamics
Chemistry Lecture 14
The First Law of Thermodynamics
Energy transferred
from surroundings
to system
Energy transferred
from system
to surroundings

27. Heat & Enthalpy Changes
Chemistry Lecture 14
Heat & Enthalpy Changes
Enthalpy: amount of heat energy possessed by a substance. Enthalpy change corresponds to the heat change of the system at constant pressure:
Endothermic reaction: heat is added (positive value).
Exothermic reaction: heat is withdrawn (negative value).
DH = qp
DH = Hfinal - Hinitial

28. Enthalpies of Reaction
Chemistry Lecture 14
Enthalpies of Reaction
Enthalpy change: the sum of the absolute enthalpies of the products minus the absolute enthalpies of the reactants.
2 H2(g) O2(g) H2O(l) energy
2 H2(g) O2(g) H2O(l) DH = kJ
DH = H(products) - H(reactants)

29. Enthalpies of Reaction
Chemistry Lecture 14
Enthalpies of Reaction
Enthalpy is an extensive property. The magnitude of DH is proportional to the amount of reactant consumed.
CH4(g) O2(g) CO2(g) H2O(g)
DH = kJ
2 CH4(g) O2(g) CO2(g) H2O(g)
DH = kJ

30. Enthalpies of Reaction
Chemistry Lecture 14
Enthalpies of Reaction
The enthalpy change is equal in magnitude but opposite in sign to DH for the reverse reaction.
CH4(g) O2(g) CO2(g) H2O(g)
DH = kJ
CO2(g) H2O(g) CH4(g) O2(g)
DH = kJ

31. Enthalpies of Reaction
Chemistry Lecture 14
Enthalpies of Reaction
The enthalpy change depends on the states of
the reactants and the products.
CH4(g) O2(g) CO2(g) H2O(g)
DH = kJ
CH4(g) O2(g) CO2(g) H2O(l)
DH = kJ
2 H2O(g) H2O(l)
DH = -88 kJ

32. Enthalpies of Reaction Ammonium nitrate can decompose by the reaction:
Chemistry Lecture 14
Enthalpies of Reaction
Question:
Ammonium nitrate can decompose by the reaction:
NH4NO3(s) N2O(g) H2O(g) DH = kJ
Calculate the quantity of heat produced when 2.50 g of NH4NO3 decomposes at constant pressure.

33. Enthalpies of Reaction
Chemistry Lecture 14
Enthalpies of Reaction
Answer:
We know the heat produced from 1 mole of NH4NO3.
Calculate moles in 2.50 g and determine the heat produced by that amount.
heat = (2.50 g) =
-1.16 kJ

34. heat transferred = q = C x DT
Chemistry Lecture 14
Calorimetry
Calorimetry: The measurement of heat flow. Measurements are made using a calorimeter.
Recall that: The heat capacity of an object, C, is the amount of heat energy required to raise its temperature by 1°C,
In constant-pressure calorimetry, the pressure remains constant because the apparatus is open to the atmosphere. At constant pressure, the heat change of the reaction is the enthalpy change, DH.
heat transferred = q = C x DT

35. Calorimetry Question:
Chemistry Lecture 14
Calorimetry
Question:
When 50 mL of 1.0 M HCl(aq) and 50 mL of 1.0 M NaOH(aq) are mixed in a constant-pressure calorimeter, the temperature increases from 21.0 to 27.5 oC. Calculate DH for this reaction assuming a specific heat of 4.18 J/g-oC and a density of 1.0 g/mL for the final solution.

36. Calorimetry Answer: Recall, C = q = C x m x DT
Chemistry Lecture 14
Calorimetry
Answer:
Recall, C =
q = C x m x DT
Since the total solution is 100 mL and the density is 1.0 g/mL,
then m = (100 mL) (1.0 g/mL) = 100 g
DT = ( oC) = 6.5 oC
and C = J/g-oC

37. Since moles in solution were (0.050 L)(1.0 M ) = 0.050 mol
Chemistry Lecture 14
Calorimetry
Answer:
then qp = (4.18 J/g-oC) (100 g) (6.5 oC)
= J =
Since moles in solution were (0.050 L)(1.0 M ) = mol
DH = =
-2.7 kJ
-54 kJ/mol

38. DE = qevolved = -Ccalorimeter x DT
Chemistry Lecture 14
Bomb Calorimetry
In bomb calorimetry, the apparatus is sealed and the experiment is a constant-volume process. The heat change of the reaction is the internal energy change.
DE = qevolved = -Ccalorimeter x DT

39. Chemistry Lecture 14
Bomb Calorimetry

40. A Student “Bomb” Calorimetry
Chemistry Lecture 14
A Student “Bomb” Calorimetry
Thermometer
Cardboard or
Styrofoam Lid
Nested
Styrofoam Cups
Exothermic Reaction
Occurs in Solution

41. N2H4(l) + O2(g) N2(g) + 2 H2O(g)
Chemistry Lecture 14
Bomb Calorimetry
Question:
Hydrazine, N2H4, and its derivatives are widely used as rocket fuels.
N2H4(l) O2(g) N2(g) H2O(g)
When 1.00 g of hydrazine is burned in a bomb calorimeter, the temperature of the calorimeter increases by 3.51 oC. If the calorimeter has a heat capacity of
5.510 kJ/oC, what is the quantity of heat evolved?

42. Bomb Calorimetry Answer: Recall that:
Chemistry Lecture 14
Bomb Calorimetry
Answer:
Recall that:
qevolved = ( kJ/oC) x (3.51 oC)
qevolved = kJ
Calculate this on a per mole basis:
qevolved = =
DE = qevolved = -Ccalorimeter x DT
-618 kJ/mol

43. Term Test #1 Friday October 12th, 2001 6:30 P.M.
Chemistry Lecture 14
Term Test #1
Friday October 12th, 2001
6:30 P.M.

44. A to H Education Gym I to Z Rm 104 Odette Bldg
Chemistry Lecture 14
A to H Education Gym
I to Z Rm 104 Odette Bldg

45. Contents: SIX “Problems”!!
Chemistry Lecture 14
Duration: 75 minutes
Contents: SIX “Problems”!!
Covers Material From Chapters 4 - 6
A Simple Periodic Table & ALL Required Constants will be Supplied

46. Chapter 6: Enthalpy Changes for Chemical Reactions
Chemistry Lecture 16
Chapter 6: Enthalpy Changes for Chemical Reactions
Chapter Highlights
energy transfer, specific heat and heat transfer
heat of fusion & heat vapourization
exothermic & endothermic reactions
first law of thermodynamics
enthalpy changes
calorimetry
Hess’s Law

47. Chemistry Lecture 16
Hess’s Law
If a reaction is carried out in a series of steps, DH for the reaction will be equal to the sum of the enthalpy changes for each step.
For example:
(1) CH4(g) O2(g) CO2(g) H2O(g)
DH = kJ
(2) 2 H2O(g) H2O(l)
DH = -88 kJ

48. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)
Chemistry Lecture 16
Hess’s Law
Add: (1) + (2)
(1) + (2) CH4(g) O2(g) H2O(g)
CO2(g) H2O(l) H2O(g)
DH = ( ) kJ = kJ
Net Equation:
CH4(g) O2(g) CO2(g) H2O(l)
DH = kJ

49. Hess’s Law Question: Calculate DH for the reaction:
Chemistry Lecture 16
Hess’s Law
Question:
Calculate DH for the reaction:
2 C(s) H2(g) C2H2(g)

50. Hess’s Law Question: Given the following:
Chemistry Lecture 16
Hess’s Law
Question:
Given the following:
(1) C2H2(g) O2(g) CO2(g) H2O(l)
DH = kJ
(2) C(s) O2(g) CO2(g)
DH = kJ
(3) H2(g) O2(g) H2O(l)
DH = kJ

51. Chemistry Lecture 16
Hess’s Law
Answer:
Step 1: Target equation has C2H2(g) as a product so we reverse the first equation and change the sign on DH:
-(1) 2 CO2(g) H2O(l) C2H2(g) O2(g)
DH = kJ

52. Chemistry Lecture 16
Hess’s Law
Step 2: Target equation has 2 C(s) as a reactant so multiply the second equation x 2:
2 x (2) 2 C(s) O2(g) CO2(g)
DH = kJ

53. Chemistry Lecture 16
Hess’s Law
Step 3: Target equation has H2(g) as a reactant so we keep the third equation as is:
(3) H2(g) O2(g) H2O(l)
DH = kJ

54. Hess’s Law Step 4: Add the three equations from steps 1,2 and 3:
Chemistry Lecture 16
Hess’s Law
Step 4: Add the three equations from steps 1,2 and 3:
-(1) 2 CO2(g) H2O(l) C2H2(g) O2(g)
2 x (2) 2 C(s) O2(g) CO2(g)
(3) H2(g) O2(g) H2O(l)
Total
2 CO2(g) + H2O(l) C(s) O2(g) H2(g) O2(g)
C2H2(g) O2(g) CO2(g) H2O(l)

55. Hess’s Law Net 2 C(s) + H2(g) C2H2(g) -(1) DH = +1299.6 kJ
Chemistry Lecture 16
Hess’s Law
Net
2 C(s) H2(g) C2H2(g)
-(1) DH = kJ
2 x (2) DH = kJ
(3) DH = kJ
Total
DH = kJ

56. Enthalpies of Formation
Chemistry Lecture 16
Enthalpies of Formation
We can calculate enthalpy associated with many changes.
Vapourization (DHvap for converting liquids to gas)
Fusion (DHfus for melting solids)
Combustion (DHcom for combusting in oxygen)
Enthalpy of the reaction that forms a substance from its constituent elements is called
the enthalpy of formation, DHf, or the heat of formation.
2 C(s) H2(g) C2H2(g)
DHf = kJ

57. Enthalpies of Formation C(s) diamond graphite fullerene
Chemistry Lecture 16
Enthalpies of Formation
Standard state: the state that is most stable for the substance at the temperature of interest, usually 298 K and 1 atm.
C(s) diamond graphite fullerene
Standard enthalpy of formation, DH°f, is the enthalpy of the reaction that forms the substance from its constituent elements in their standard states.
2 C(s) H2(g) C2H2(g)
DHof = kJ

58. S S Enthalpies of Formation
Chemistry Lecture 16
Enthalpies of Formation
Enthalpies of reaction can be calculated by applying Hess's law to the enthalpies of formation of the participants:
DH°rxn = [nDH°f(products)] [mDH°f(reactants)] S S

59. Enthalpies of Formation
Chemistry Lecture 16
Enthalpies of Formation
The standard enthalpy of formation for ethanol, C2H5OH, is:
2 C(graphite) H2(g) O2(g) C2H5OH(l)
DH°f = kJ
By definition, DH°f = 0 for the standard state of an element.
Thus, DH°f = 0 for C(graphite)
DH°f = 0 for H2(g)
DH°f = 0 for O2(g)

60. Enthalpies of Formation
Chemistry Lecture 16
Enthalpies of Formation
Question:
Calculate DH for the combustion of propane, C3H8(g) using tables of standard enthalpies of formation.

61. S S Enthalpies of Formation Answer:
Chemistry Lecture 16
Enthalpies of Formation
Answer:
C3H8(g) O2(g) CO2(g) H2O(l)
DH°rxn = [nDH°f(products)] [mDH°f(reactants)]
DH°rxn = [3DH°f(CO2) + 4DH°f(H2O)] - [DH°f(C3H8) + 5DH°f(O2)]
= [3( kJ/mol) + 4( kJ/mol)] -
[( kJ/mol) + 5(0.0 kJ/mol)]
= (-2324 kJ) - (-103.8) = S S
-2220 kJ/mol

62. Enthalpies of Formation
Chemistry Lecture 16
Enthalpies of Formation
Question:
The standard enthalpy change for the reaction
CaCO3(s) CaO(s) CO2(g) is kJ.
Calculate the standard enthalpy of formation of CaCO3(s) from the standard enthalpies of formation of CaO(s) and CO2(g); kJ and kJ respectively.

63. Enthalpies of Formation
Chemistry Lecture 16
Enthalpies of Formation
Answer:
DH°rxn = [nDH°f(products)] [mDH°f(reactants)]
DH°rxn = [DH°f(CaO) + DH°f(CO2)] - [DH°f(CaCO3)]
178.1 kJ = [( kJ/mol) + ( kJ/mol)] - [DH°f(CaCO3)]
[DH°f(CaCO3)] = kJ kJ kJ
[DH°f(CaCO3)] = S
kJ/mol

64. Textbook Questions From Chapter #6
Chemistry Lecture 14
Textbook Questions From Chapter #6
Specific Heat: , 18, 24
Changes of State: 30
Enthalpy: , 38
Hess’s Law: , 42, 44, 48, 54
Calorimetry:
General & Conceptual 70, 78