1. Tro, Chemistry: A Molecular Approach

2. A LOOK AHEAD 1-The study of the nature and the different types of energy. 2- The study of heat change in chemical reactions, which is thermo chemistry . The vast majority of reactions are either endothermic (absorbing heat) or exothermic (releasing heat). 3-The thermo chemistry is a part of a broad subject called the first law of thermodynamics, which is based on the law of conservation of energy. The change in internal energy can be expressed in terms of the changes in heat and work done of a system. 4- The measure of the heat of reaction or calorimetry under constant-volume and constant-pressure, and the meaning of specific heat and heat capacity, quantities used in experimental work.
Tro, Chemistry: A Molecular Approach

3. A LOOK AHEAD 5- Knowing the standard enthalpies of formation of reactants and products enables us to calculate the enthalpy of a reaction. Discuss ways to determine these quantities either by the direct method from the elements or by the indirect method, which is based on Hess’s law of heat summation. 6- Finally, the study the heat changes when a solute dissolves in a solvent (heat of solution) and when a solution is diluted (heat of dilution). Every chemical reaction obeys two fundamental laws; the law of conservation of mass and the law of conservation of energy. The mass relationships between reactants and products are discussed .
Tro, Chemistry: A Molecular Approach

4. Thermochemistry 6. 1 The Nature of Energy and Types of Energy. 6
Thermochemistry 6.1 The Nature of Energy and Types of Energy Energy Changes in Chemical Reactions. 6.3 Introduction to Thermodynamics. 6.4 Enthalpy of Chemical Reactions. 6.5 Calorimetric. 6.6 Standard Enthalpy of Formation and Reaction. 6.7 Heat of Solution and Dilution.
Tro, Chemistry: A Molecular Approach

5. 6.1 The Nature of Energy and Types of Energy Energy is a much-used term that represents a rather abstract concept. For instance when we feel tired, we might say we haven’t any energy; and we read about the need to find alternatives to nonrenewable energy sources. Unlike matter, energy is known and recognized by its effects.
Tro, Chemistry: A Molecular Approach

6. Energy is defined as the capacity to do work
Energy is defined as the capacity to do work Work = force x distance Chemists define work as directed energy change resulting from a process. Kinetic energy: is the energy produce by a moving object : is one form of energy that is of particular interest to chemists. Other include radiant energy, thermal energy, chemical energy and potential energy.
Tro, Chemistry: A Molecular Approach

7. Tro, Chemistry: A Molecular Approach

8. Chemical energy can be considered a form of potential energy because it is associated with the relative positions and arrangements of atoms within a given substance.
Tro, Chemistry: A Molecular Approach

9. 6.2 Energy Changes in Chemical Reactions All Chemical reactions absorb or produce (release) energy in the form of heat.
Tro, Chemistry: A Molecular Approach

10. Tro, Chemistry: A Molecular Approach

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12. Tro, Chemistry: A Molecular Approach

13. Tro, Chemistry: A Molecular Approach

14. 6.3 Introduction to Thermodynamics Thermo chemistry is part of a broader subject called thermodynamics, which is the scientific study of the interaconversion of heat and other kinds of energy. In thermodynamics, we study changes in the state of a system, which is defined by the values of all relevant macroscopic properties, composition, energy, temperature, pressure and volume which called state functions -- properties that are determined by the state of the system, regardless of how that condition was achieved. The magnitude of change in any state function depends only on the initial and final states of the system and not on how the change is accomplished.
Tro, Chemistry: A Molecular Approach

15. Tro, Chemistry: A Molecular Approach

16. The First Law of Thermodynamics The first law of thermodynamics is based on the law of conversation of energy.
First Law of Thermodynamics: Energy cannot be Created or Destroyed
the total energy of the universe cannot change
though you can transfer it from one place to another
DEuniverse = 0 = DEsystem + DEsurroundings
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17. Tro, Chemistry: A Molecular Approach

18. Example Consider the reaction between 1 mole of sulfur and 1 mole of oxygen gas to produce 1 mole of sulfur dioxide: S(s) + O2(g) SO2(g) This system composed of the reactant molecules S and O2 and the product molecules SO2. We don’t know the internal energy of the reactants and the products, but we can measure the change in energy content, ΔΕ, ΔΕ = E (product) – E (reactant) = energy content of 1 mol SO2(g) – energy content of [1 mol S(s) + 1 mol O2(g)]
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19. First Law of Thermodynamics
*Conservation of Energy *For an exothermic reaction, “lost” heat from the system goes into the surroundings *two ways energy “lost” from a system, 1- converted to heat, q 2- used to do work, w *Energy conservation requires that the energy change in the system equal the heat released + work done DE = q w DE = DH + PDV DE is a state function * internal energy change independent of how done
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20. Tro, Chemistry: A Molecular Approach

21. Work and Heat work can be defined as force F multiplied by distance d: w = Fd One way to illustrate mechanical work is to study the expansion or compression of a gas (see Fig.(6.5)). The work done by the surroundings is: w = - PΔV where ΔV, the change in volume, is given by Vf - Vi
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22. Worked Example 6.1

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24. Tro, Chemistry: A Molecular Approach

25. Heat (q) is the other component of internal energy, heat also is not a state function, like work, because they are not properties of a system. They manifest themselves only during a process (during a change). Thus, their values depend on the path of the process and vary accordingly. We cannot write Δq = qf - qi . But the sum of work and heat is equal to ΔΕ and as we know that E is a state function. Thus, if changing the path from the initial state to the final state increases the value of q, it will decrease the value of w by the same amount and vice versa.
Tro, Chemistry: A Molecular Approach

26. Tro, Chemistry: A Molecular Approach

27. 6.4 Enthalpy of Chemical Reactions - How the first law of thermodynamics can be applied to processes carried out under different conditions? If a chemical reaction is run at constant volume, then : ΔV = 0 and no P-V work will result from this change. from the equation: Δ E = q + w, it follows that: we add the subscript v to that this is a constant volume- process
Δ E = q – PΔV = qv
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28. Tro, Chemistry: A Molecular Approach

29. ΔH = ΔΕ + PΔV at constant pressure process, qp = ΔH, although q is not a state function, the heat change at constant pressure is equal toΔ H because the path is defined, and therefore it can have only a specific value. If the reaction occurs under constant volume(V=0), the heat change, qv, is equal to ΔΕ, and when the reaction is carried out at constant pressure(P=0), the heat change, qp, is equal to ΔH.
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30. * The two factors that determine the thermodynamic favorability are the enthalpy and the entropy. * The enthalpy is a comparison of the bond energy of the reactants to the products. * bond energy = amount needed to break a bond. DH * The entropy factors relates to the randomness/orderliness of a system DS * The enthalpy factor is generally more important than the entropy factor
Tro, Chemistry: A Molecular Approach

31. Enthalpy of Reactions Because most of reactions are constant pressure processes, we can equate the heat change to the change in enthalpy. For any reaction such as: A (reactant) B (product) the change in enthalpy, called the enthalpy of reaction, ΔH = H (products) – H (reactants)
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32. ,
Tro, Chemistry: A Molecular Approach

33. Enthalpy. related to the internal energy DH generally kJ/mol
Enthalpy * related to the internal energy DH generally kJ/mol *stronger bonds = more stable molecules * if products more stable than reactants, energy released exothermic DH = negative * if reactants more stable than products, energy absorbed endothermic DH = positive *The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions. Hess’ Law DH°rxn = S(DH°prod) - S(DH°react)
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34. Entropy
entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases, S
S generally J/mol
S = k ln W
k = Boltzmann Constant = 1.38 x J/K
W is the number of energetically equivalent ways, unitless
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35. Thermochemical Equations At 0̊ C and a pressure of 1 atm, ice melts to form liquid water, measurements show that for every mole of ice converted to liquid water under these conditions, 6.01 kilojoules(kJ) of heat energy re absorbed by the system(ice). P = constant, the heat change is equal to enthalpy change (ΔH).
Tro, Chemistry: A Molecular Approach

36. Tro, Chemistry: A Molecular Approach

37. Tro, Chemistry: A Molecular Approach

38. Tro, Chemistry: A Molecular Approach

39. Worked Example 6.3

40. A comparison of ΔH and ΔE In the thermo chemical reaction : What is the relation between ΔH and ΔΕ 2Na(s) + 2H2O NaOH(aq) +H2(g) ΔH = kJ/mol when 2 moles of Na react with an excess of water(H2O), kJ of heat are given off. H2 gas must push air to enter the atmosphere. Some of the energy produced by the reaction is used to do work of pushing back a volume of air (ΔV) against atmospheric pressure (P), see figure 6.7. The change in internal energy can be calculated: ΔΕ = ΔH - PΔV
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41. Tro, Chemistry: A Molecular Approach

42. This calculation shows that ΔΕ and ΔH are approximately the same
This calculation shows that ΔΕ and ΔH are approximately the same. The reason ΔH is smaller than ΔΕ in magnitude is that some of the internal energy released is used to do gas expansion work, so less heat is involved. For reactions that do not involve gases, ΔV is usually very small and so ΔΕ is practically the same as ΔH . Another way to calculate the internal energy change of a gaseous reaction is to assume ideal gas behavior and constant temperature: ΔΕ = ΔH - Δ(PV) = ΔH - Δ(nRT) = ΔH – RTΔn where Δn = number of moles of products – number of moles of reactants
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43. Worked Example 6.4

44. Specific Heat and Heat Capacity The specific heat of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree celsius (J/g. ̊ C).
Tro, Chemistry: A Molecular Approach

45. Tro, Chemistry: A Molecular Approach

46. Tro, Chemistry: A Molecular Approach

47. See example 6.5 page 240
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48. Worked Example 6.5

49. Worked Example 6.6

50. Costant – pressure Calorimetry It is used to determine the heat changes for noncombustion reactions, as in figure 6.9.
Tro, Chemistry: A Molecular Approach

51. Tro, Chemistry: A Molecular Approach

52. Tro, Chemistry: A Molecular Approach

53. Worked Example 6.7

54. Worked Example 6.8

55. Tro, Chemistry: A Molecular Approach

56. 6.6 Standard Enthalpy of Formation Reaction
Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f
Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f
The standard enthalpy of formation of any element in its most stable form is zero.
DH0 (O2) = 0 f
DH0 (C, graphite) = 0 f
DH0 (O3) = 142 kJ/mol f
DH0 (C, diamond) = 1.90 kJ/mol f
6.6

57. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.
rxn
aA + bB cC + dD
DH0
rxn
dDH0 (D) f
cDH0 (C) = [ + ] -
bDH0 (B)
aDH0 (A)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)
6.6

58. C (graphite) + 1/2O2 (g) CO (g)
CO (g) + 1/2O2 (g) CO2 (g)
C (graphite) + O2 (g) CO2 (g)
6.6

59. 6.6

60. Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
rxn
S(rhombic) + O2 (g) SO2 (g) DH0 = kJ
rxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ
rxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
rxn
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
2S(rhombic) + 2O2 (g) SO2 (g) DH0 = x2 kJ
rxn +
CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = kJ
rxn
C(graphite) + 2S(rhombic) CS2 (l)
DH0 = (2x-296.1) = 86.3 kJ
rxn
6.6

61. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
DH0
rxn =
[ 12x– x–187.6 ] – [ 2x49.04 ] = kJ
-5946 kJ
2 mol
= kJ/mol C6H6
6.6

62. Worked Example 6.9

63. Worked Example 6.10

64. Chemistry in Action: Bombardier Beetle Defense
C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2H2O (l) DH0 = ?
C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g) DH0 = 177 kJ/mol
H2O2 (aq) H2O (l) + ½O2 (g) DH0 = kJ/mol
H2 (g) + ½ O2 (g) H2O (l) DH0 = -286 kJ/mol
DH0 = – 286 = -204 kJ/mol
Exothermic!

65. DHsoln = Shown - Hcomponents
The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.
DHsoln = Shown - Hcomponents
Which substance(s) could be used for melting ice?
Which substance(s) could be used for a cold pack?
6.7

66. 6.7 Heat of Solution and Dilution Enthalpy changes occure as well when a solute dissolves in a solvent or when a solution is diluted. Heat of Solution The heat of solution, or enthalpy of solution, ΔHsoln, is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. The quantity ΔHsoln represents as: ΔHsoln = Hsoln - Hcomponents ΔHsoln is positive for endothermic (heat-absorbing) reactions and negative for exothermic (heat-generating) reactions.
Tro, Chemistry: A Molecular Approach

67. Heat of Dilution Dilution is more solvent is added to lower the overall concentration of the solute, additional heat is usually given off or absorbed. The heat of dilution is the heat change associated with the dilution process. If a certain solution process is endothermic and the solution subsequently diluted, more heat will be absorbed by the same solution from the surrounding. The converse holds true for an exothermic solution process– more heat will be liberated if additional solvent is added to dilute the solution.
Tro, Chemistry: A Molecular Approach

68. Tro, Chemistry: A Molecular Approach