Download presentation

1
**CHEMICAL THERMODYNAMICS**

AP Chemistry Chapter 19 Notes CHEMICAL THERMODYNAMICS

2
**1st Law of Thermodynamics Energy in the universe is conserved**

REVIEW 1st Law of Thermodynamics Energy in the universe is conserved

3
**value depends on how process takes place (i.e. q, w)**

REVIEW Path Function value depends on how process takes place (i.e. q, w)

4
heat q = mCpT

5
REVIEW State Function value independent of path - is a defined reference or zero point (i.e. H)

6
**Hf0 of elements in natural form at 250C**

Enthalpy - H zero point Hf0 of elements in natural form at 250C

7
Enthalpy - H types: Hrxn = nHf0(prod) - nHf0(react)

8
Enthalpy - H types: Hfus , Hvap

9
**a measure of randomness or disorder of a system**

ENTROPY “S” a measure of randomness or disorder of a system

10
**Entropy is NOT conserved. The universe seeks maximum disorder.**

11
2nd Law of Thermodynamics for a spontaneous process, entropy increases

12
**Entropy - state function**

zero reference: S = 0 for a perfect crystal at absolute zero

13
**Ssolid < Sliquid << Sgas**

14
S0 = standard entropy of elements & cmpds at P = 1 atm; T = 250C units = J/K mol [Appendix L, text]

15
Calculate S using Hess’ Law S(rxn) = nS0(prod) - nS0(react)

16
Example: Ca(s) + C(gr) + 3/2 O2 CaCO3(s)

17
Suniv = Ssys + Ssurr if Suniv > 0 process spontaneous

18
Suniv = Ssys + Ssurr if Suniv < 0 process ?

19
Suniv = Ssys + Ssurr if Suniv = 0 process ?

20
to relate S and H consider

21
**where the water is the system & everything else is the surroundings**

H2O(s) H2O(l) where the water is the system & everything else is the surroundings entropy of water increases, therefore Ssys > 0 the vaporization of water is an endothermic process, therefore heat is flowing from the surroundings to the system & the random motion of the atoms in the surroundings decrease. thus, Ssurr < 0 so, Suniv = Ssys(+) + Ssurr(-) ; which predominates? we know that above 1000C, water spontaneously evaporates and that below 1000C, water spontaneously condenses, thus TEMPERATURE must have an effect on relative importance of Ssys and Ssurr .

22
**Temperature Dependence S = J/mol (1/T) Ssurr = -H/T**

introduces central idea that the entropy changes in the surroundings are primarily determined by heat flow. exothermic process increases the entropy of the surroundings, thus is an important driving force for spontaneity. Remember that a system tends to undergo changes that lower its energy. The SIGN on Ssurr depends on the direction of the heat flow. The MAGNITUDE of Ssurr depends on the temperataure.

23
**Ssys Ssurr Suniv spon**

24
**Ssys Ssurr Suniv spon**

25
**Ssys Ssurr Suniv spon**

26
**Ssys Ssurr Suniv spon + + + Y - - - N + - ? D - +**

It will be spontaneous if Ssys has a larger magnitude larger than Ssurr.

27
**Ssys Ssurr Suniv spon + + + Y - - - N + - ? D - + ? D**

It will be spontaneous if Ssurr has a larger magnitude larger than Ssys.

28
For a phase change:

29
GIBB’S FREE ENERGY G

30
**most abstract of thermodynamic state functions**

31
w w = reversible work w2 P V

32
**G = w + PV Definition: w: reversible work PV: pressure/volume work**

isothermal, reversible path

33
G = w + PV + VP at constant P P = 0 so VP = 0 G = w + PV

34
G = w + PV + VP at constant P G = w + PV at constant V V = 0 so PV = 0 G = w = useful work

35
**must measure G over a process**

G cannot be measured must measure G over a process

36
ZERO REFERENCE G = 0 for elements in stable form under Standard Thermodynamic Conditions T = 25oC ; P = 1 atm

37
**Gf0 = standard Free Energy of Formation from the elements**

Appendix L, text

38
G follows Hess’ Law: G0 (rxn) = nGf0(p) - nGf0(r)

39
Summary of Laws of Thermodynamics Zeroth Law: Heat Gain = Heat Loss

40
Summary of Laws of Thermodynamics First Law: Law of Conservation of Energy

41
Summary of Laws of Thermodynamics Second Law: Defines Entropy

42
Summary of Laws of Thermodynamics Third Law: Defines Absolute Zero

43
GIBB’S HELMHOLTZ EQUATION

44
combine G H T

45
G = -aT + H G, H are state functions, thus “a” must be a state function G = -S(T) + H

46
Gibb’s Helmholtz Equation G = H - TS

47
**Units on the State Functions**

51
**thus, a process is spontaneous**

if and only if G is negative

52
**enthalpy (minimum energy)**

Spontaneity controlled by enthalpy (minimum energy)

53
**enthalpy entropy (maximum disorder)**

Sponaneity controlled by enthalpy entropy (maximum disorder)

54
Sponaneity controlled by enthalpy entropy both

55
Predict Spontaneity IF H(-) and S(+) G = -H - T(+S) G < 0, => spontaneous

56
Predict Spontaneity IF H(+) and S(-) G = +H - T(-S) G > 0, => NOT spontan

57
**Summary of Spontaneity**

H S G Spont. yes no or ? or ?

58
Uses of the Gibb’s Helmholtz Equation

59
1. Find the molar entropy of formation for ammonia.

60
**2. Elemental boron, in thin fibers, can be made from a boron halide:**

BCl3(g) + 3/2 H2(g) ->B(s) + 3HCl(g)

61
Calculate: H0, S0 and G0. Spontaneous? Driving force?

62
3. Using thermodynamic information, determine the boiling point of bromine.

63
Thermodynamic Definition of Equilibrium Geq = 0

64
by definition G = H - TS & H = E + PV

65
**thus, G = E + PV - TS take derivative of both sides**

dG = dE + PdV + VdP - TdS - SdT

66
**for a reversible process**

TdS = q

67
**derivative used for state function while **

partial derivative used for path function

68
if the only work is PV work of expansion PdV = w

69
First Law of Thermodynamics E = q - w dE = q - w = 0

70
thus q = w or TdS = PdV

71
by substitution dG = 0 + PdV + VdP - PdV - SdT or dG = VdP - SdT

72
let’s assume we have a gaseous system at equilibrium, therefore, examine Kp at that constant temperature

73
at constant T SdT = 0 thus dG = VdP

74
but

78
if “condition 2” is at standard thermodynamic conditions, then G2 = G0 and P2 = 1 atm

79
thus

80
determine G for aA + bB cC + dD where all are gases

81
G = Gprod - Greact = cGC + dGD - aGA - bGB

82
but and likewise for the others

83
and

84
but

85
Thus, DG = DG0 + (RT) ln Q

86
But aA + bB cC + dD G = 0

87
thus or in general

88
**THERMODYNAMICS & EQUILIBRIUM**

89
**information, determine**

3. Using thermodynamic information, determine the boiling point of bromine.

90
**Thermodynamics and Keq**

FACT: Product-favored systems have Keq > 1.

91
**Thermodynamics and Keq**

Therefore, both ∆G˚rxn and Keq are related to reaction favorability.

92
**Thermodynamics and Keq**

Keq is related to reaction favorability and thus to ∆Gorxn. The larger the value of K the more negative the value of ∆Gorxn

93
**Thermodynamics and Keq**

∆Gorxn= - RT lnK where R = J/K•mol

94
**Thermodynamics and Keq**

∆Gorxn = - RT lnK Calculate K for the reaction N2O4 2 NO2 ∆Gorxn = +4.8 kJ K = 0.14 When ∆G0rxn > 0, then K < 1

95
∆G, ∆G˚, and Keq ∆G is change in free energy at non-standard conditions. ∆G is related to ∆G˚ ∆G = ∆G˚ + RT ln Q where Q = reaction quotient

96
**∆G, ∆G˚, and Keq When Q < K or Q > K, reaction is spontaneous.**

When Q = K reaction is at equilibrium When ∆G = 0 reaction is at equilibrium Therefore, ∆G˚ = - RT ln K

97
**Product favored reaction**

∆G, ∆G˚, and Keq Product favored reaction –∆Go and K > 1 In this case ∆Grxn is < ∆Gorxn , so state with both reactants and products present is MORE STABLE than complete conversion.

98
**∆G, ∆G˚, and Keq Product-favored reaction. 2 NO2 N2O4**

∆Gorxn = – 4.8 kJ Here ∆Grxn is less than ∆Gorxn , so the state with both reactants and products present is more stable than complete conversion.

99
**Thermodynamics and Keq Overview**

∆Gorxn is the change in free energy when reactants convert COMPLETELY to products.

100
**Keq is related to reaction favorability.**

When ∆Gorxn < 0, reaction moves energetically “downhill”

101
**4. For the following reaction, calculate the temperature at which the reactants are favored.**

102
**THERMODYNAMICS OF CHEMICAL REACTIONS**

103
**5. How much useful work can be obtained from an engine fueled with 75**

5. How much useful work can be obtained from an engine fueled with 75.0 L of hydrogen at 10 C at 25 atm?

104
**reacting silver with water.**

6. The reaction to split water into hydrogen and oxygen can be promoted by first reacting silver with water. 2 Ag(s) + H2O(g) Ag2O(s) + H2(g) Ag2O(s) 2 Ag(s) + 1/2 O2(g)

105
**Calculate H0, S0 and G0 for each reaction.**

106
**Combine the reactions and calculate H0 and G0 for the combination.**

Is the combination spontaneous?

107
**At what temperature does the second reaction become spontaneous?**

108
**7. The conversion of coal into hydrogen for fuel is:**

C(s) + H2O(g) CO(g) + H2(g)

109
Calculate G0 and Kp at 250C. Is the reaction spontaneous?

110
**At what temperature does the reaction become spontaneous?**

Calculate the temperature at which K = 1.0 x 10-4.

111
8. The generation of nitric acid in the upper atmosphere might destroy the ozone layer by: NO(g) + O3(g) NO2(g) + O2(g)

112
**Calculate G0 (reaction)**

and K at 250C.

Similar presentations

OK

Ch. 19: Chemical Thermodynamics (Thermochemistry II) Chemical thermodynamics is concerned with energy relationships in chemical reactions. - We consider.

Ch. 19: Chemical Thermodynamics (Thermochemistry II) Chemical thermodynamics is concerned with energy relationships in chemical reactions. - We consider.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on law against child marriages Ppt on earthquake in bhuj Research presentation ppt on caterpillar Free download ppt on flow of control in c++ Ppt on product specification for food Ppt on diodes Ppt on blind stick with sensors Ppt on solar power system Ppt on census and sample methods of collection of data Ppt on private labels in india