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CHEM 160 General Chemistry II Lecture Presentation Chemical Thermodynamics. Chapter 19. Thermodynamics. Thermodynamics study of energy and its transformations heat and energy flow Helps determine natural direction of reactions

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thermodynamics
Thermodynamics
  • Thermodynamics
    • study of energy and its transformations
      • heat and energy flow
    • Helps determine natural direction of reactions
      • Allows us to predict if a chemical process will occur under a given set of conditions
  • Organized around three fundamental laws of nature
    • 1st law of thermodynamics
    • 2nd law of thermodynamics
    • 3rd law of thermodynamics

Chapter 19

first law of thermodynamics
First Law of Thermodynamics
  • 1st Law:
    • Energy can be neither created nor destroyed
      • Energy of the universe is constant
  • Important concepts from thermochemistry
    • Enthalpy
    • Hess’s law
  • Purpose of 1st Law
    • Energy bookkeeping
      • How much energy?
      • Exothermic or endothernic?
      • What type of energy?

Chapter 19

spontaneous vs nonspontaneous
Spontaneous vs Nonspontaneous
  • Spontaneous process
    • Occurs without outside assistance in the form of energy (occurs naturally)
      • Product-favored at equilibrium
      • May be fast or slow
      • May be influenced by temperature
  • Nonspontaneous process
    • Does not occur without outside assistance
      • Reactant-favored at equilibrium
  • All processes which are spontaneous in one direction cannot be spontaneous in the reverse direction
    • Spontaneous processes have a definite direction
    • Spontaneous processes are irreversible

Chapter 19

spontaneous vs nonspontaneous5
Spontaneous vs. Nonspontaneous
  • Spontaneous Processes
    • Gases expand into larger volumes
    • H2O(s) melts above 0C
    • H2O(l) freezes below 0C
    • NH4NO3 dissolves spontaneously in H2O
    • Steel (iron) rusts in presence of O2 and H2O
    • Wood burns to form CO2 and H2O
    • CH4 gas burns to form CO2 and H2O
  • Reverse processes are not spontaneous!
    • nonspontaneous

Chapter 19

spontaneous vs nonspontaneous6
Spontaneous vs. Nonspontaneous
  • Spontaneous Processes
    • H2O(s) melts above 0C (endothermic)
    • NH4NO3 dissolves spontaneously in H2O (endothermic)
    • Steel (iron) rusts in presence of O2 and H2O (exothermic)
    • Wood burns to form CO2 and H2O (exothermic)
    • CH4 gas burns to form CO2 and H2O (exothermic)
  • Heat change alone is not enough to predict spontaneity because energy is conserved
    • Many, but not all, spontaneous processes are exothermic

Chapter 19

factors that favor spontaneity
Factors That Favor Spontaneity
  • Two thermodynamic properties of a system are considered when determining spontaneity:
    • Enthalpy, H
      • Many, but not all, spontaneous processes tend to be exothermic as already noted
    • Entropy, S (J/K)
      • Measure of the disorder of a system
      • Many, but not all, spontaneous processes tend to increase disorder of the system

Chapter 19

entropy
Entropy
  • Entropy, S (J/K)
    • describes # of ways the particles in a system can be arranged in a given state
      • More arrangements = greater entropy
    • S = heat change/T = q/T
    • S = Sfinal - Sinitial
      •  S > 0 represents increased randomness or disorder

Chapter 19

patterns of entropy change
Patterns of Entropy Change
  • For the same or similar substances: Ssolid < Sliquid < Sgas

solid

vapor

Chapter 19

liquid

patterns of entropy change10
Patterns of Entropy Change

Particles farther apart, occupy larger volume of space; even more positions available to particles

Rigidly held particles; few positions available to particles

Particles free to flow; more positions available for particles

solid

vapor

Chapter 19

liquid

patterns of entropy change11
Patterns of Entropy Change

least ordered

less ordered

most ordered

solid

vapor

Chapter 19

liquid

patterns of entropy change12
Patterns of Entropy Change
  • Solution formation usually leads to increased entropy

particles more disordered

solvent

solute

solution

Chapter 19

patterns of entropy change13
Patterns of Entropy Change
  • Chemical Reactions
    • If more gas molecules produced than consumed, S increases. (Srxn > 0)
    • If only solids, ions and/or liquids involved, S increases if total # particles increases.

Chapter 19

slide14

fewer arrangements possible so lower entropy

more arrangements possible so higher entropy

Chapter 19

patterns of entropy change15
Patterns of Entropy Change
  • Increasing temperature increases entropy

System at T1 (S1)

System at T2 (T2 > T1) (S2 > S1)

Chapter 19

patterns of entropy change16
Patterns of Entropy Change

more energetic molecular motions

less energetic molecular motions

System at T1 (S1)

System at T2 (T2 > T1) (S2 > S1)

Chapter 19

third law of thermodynamics
Third Law of Thermodynamics
  • If increasing T increases S, then the opposite should be true also.
    • Is it possible to decrease T to the point that S is zero?
    • At what T does S = 0?
    • If entropy is zero, what does that mean?

Chapter 19

third law of thermodynamics18
Third Law of Thermodynamics
  • 3rd Law of Thermodynamics
    • Entropy of a perfect crystalline substance at 0 K is zero
      • No entropy = highest order possible
      • Why?
  • Purpose of 3rd Law
    • Allows S to be measured for substances
      • S = 0 at 0 K
    • S = heat change/temperature = q/T
      • S = standard molar entropy

Chapter 19

slide19

50

40

Standard entropy,S° (J/K)

30

20

10

0

50

100

150

200

250

300

Chapter 19

Temperature (K)

slide20

50

Gas

Liquid

40

Standard entropy,S° (J/K)

30

Solid

20

10

0

50

100

150

200

250

300

Chapter 19

Temperature (K)

entropy versus probability
Entropy versus Probability
  • Systems tend to move spontaneously towards increased entropy. Why?
    • Entropy is related to probability
    • Disordered states are more probable than ordered states
  • S = k(lnW)
    • k (Boltzman’s constant) = 1.38 x 10-23 J/K
    • W = # possible arrangements in system

Chapter 19

slide23

Isothermal Gas Expansion

Consider why gases tend to isothermally expand into larger volumes.

Gas Container = two bulbed flask

Ordered State

Gas Molecules

Chapter 19

slide24

Isothermal Gas Expansion

Gas Container

Ordered State

S = k(ln 1) = (1.38 x 10-23 J/K)(0) = 0 J/K

Chapter 19

slide32

Disordered States

More probable that the gas molecules will disperse between two halves than remain on one side

Chapter 19

slide33

Disordered States

Driving force for expansion is entropy (probability); gas molecules have a tendency to spread out

Chapter 19

slide34

Disordered States

S = k(ln 7) = (1.38 x 10-23 J/K)(1.95) = 2.7 x 10-23 J/K

Chapter 19

slide35

Total Arrangements

Stotal = k(ln 23) = k(ln 8) = (1.38 x 10-23 J/K)(1.79) = 2.9 x 10-23 J/K

Chapter 19

2nd law of thermodynamics
2nd Law of Thermodynamics
  • 2nd Law of Thermodynamics
    • The entropy of the universe increases in any spontaneous process (Suniv > 0)
    • Increased disorder in the universe is the driving force for spontaneity

Chapter 19

2nd law of thermodynamics37
2nd Law of Thermodynamics
  • 2nd Law of Thermodynamics
    • Suniv = Ssysyem + Ssurroundings
    • Suniv > 0, process is spontaneous
    • Suniv = 0, process at equilibrium
    • Suniv < 0, process is nonspontaneous
      • Process is spontaneous in reverse direction

Chapter 19

2nd law of thermodynamics38
2nd Law of Thermodynamics
  • Spontaneous Processes
    • H2O(s) melts above 0C (endothermic)
    • Steel (iron) rusts in presence of O2 and H2O (exothermic)
      • 4Fe(s) + 3O2(g)  2Fe2O3(s)
    • CH4 gas burns to form CO2 and H2O (exothermic)
      • CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
  • Each process increases Suniverse.

Chapter 19

2nd law of thermodynamics39
2nd Law of Thermodynamics
  • To determine Suniv for a process, both Ssystem and Ssurroundings need to be known:
    • Ssysyem
      • related to matter dispersal in system
    • Ssurroundings
      • determined by heat exchange between system and surroundings and T at which it occurs
        • Sign of Ssurroundings depends on whether process in system is endothermic or exothermic
          • Why?
        • Magnitude of Ssurroundings depends on T
      • Ssurroundings = -Hsystem/T

Chapter 19

entropy changes in a system
Entropy Changes in a System

For any reaction:

S°rxn = nS°(products) - mS°(reactants)

Where n and m are stoichiometric coefficients

Chapter 19

example 1 1 on example problem handout
Example 1(1 on Example Problem Handout)

Using standard molar entropies, calculate S°rxn for the following reaction at 25°C:

2SO2(g) + O2(g) --> 2SO3(g)

S° = 248.1 205.1 256.6 (J · K-1mol-1)

(Ans.: -187.9 J/K)

Chapter 19

gibbs free energy g
Gibbs Free Energy, G

2nd law: Suniv = Ssys + Ssurr = Ssys - Hsys/T

Rearrange (multiply by -T)

-TSuniv = -TSsys + Hsys = Hsys - TSsys

-TSuniv = G (-TSuniv = G)

G = Gibbs free energy (J or kJ)

G = H - TS

and

G = Hsys - TSsys

G° = H°sys - TS°sy (if at standard state)

Chapter 19

gibbs free energy
Gibbs Free Energy
  • Summary of Conditions for Spontaneity
    • G < 0
    • G > 0
    • G = 0

Chapter 19

gibbs free energy44
Gibbs Free Energy
  • Summary of Conditions for Spontaneity
    • G < 0
      • reaction is spontaneous in the forward direction
        • (Suniv > 0)
    • G > 0
    • G = 0

Chapter 19

gibbs free energy45
Gibbs Free Energy
  • Summary of Conditions for Spontaneity
    • G < 0
      • reaction is spontaneous in the forward direction
        • (Suniv > 0)
    • G > 0
      • reaction is nonspontaneous in the forward direction
        • (Suniv < 0)
    • G = 0

Chapter 19

gibbs free energy46
Gibbs Free Energy
  • Summary of Conditions for Spontaneity
    • G < 0
      • reaction is spontaneous in the forward direction
        • (Suniv > 0)
    • G > 0
      • reaction is nonspontaneous in the forward direction
        • (Suniv < 0)
    • G = 0
      • system is at equilibrium
        • (Suniv = 0)

Chapter 19

example 2 2 in example problem handout
Example 2(2 in Example Problem Handout)

For a particular reaction, Hrxn = 53 kJ and Srxn = 115 J/K. Is this process spontaneous a) at 25°C, and b) at 250°C? (c) At what temperature does Grxn = 0?

(ans.: a)G = 18.7 kJ, nonspontaneous; b) –7.1 kJ, spontaneous; c) 460.9 K or 188C)

Chapter 19

gibbs free energy and temperature
Gibbs Free Energy and Temperature
  • G = H - TS
    • S > 0 H < 0
      • Spontaneous at all T
    • S < 0 H > 0
      • Not spontaneous at any T
    • S > 0 H > 0
      • Spontaneous at high T; nonspontaneous at low T
    • S < 0 H < 0
      • Spontaneous at low T; nonspontaneous at high T

Chapter 19

gibbs free energy and temperature49
Gibbs Free Energy and Temperature
  • Spontaneous Processes
    • H2O(s) melts above 0C (endothermic)
      • H > 0, S > 0
    • Steel (iron) rusts in presence of O2 and H2O at 25 C (exothermic)
      • 4Fe(s) + 3O2(g)  2Fe2O3(s)
      • H < 0, S < 0
  • G < 0 for each process
    • T determines sign of G: G = H -TS

Chapter 19

calculating free energy changes
Calculating Free Energy Changes

For any reaction:

G°rxn = Gf°(products) - mGf°(reactants)

Where n and m are stoichiometric coefficients and Gf° = standard free energy of formation

Chapter 19

example 3 3 on example problem handout
Example 3(3 on Example Problem Handout)

Calculate the standard free energy change for the reaction at 25°C:

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)

G°f = -50.8 0 -394.4 -237.4 (kJ/mol)

Chapter 19

gibbs free energy52
Gibbs Free Energy
  • What does Gibbs free energy represent?
    • For a spontaneous process:
      • Maximum amount of energy released by the system that can do useful work on the surroundings
      • Energy available from spontaneous process that can be used to drive nonspontaneous process
    • For a nonspontaneous process:
      • Minimum amount of work that must be done to force the process to occur
      • In actuality, Gspont > Gnonspont

Chapter 19

gibbs free energy53
Gibbs Free Energy
  • Conversion of rust to iron
    • 2Fe2O3 4Fe + 3O2 G = 1487 kJ (NS)
  • To convert iron to rust, G must be provided from spontaneous rxn
    • 2Fe2O3 4Fe + 3O2 G = 1487 kJ (NS)
    • 6CO + 3O2 6CO2G = -1543 kJ (S)

Chapter 19

gibbs free energy54
Gibbs Free Energy
  • Conversion of rust to iron
    • 2Fe2O3 4Fe + 3O2 G = 1487 kJ (NS)
  • To convert iron to rust, G must be provided from spontaneous rxn
    • 2Fe2O3 4Fe + 3O2 G = 1487 kJ (NS)
    • 6CO + 3O2 6CO2G = -1543 kJ (S)
    • 2Fe2O3 + 6CO  4Fe + 6CO2  G = -56 kJ (S)
      • Reactions are “coupled”

Chapter 19

gibbs free energy55
Gibbs Free Energy
  • Many biological rxns essential for life are NS
    • Spontaeous rxns used to “drive” the NS biological rxns
  • Example: photosynthesis
    • 6CO2 + 6H2O  C6H12O6 + 6O2  G > 0
    • What spontaneous rxns drive photosynthesis?

Chapter 19

slide56

sun

big ball of G

= G released

high free energy

DGsurr < 0

proteins, cells etc., lower S

C6H12O6, O2

ATP

NS

Sp

NS

Sp

NS

CO2, H2O

amino acids, sugars, etc., higher S

ADP

photosynthesis

solar nuclear reactions

low free energy

Chapter 19

free energy and equilibrium
Free Energy and Equilibrium
  • G (Nonstandard State)
      • G = G° + RTlnQ
        • R = gas constant (8.314 J/K·mol)
        • T = temperature (K)
        • Q = reaction quotient

Chapter 19

example 4 4 on example problem handout
Example 4(4 on Example Problem Handout)

Calculate Grxn for the reaction below:

2A(aq) + B(aq)  C(aq) + D(g)

if G°rxn = 9.9 x 103 J/mol and (a) [A] = 0.8 M, [B] = 0.5 M, [C] = 0.05 M, and PD = 0.05 atm, and (b) [A] = 0.1 M, [B] = 1 M, [C] = 0.5 M, and PD = 0.5 atm. Is the reaction spontaneous under these conditions?

(ans.: a) –2121 J, spontaneous; b) 17875 J, nonspontaneous)

Chapter 19

free energy and equilibrium59
Free Energy and Equilibrium
  • At equilibrium:
    • Grxn = 0 and Q = K
      • 0 = G°rxn + RTlnK
      • G°rxn = -RTlnK

Chapter 19

example 5 5 on example problem handout
Example 5(5 on Example Problem Handout)

Calculate G°rxn for the ionization of acetic acid, HC2H3O2 (Ka = 1.8 x 10-5) at 25°C. Is this reaction spontaneous under standard state conditions?

(ans.: 27 kJ)

Chapter 19

example 6 6 on example problem handout
Example 6(6 on Example Problem Handout)

Calculate G° for the neutralization of a strong acid with a strong base at 25°C. Is this process spontaneous under these conditions?

For the reaction below, K = 1.0 x 1014.

H+ + OH- H2O

(ans.: -80 kJ)

Chapter 19

example 7
Example 7
  • Calculate Keq for a reaction at (a) 25°C and (b) 250°C if H°rxn = 42.0 kJ and S°rxn = 125 J/K. At which temperature is this process product favored?

(ans.: a) 0.15; b) 216; 250C)

Chapter 19

slide63

equilibrium position

Pure reactants

Pure products

Spontaneous Reaction

DGrxn < 0

Gproducts

Greactants

extent of reaction

Chapter 19

slide64

Spontaneous Reaction

At equilibrium, any change requires an uphill climb in energy, DGrxn > 0

Gproducts

Greactants

Pure reactants

Pure products

extent of reaction

Chapter 19

calculating g for processes
Calculating G for Processes
  • G may be calculated in one of several ways depending on the information known about the process of interest
    • G = H - TS; G° = H° - TS°
    • G°rxn = Gf°(products) - mGf°(reactants)
    • G = G° + RTlnQ
    • G°rxn = -RTlnK

Chapter 19

entropy and life processes
Entropy and Life Processes
  • If the 2nd law is valid, how is the existence of highly-ordered, sophisticated life forms possible?
    • growth of a complex life form represents an increase in order (less randomness)
      • lower entropy

Chapter 19

entropy and life processes67
Entropy and Life Processes

Organisms “pay” for their increased order by increasing Ssurr.

Over lifetime, Suniv > 0.

CO2

H2O

heat

Chapter 19