1. Calendar 13-14 M 1/27 Sec 10.1-3:Thermodym, Phase D
B-1 1/29 Sec 10.3: Calorim, DHfo, Hess Law
B-1 1/31 Lab
M 2/3 Sec Entropy and DS
B-1 2/5 Quiz DH, DS Sec 11.3: DG, Equil temp calcs
B-2 2/7 Review
M 2/10 Test

2. Calendar 14-15 B-2 1/23-4 Sec 10.1-3:Thermodym Intro
M 1/26 Sec /11.3: Calorim, Phase D Demo lab w/ phase thermodynamics
B-1 1/27-8 Sec 10.3 Calcs of DH DHfo, Hess Law
B-2 1/29-30 Quiz, Rev of DH calcs,
M = B-1 2/2-3 Lab Hess
Wed 2/4 Sec Entropy and DS, DG
B-2 2/5-6 Quiz DS, DG: Review
M 2/9 Test

4. Driving Forces What causes chemical reactions to occur?
Nature drives things toward lower energy
Enthalpy = total energy content of a system
Nature likes things to be random, or more disorganized
Entropy = disorder in a system

5. Entropy = S A measure of the disorder in a system
Which has a greater entropy
A chess board before the 1st move or in the middle of a game
A jigsaw puzzle that is put together, or the same puzzle unassembled in its box
Humpty Dumpty before or after his tragedy
Nature tends to move spontaneously to a more disorganized state

6. Thermodynamics Study of energy, heat, and its flow
Energy = ability to do work
Push or pull a mass
Heat = q = a form of energy
Heat content is a measure of total Ek in a substance
Temp = meas of average movement (Ek)
Temp and Ek are directly relation
Ek = ½mv2
Temp diff determines direction of heat flow

7. System vs. Surrounding System = reaction or process you are examining
Surroundings = rest of the universe
Energy is exchanged between the system and the surroundings
Endothermic vs Exothermic reactions
Diagrams follow

8. Enthalpy = H The total energy content of a system
Sum Ek + Ep
Cannot be measured
Can meas Ek but not Ep
Ek = ½mv2
Can determine the change in heat content of a system = DH
Change in Enthalpy = DH = Hprod-Hreact

9. Thermochemical Equation
Exothermic
CH4(g) + 1½O2(g) CO2(g) + H2O(g) KJ
CH4(g) + 1½O2(g) CO2(g) + H2O(g) DH = -890 KJ
Endothermic
483.6 KJ + H2O(g) H2(g) + O2(g)
H2O(g) H2(g)+ O2(g) DH = KJ
Thermo chemical equation show the reactants and product
with the energy change that occurs

10. Endothermic Rx Cure

11. Exothermic Rx Cure

12. Calorimetry The measurement of heat flow = DH
We can catch the energy going into or out of a system
Use a calorimeter
Highly insulated device used to measure the amount of heat moving into or out of the system
Water is usually used in the cal is used to absorb energy from, or give off energy to the system

13. Heat Capacity = Cp
To measure heat flow into or out of the system we must know how much heat the material in the calorimeter can hold
Called heat capacity
Unique for each substance
Specific Heat Capacity = Cp
Amount heat (q) required to increase the temp of 1 g of a subst 1˚C
Water has a HUGE heat capacity
Units of J/g°C or J/g K

16. Coffee Cup Calorimeter
qH2O = mH2O x DTH2O x CpH2O
qH2O = DH = energy absorbed or given off by the water in the calorimeter
mH2O = mass (g) of the water
1 mL H2O = 1 gram H2O
DTH2O = change in temp
Tfinal – Tinitial
Cp = specific ht capacity of the water in the calorimeter
CpH2O = 4.18 J/g˚C

18. Sample Problem
1. A reaction occurs in a calorimeter that contains g of water. If the temp of the water goes from 25.0°C to 55.0°C, calculate the energy exchanged.

19. q(H2O) = m(H2O) x Dt(H2O) x 4.18 J/g⁰C
How much energy is needed to increase the temperature of 50.0 g of water 15.0⁰C?
How much energy (joules) is needed to increase the temp of mL water from 40.0⁰C to 60.0⁰C?
Haw many grams of water can be heated from 20.0⁰C to 75⁰C using Joules?
A mL sample of water, at 20.5 ⁰C, absorbs 9985 J of energy. Determine the final temp of the water.

20. q = m x (tf – ti) x 4.18 J/g⁰C q = (tf – ti) m x 4.18 J/g⁰C q
A mL sample of water, at 20.5 ⁰C, absorbs 9985 J of energy. Determine the final temp of the water.
q = m x (tf – ti) x 4.18 J/g⁰C
q = (tf – ti)
m x 4.18 J/g⁰C q
+ ti = tf

21. q of the Reaction
The heat lost by the reaction is gained by the H2O in the calorimeter
Calc q for the water in the calorimeter. The value of DHrx for the reaction is = to –qH2O
-qH2O = qrx = DHrx

22. Sample Problem
A reaction is carried out in a coffee cup calorimeter that contains g of water at 23.5˚C. If the final temp of the water becomes 20.0˚C, determine the value of DH for the reaction.
qH2O = mH2O x DTH2O x CpH2O
= g x (20.0˚C – 23.5˚C) x 4.18 J/g˚C
= x 103J
DHrx = -qH2O = x 103 J or KJ

23. Sample Problem
An acid and a base are mixed in a coffee-cup calorimeter. The calorimeter contains mL of water at 22.0˚C. A neutralization reaction occurs that increased the temp to 28.0˚C. What is DH for the reaction?
qH2O = mH2O x DTH2O x CpH2O
= g x (28.0˚C – 22.0˚C) x 4.18 J/g˚C
= 3.34 x 103 J or 3.34 KJ
DHrx = -qH2O = x 103 J or -3.34KJ

24. Calculation of C
Can calc the value of C for any substance through calorimetry
Heat the metal in a hot water bath and drop it in calorimeter: the heat picked up from the metal goes into the water
Use the calorimetry equation to determine the amount of energy transferred by the substance
qH2O = -qPb = mPb x DTPb x CpPb

25. A 90.0 g sample of metal is placed in a calorimeter
that contains g of water at 25.5˚C. If the
temp of the water increases to 37.5 ˚C, determine the
Amount of energy transferred in the process.
Hint: q(H2O) = m (H2O) x Dt (H2O) x C (H2O)
and Dt (H2O) = (tf – ti)

26. Phase Changes Chap 11.3 Solid Liquid Gas
Blue = endo
Red = exo
sublimation
fusion
vaporization
Solid
Liquid
Gas
crystallization
condensation
deposition
Each phase change requires a specific amount of energy
Molar Heat of fusion water = 6.01 KJ/mol
Molar Heat vaporization water = 40.7 KJ/mol
What are the molar heats of condensation and crystallization???

27. Lab Demo Heat Fusion Ice
How much energy does it take to melt a gram of ice? A mole of ice?
Place ice in warm water in a coffee-cup calorimeter
Measure
1- temp change for water in calorimeter (Dt)
2- how many grams (or mole)

28. Phase Change Sample Problem
How much energy is required to melt 27.0 g of ice at 0°C?
q = molesH2O x Molar Ht FusionH2O
Mol Htfus = 6.01 KJ/mol
Hint: 1 - calc the moles of water (mol=g/mm)
2 - multiply moles times 6.01 KJ/mol
27.0 g H2O = mol H2O
18.0 g/mol H2O
q = 1.50 mol water x KJ = 9.01 KJ of is needed
1 mol water

29. Phase Change Sample Problem
How much energy is required to increase the temp of 27.0 g of water from zero to degrees C? q(H2O) = m (H2O) x Dt (H2O) x C (H2O) = 27.0 g x 100.0°C x 4.18 J/g°C = J = KJ

30. Phase Change Sample Problem
How much energy is required to vaporize 27.0 g of water to steam, at 100.0°C?
q = molesH2O x Molar Ht VapH2O
Mol Htvap = 40.7 KJ/mol
Hint: - 1 calculate the moles of water (mol=g/mm)
- 2 multiply moles times 40.7 KJ/mol
27.0 g H2O = mol H2O
18.0 g/mol H2O
q = 1.50 mol water x KJ = 61.1 KJ
1 mol water

31. Phase Change Sample Problem
How much energy is required to take 27.0 g of ice at 0°C and change it to steam at °C?
q = molesH2O x Molar Ht FusionH2O
qH2O = mH2O x DTH2O x CpH2O
q = molesH2O x Molar Ht VapH2O
= 9.1 KJ KJ KJ
= 81.5 KJ

32. Phase Change Sample Problem
How much energy is required to take 63.0 g of ice at 0°C and change it to steam at °C?
q = molesH2O x Molar Ht FusionH2O
(63.0/18.0) mol x 6.01 KJ/mol = q1
qH2O = mH2O x DTH2O x CpH2O
63.0 g x 100.0°C x 4.18 J/g°C = q2
q = molesH2O x Molar Ht VapH2O
(63.0/18.0) mol x 40.7 KJ/mol = q3

33. Phase Change Math
Molar heats of Fusion (etc) are used to calc energy change for phase change, (at a constant temp).
q = moles x Heat Fusion (etc)
Calorimetry is used to calc energy exchanged went a substance temp increases
qH2O = mH2O x DTH2O x CpH2O

34. qsteam = msteam x DTsteam x Cpsteam
qH2O = molH2O x HvapH2O
qH2O = mH2O x DTH2O x CpH2O
qice = molice x Hfusice
qice = mice x DTice x Cpice

35. Phase Change Sample Problem
How much energy is given off when 27.0 g of steam condenses to water at 100.0°C?
q = molesH2O x Molar Ht VapH2O
Mol Htvap = 40.7 KJ/mol
Hint: - 1 calculate the moles of water (mol=g/mm)
- 2 multiply moles times 40.7 KJ/mol
27.0 g H2O = mol H2O
18.0 g/mol H2O
q = 1.50 mol water x KJ = 61.1 KJ is given off
1 mol water

36. 3 Ways to Determine DH 1. Calorimetry . DH° = -qH2O 2. DH°f tables
3. Hess’s Law of Heat Summation
Rules of thermodynamics

37. Rules of Thermodynamics
DH is proportional to the amount of reactant or product
Reverse reactions have opposite sign
Hess’s Law of Heat Summation

38. DH is proportional to the amount of Reactant or Product
Based on stoichiometric coefficients
Must have a balanced thermochemical equation
CH O CO H2O DH = -890 KJ
How many KJ are given off by the complete combustion of 2 moles of CH4?
2 mol CH4 x KJ = KJ
1 mol CH4

39. Sample Problem How many KJ of energy are produced by burning 1
Sample Problem How many KJ of energy are produced by burning 1.25 g of CH4 according to the rx: CH O CO H2O DH = -890 KJ How many KJ of energy are produced by burning 1.25 g of O2?

40. Reverse Reactions
DH for reverse reactions have the same value, but the opposite sign
Reactants Products DH = (-)
Products Reactants DH = (+)

41. Reverse Reactions Have Opposite Signs on DH
CH O CO H2O DH = -890 KJ
CO H2O CH O2 DH = +890 KJ

42. Hess's Law of Heat Summation
In a multi-step reaction the sum of the DH values for all the steps equals DH for the overall reaction
DH1 + DH2 = DH3
A + B --> C DH1 = -20 KJ
B + C ---> D DH2 = -40 KJ
A + 2B --> D DH3 = ?
= -60 KJ

43. Hess's Law of Heat Summation
Can use Hess's Law to measure DH for a reaction that is too hard to run in the lab
C + ½O2 ---> CO DH1 = ?
CO + ½O2 ---> CO2 DH2 = -283 KJ
C + O2 ---> CO DH3 = -393 KJ
DH1 + DH2 = DH3
DH1 = DH3 - DH2
DH1 = (-393 KJ) - (-283 KJ)
= -110 KJ

44. Use the following 2 equations to determine DH for the overall reaction: Sn (s) + 2 Cl2 (g) --- > SnCl4 (aq) Sn (s) + Cl2(g) ----> SnCl2(s) kJ/mol SnCl2 (s) + Cl2 (g) ----> SnCl4(aq) kJ/mol

45. Mn (s) + 02 (g) ---> MnO2 (s)
MnO2 (s) + Mn (s) --- > 2 MnO (s) kJ
2 MnO2 (s) --- > 2 MnO (s) + 02 (g) kJ

46. SnO2 (s) + 2H2 (g) --- > Sn (s) + 2 H20 (l)
Sn (s) + 02 (g) --- > SnO2 (s) kJ
H2 (g) + ½ 02 (g) --- > H20 (l) kJ

47. 2 Mg (s) + SiCl4 (1) --- > Si (s) + 2 MgCl2 (s)
SiCl4 (l) --- > Si (s) + 2 Cl2 (g) kJ
Mg (s) + Cl2 (g) --- > MgCl2 (s) kJ

48. N2 + 2 H2 --> N2H4 H2 + ½ 02 --> H20 -242 kJ
N H20 --> N2H kJ

49. CH > C H20
C + 2 H2 ---> CH kJ
C + O > CO kJ
H2 + ½ O > H2O kJ

50. Find DH° for the reaction 2H2(g) + 2C(s) + O2(g) --> C2H5OH(l) using the following thermochemical data. C2H5OH (l) + 2 O2 (g) --> 2 CO2 (g) + 2 H2O (l) DH = kJ C (s) + O2 (g) --> CO2 (g) DH = kJ H2 (g) + ½ O2 (g) --> H2O (l) DH = kJ

51. H2(g) + Br2(g) → 2 HBr(g) ΔH = −72 kJ
H2(g) → 2 H(g) ΔH = 436 kJ
Br2(g) → 2 Br(g) ΔH = 224 kJ
Calculate the enthalpy change for
H(g) + Br(g) → HBr(g)

52. Enthalpies of Formation = DHf° Cu(s) + Cl2(g) ---> CuCl2(s) DH = -220 KJ
The energy change DH⁰, when 1 mol of a substance is formed from its elements in their normal (standard) state (the ° means 25°C and normal pressure)
Elements (including diatomic like H2, N2, Cl2) are arbitrarily assigned DHf° = 0
DHf° = a relative measure of how stable compounds are
A (-) DHf° means compound is more stable than its elements
A (+) DHf° means compound is less stable than its elements

53. Enthalpies of Formation
DH°rx = SDHf°(products) - SDHf°(reactants)
DHf° values are on table A-11 pg 833
They are used to calculate, NOT measure, the change in enthalpy, DH°, for a chemical reaction that is hard or impossible to carry out in the lab.
Sample prob E pg 356 Do pract prob #1,2 pg 356

55. Sample Problem Calculation DH° for the reaction:
C3H8(g) + 5O2 ---> 3CO2(g) + 4H2O(l)
= DHf° KJ/mol
DH°rx = SDHf°(products) - SDHf°(reactants)
DH°rx = [3mol(-393.5)KJ/mol + 4 mol(-285.8)KJ/mol]- [1mol(-103.8)KJ/mol + 0]
= KJ

56. Entropy S Randomness or disorder in a system
Entropy is ever increasing
Molar entropy is entropy possessed by
1 mole of a substance
Nature favors increases in entropy
Reactions that increase entropy occur more easily
Elements have a value for entropy

57. Entropy S = Randomness As temp entropy Sol Liq Gas entropy goes up
A solution has more entropy than its separate components
NaCl water solution vs water and solid NaCl
Reactions that produce more moles of gases increase entropy
Is DS (+) or (-) for:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
2CH3OH(g) + 3O2(g)  2CO2(g) + 4H2O(g)

58. Entropy Change = DS° Calculted in the same manner as DH°
DS°rx = SDS°(products) - SDS°(reactants)
Sample prob F pg 361

59. Calc DS° for: 2CH3OH(g) + 3O2(g)  2CO2(g) + 4H2O(g)
DS°J/molK =
DS° = [2 mol (213.8KJ/molK) + 4 mol (188.7KJ/molK)] [2 mol (127.2 KJ/molK) + 3 mol (205.0 KJ/molK)]
= 313 KJ/K
Does the entropy change, DS, favor a spontaneous reaction or not???

60. What determines if a reaction or process occurs spontaneously?
What energy change favors a spontaneous reaction?
(-) change in energy
What entropy change favors a spontaneous reaction?
(+) positive change in entropy
Neither DH or DS alone, can predict if a rx will be spontaneous

61. Gibb's Free Energy = G
Determines whether a reaction is spontaneous or not spontaneous at a specific temperature
If DG is (-) reaction is spontaneous
If DG is (+) reaction is NOT spontaneous
If DG = 0 reaction is just beginning
Called equilibrium

62. Gibb's Free Energy = G Can be calculated in 2 ways
1-From free energy tables
Must be at standard conditions to use:
25˚C and pressure at surface of the earth
DG°rx = SDG°(products) - SDG°(reactants)
2-From DH and DS values at temperatures other than 25˚C
DG = DH – TDS

63. Gibb's Free Energy = G Calc DG˚ for H2(g) + CO2(g)  H2O(g) + CO(g)
(KJ/mol)
DG°rx = SDG°(products) - SDG°(reactants)
= [1 mol (-228.6KJ/mol) + 1 mol (-137.2KJ/mol] – [0 + 1mol ( KJ/mol)]
= 28.6 KJ
Is the rx spontaneous???

64. DG at Temp other than 25˚C Calc DG at 100˚C for the reaction:
H2(g) + CO2(g)  H2O(g) + CO(g) DH DS
DH˚ = [(1 x ) + (1 x )] – [(1 x ) + 0]
= 41.2 KJ
DS˚ = [(1 x 188.7) + (1 x 197.2)] – [(1 x 130.7) + (1 x 213.8)]
= 41.4 J/K
= KJ/K
(Move decimal 3 places to convert J/K to KJ/K)

65. DG = DH – TDS
Is a reaction spontaneous at 100˚C if DH = 41.2 KJ and DS = 41.4 J/K
DG = DH – TDS
= 41.2KJ – ( K) KJ/K
= 25.8 KJ is the reaction spontaneous?

66. Determination of Equilibrium Temp
The temp (K) at which the rx just begins
Equilibrium
T = DH = KJ = 995 K
DS KJ/K
ºC = K or ºC = K – 273
995 – 273 = 722 ºC

67. Gibb’s Free Energy At what temp does
H2O(l)  H2O(g) becomes spontaneous
DH = ( KJ/mol) – ( KJ/mol) = 44.0 KJ/mol
DS = (188.7 J/mol K) – (70.0 J/mol K) = J/mol K
T = DH = KJ/mol = 370 K
DS KJ/mol K
ºC = K - 273
= 370 K – 273 = 97 ºC

68. Affects of Signs of DH and DG
If DH is (-) and DS is (+)
Rx is always spontaneous
If DH is (+) and DS is (-)
Rx is never spontaneous
If DH is (+) and DS is (+)
Increase temp to make rx spontaneous
If DH is (-) and DS is (-)
Decreases temp to make rx spontaneous

69. Initial water temp ºC
final water temp ºC
Change in water temp ºC
Final water volume mL
Initial water volume mL
Volume of melt mL
mass ice melted g
Heat released by cooling water J
J/g ice melted (heat of fusion) J/g
kJ/mol ice melted (molar heat of fusion) kJ/mol

70. How much energy is lost when 125
How much energy is lost when g of water in a calorimeter changes from 98.6°C to 18.9°C?

71. CaO + CO2 --------> CaCO3 ΔHº = -178 kJ
CaO + H2O(l) > Ca(OH)2 ΔHº = -65 kJ
Ca(OH)2 + CO > CaCO3 + H2O(l)

72. SOCl2 + NiO --------> SO2 + NiCl2 150 kJ
SOCl2 + H2O > SO2 + 2HCl kJ
NiO + 2HCl -----> NiCl2 + H2O(g)