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Chapter 19 Entropy and Free Energy

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Chapter 19 Entropy and Free Energy Objectives: Define entropy and spontaneity. Predict whether a process will be spontaneous. Describe free energy. – PowerPoint PPT presentation

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Title: Chapter 19 Entropy and Free Energy


1
Chapter 19Entropy and Free Energy
  • Objectives
  • Define entropy and spontaneity.
  • Predict whether a process will be spontaneous.
  • Describe free energy.
  • Describe the relationship between DG, K, and
    product favorability.

2
Introduction
  • How to predict if a reaction can occur, given
    enough time?
  • THERMODYNAMICS
  • How far will it proceed? (K)

How to predict if a reaction can occur at a
reasonable rate? KINETICS (Ea)
3
Thermodynamics
  • Energy relationships
  • 1st law of thermodynamics
  • Energy is conserved energy is not created nor
    destroyed
  • DE q w
  • qp DH Enthalpy _______ transferred in a
    process at constant pressure
  • 2nd law of thermodynamics
  • DS Entropy increases in ________ processes
  • Direction of the process

4
  • Lighter reaction
  • C4H8 6 O2 4 CO2 4 H2O
  • Burning candle
  • Drop a pen
  • Gas expansion
  • Heat transfer

non spontaneous
spontaneous
5
SpontaneousProcesses
  • Proceed on its own without outside assistance.
  • Occurs in a define _______
  • Leads to ___________
  • Maybe determined by T, P
  • The reverse process is ____________________

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Identify spontaneous processes
  • Predict whether the following are a) spontaneous
    as described, b) spontaneous in the reverse
    direction, c) in equilibrium
  • When a piece of metal heated to 150oC is added to
    water at 40oC, the water gets hotter.
  • Benzene vapor, C6H6(g), at P 1atm, condenses to
    liquid benzene at the normal boiling point of
    benzene, 80.1oC
  • Water at room temperature decomposes into H2(g)
    and O2(g)
  • AgCl(s) Ag(aq) Cl- (aq) K
    1.8 x 10-10

8
Reversible vs Irreversible Processes
  • Sadi Carnot (French engineer -1824)
  • Efficiency of heat to work (steam engines)
  • Significant amount of heat is lost to
    surroundings
  • Rudolph Clausius (German physicist 1924)
  • Ratio of heat in an ideal engine and temperature
    at which is delivered (q/T)
  • Entropy
  • Amount of work extracted from spontaneous
    processes depends on the manner in which the
    process is carried (pathway)

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Reversible vs Irreversible Processes
  • Reversible a process which reverse direction
    whenever an infinitesimal change is made in some
    property of the system. Often at equilibrium.
  • Here entropy can be obtained at any T by
    measuring the heat required rise the temperature
    from 0K, with the slow addition of heat in very
    small amounts.
  • DS qrev/T
  • Any spontaneous process is Irreversible, they
    often involve non equilibrium conditions in
    order to reverse the surroundings must do some
    work on the system (so the surroundings change).

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12
Entropy
  • Associated with the ___________ in a system
  • Associated with the extent to which energy is
    ___________among the various motions of molecules
    of the system.
  • Related to heat transfer and __________.

13
Entropy
  • S, is a ________ function (like internal energy,
    E, and enthalpy, H).
  • The value of S is characteristic of the state of
    a system (and a property of the bulk matter).
  • DS, change in entropy, depends only on the
    initial and final states of the system, and not
    in the path taken from one state to the other
    DS Sfinal Sinitial
  • For isothermal processes DS qrev/T
  • q heat absorbed/released
  • T temperature in K
  • Adding entropy changes total entropy

14
DS for phase changes
  • Melting of a substance at its m.p. and
    vaporization of a substance at its b.p. are
    isothermal processes.
  • Change can be achieved by adding heat to/from the
    system to/from surroundings.
  • qrev DH fusion (melting)
  • T 273 K (normal m.p. 1 atm and 273K).
  • DS fusion qrev/T DH fusion /T

15
Calculate DS
  • Calculate DSsystem and the DSsurroundings when 1
    mol of ice (size of an ice cube) melts in your
    hand.
  • Process is not reversible (different Ts).
  • DS can be calculated whether rev or irrev.
  • DHfusion H2O 6.01 kJ/mol (melting is
    endothermic processes, DH is positive).
  • DSsystem qrev/T
  • DSsurroundings qrev/T (heat gained -
    heat lost and T 37oC)
  • DStotal DSsystem DSsurroundings
  • Spontaneous or non spontaneous?
  • If Ts were about the same, process would be
    reversible overall DS 0

16
2nd Law of Thermodynamics
  • Any irreversible process results in an overall
    (increase/decrease) ___________ in entropy,
    whereas a reversible process results in no
    overall change in entropy.
  • The sum of entropy of a system entropy of the
    surroundings total entropy change DS
    universe.
  • Irreversible processes occur of their own accord
    are _________ (spontaneous/non spontaneous).
  • The total entropy of the universe increases
    (DSuniverse is __________ (positive/negative)) in
    any spontaneous process.

17
  • DSsystem decreases as Fe O2 form rust
  • DSsurroundings?
  • DStotal DSsystem DSsurroundings

18
Entropy- Molecular Interpretation
  • Ludwig Boltzmann (1844-1906)
  • Molecules store energy
  • KMT
  • Higher T, higher KE and broader distribution of
    molecular speeds
  • 3 kinds of motion translational, vibrational ,
    rotational motional energy of the molecule

19
Entropy- Molecular Interpretation
  • Molecules moving snapshot microstate
  • W number of microstates of a system
  • S k ln W
  • k Boltzmanns constant 1.38 x 10-23 J/K
  • DS k ln Wfinal k lnWinitial k ln
    Wfin/Winit
  • Change leading to increase in number of
    microstates leads to a ________
    (positive/negative) value of DS.
  • Related to probability.

20
Entropy and probability
Most often case is when energy is distributed
over all particles and to a large number of
states.
21
Entropy and probability
Probability and the locations of gas molecules.
The two molecules are colored red and blue. a)
Before the stopcock is opened, both molecules are
in the left-hand flask. b) After the stopcock is
opened, there are 4 possible arrangements of the
two molecules. The greater number of possible
arrangements corresponds to greater disorder in
the system, in general, the probability that the
molecules will stay in the original flask is
(1/2)n, where n is the number of molecules.
22
Dispersal of Energy
  • Dispersal of matter often contributes to energy
    dispersal.

23
W increases when
  • Change involves an
  • Increase in _______
  • Increase in __________
  • Increase in _______________
  • Entropy change, DS sign will be ________
  • The maximum entropy will be achieved at
    ____________ a state in which W has the
    maximum value.

24
Explain why melting of ice is spontaneous process
25
Explain formation of solutions of some ionic
solids
26
  • If energy and matter are both dispersed in a
    process, the process is (spontaneous/non
    spontaneous) _______________.
  • If only matter is dispersed, ___________________
    _______.
  • If energy is NOT dispersed, the process will
    (be/never be) ______________ spontaneous.

27
Predict the sign of DS each process occurs at
constant T
  1. Ag(aq) Cl- (aq) AgCl(s)
  2. H2O(l) H2O(g)
  3. N2(g) O2(g) 2 NO (g)
  4. CO(g) 3 H2(g) CH4(g) H2O(g)

28
3rd Law of Thermodynamics
  • Lower the temperature until there is only a
    single microstate
  • The entropy of a pure crystalline substance at
    absolute zero is ________ S (0K) ____
  • The units of the lattice have no thermal motion.
  • S k ln W
  • As T increases, S ___________
  • S solid is (larger/smaller) than S liquid
    is (larger/smaller) than S gas
  • Then, all substances have ________
    (positive/negative) entropy values at T gt 0K.

29
Entropy Change
30
Standard Molar Entropy Values
31
Standard Molar Entropy Values
32
Thermodynamics vs Kinetics
Diamond is thermodynamically (favored/not
favored) ___________ to convert to graphite,
and _______ (favored/not favored) kinetically.
33
Standard Entropy
  • So, is the entropy gained by converting it from
    a perfect crystal at 0K to standard state
    conditions (1 bar, 1 molal solution).
  • Units J/Kmol
  • Entropies of gases are ________ than those for
    liquids, entropies of liquids are __________than
    those for solids.
  • Larger molecules have a __________ entropy
    than smaller molecules, molecules with more
    complex structures have _______entropies than
    simpler molecules.

34
Entropy
  • The entropy of liquid water is _______ than the
    entropy of solid water (ice) at 0 C.

S(H2O sol) ____ S(H2O liq)
35
Entropy of Solids
  • Entropy values of solids depend on
  • Columbic attractions

So (J/Kmol) MgO 26.9 NaF 51.5
Mg2 O2-
Na F-
The larger coulombic attraction on MgO than NaF
leads to a lower entropy.
36
Which sample has the higher entropy
  • 1 mol NaCl(s) or 1 ml HCl(g) at 25oC
  • b) 2 mol HCl(g) or 1 mol HCl(g) at 25oC
  • c) 1 mol HCl(g) or 1 mol Ar(g) at 298 K
  • d) O2 (g) or O3 (g)

37
Entropy Change
  • The entropy change is the sum of the entropies of
    the products minus the sum of the entropies of
    reactants
  • DS0system S S0 (products) S S0 (reactants)

You will find DSo values in the Appendix L of
your book.
38
Calculate the standard entropy changes for the
evaporation of 1.0 mol of liquid ethanol to
ethanol vapor.
C2H5OH(l) ? C2H5OH(g)
39
Calculate the standard entropy change for
forming 2.0 mol of NH3(g) from N2(g) and H2(g)
N2(g) 3 H2(g) ? 2 NH3 (g)
40
Using standard absolute entropies at 298K,
calculate the entropy change for the system when
2.35 moles of NO(g) react at standard conditions.
  • 2 NO(g) O2(g) ? 2 NO2(g)

41
Calculate the standard entropy change for the
oxidation of ethanol vapor (CH2H5OH (g)).
42
Show that DS0univ is positive (gt0) for dissolving
NaCl in water
  • DS0univ DS0sys DS0surr
  • 1) Determine DS0sys
  • 2) Determine DS0surr
  • NaCl(s) ?NaCl (aq)

43
Classify the following as one of the four types
of Table 19.2

DH0 (kJ) DS0 (J/K) CH4
(g) 2 O2 (g) 2 H2O (l) CO2 (g)
-890 -242.8 2
FeO3(s) 3 C (graphite) 4 Fe(s)
3 CO2 (g) 467 560.7
44
Calculate the entropy change of the UNIVERSE when
1.890 moles of CO2(g) react under standard
conditions at 298.15 K.
  • Consider the reaction6 CO2(g) 6 H2O(l) ?
    C6H12O6 6O2(g)for which DHo 2801 kJ and DSo
    -259.0 J/K at 298.15 K.
  • Is this reaction reactant or product favored
    under standard conditions?

45
Gibbs Free Energy
DSuniv DSsurr DSsys DSsurr -DHsys/T DSuniv
-DHsys/T DSsys Multiply equation by T -T
DSuniv DHsys TDSsys J. Willard Gibbs
(1839-1903) DGsys -T DSuniv DGsys DHsys
TDSsys DGsys lt 0, a reaction is
spontaneous DGsys 0, a reaction is at
equilibrium DGsys gt 0, the reaction is not
spontaneous
46
Gibbs Free Energy and Spontaneity
  • J. Willard Gibbs (1839-1903)
  • Gibbs free energy, G, free energy, a
    thermodynamic function associated with the
    ________________.
  • G H TS
  • H- Enthalpy
  • T- Kelvin temperature
  • S- Entropy
  • Changes during a process DG
  • Use to determine whether a reaction is
    spontaneous.
  • DG is ___________related to the value of the
    equilibrium constant K, and hence to product
    favorability.

47
Free Energy
  • DG w max
  • The free energy represents the maximum energy
    ____________________________.
  • Example C(graphite) 2 H2 (g) CH4
    (g)
  • DH0rx -74.9 kJ DS0rx -80.7 J/K
  • DG0rx DH0 TDS0
  • -74.9 kJ (298)(-80.7)/1000 kJ
  • -74.9 kJ 24.05 kJ
  • DG0rx - 50.85 kJ
  • Some of the energy liberated by the reaction is
    needed to order the system. The energy left is
    energy available energy to do work, free
    energy.
  • DG lt 0, the reaction is _______________.

48
Calculate DGo for the reaction below at 25.0 ?C.
  • P4(s) 6 H2O(l) ? 4 H3PO4(l)
  • DG0rx DH0 TDS0

49
Standard Molar Free Energy of Formation
  • The standard free energy of formation of a
    compound, DG0f, is the free energy change when
    forming one mole of the compound from the
    component elements, with products and reactants
    in their standard states.
  • Then, DG0f of an element in its standard states
    is _________.

50
Gibbs Free Energy
  • DG0rxn is the increase or decrease in free
    energy as the reactants in their standard states
    are converted completely to the products in their
    standard states.
  • Complete reaction is not always observed.
  • Reactions reach an ______________.
  • DG0system S G0 (products) S G0 (reactants)

51
Calculating DG0rxn from DG0f
  • DG0system S G0 (products) S G0
    (reactants)
  • Calculate the standard free energy change for
    the oxidation of 1.0 mol of SO2 (g) to form SO3
    (g).
  • DG0system

DGf0 (kJ/) SO2(g) -300.13 SO3(g) -371.04
52
Free Energy and Temperature
  • G H TS
  • G is a function of T, DG will change as T
    changes.
  • Entropy-favored and enthalpy-disfavored
  • Entropy-disfavored and enthalpy-favored

53
Changes in DG0 with T
54
Consider the reaction below. What is DG0 at 341.4
K and will this reaction be product-favored
spontaneously at this T?
  • CaCO3(s) ? CaO(s) CO2(g)
  • Thermodynamic values
  • DHf0 (kJ/mol) S0 (kJ/Kmol)
  • CaCO3(s) -1206.9 0.0929
  • CaO(s) -635.1 0.0398
  • CO2(g) -393.5 0.2136

55
Estimate the temperature required to decompose
CaSO4(s) into CaO(s) and SO3(g).
Thermodynamic values DHf0 (kJ/mol) S0
(J/Kmol) CaSO4(s) -1434.52 106.50 CaO(s)
-635.09 38.20 SO3(g)
-395.77 256.77
  • CaSO4(s) CaO(s) SO3(g)
  • DH0sys S H0 (products) S H0 (reactants)
  • DH0sys S S0 (products) S S0 (reactants)

56
For the reaction 2H2O(l) ? 2H2(g) O2(g)DGo
460.8 kJ and DHo 571.6 kJ at 339 K and 1 atm.
  • This reaction is (reactant,product) _____________
    favored under standard conditions at 339 K.
  • The entropy change for the reaction of 2.44 moles
    of H2O(l) at this temperature would be
    _________J/K.
  • DGorxn DHorxn - DT Sorxn
  • DSo (DHo - DGo)/T

57
DG0, K, and Product Favorability
  • Large K product favored
  • Small K reactant favored
  • At any point along the reaction, the reactants
    are not under standard conditions.
  • To calculate DG at these points
  • DG DG0 RT ln Q
  • R Universal gas constant
  • T - Temperature (kelvins)
  • Q - Reaction quotient

58
DG0, K, and Product Favorability
  • DG DG0 RT ln Q
  • For a A b B c C d D
  • Q Cc Dd
  • Aa Bb
  • DG of a mixture of reactants and products is
    determined by DG0 and Q.
  • When DG is negative (descending) the reaction
    is _____________ . At equilibrium (no more change
    in concentrations), DG 0.
  • 0 DG0 RT ln K (at equilibrium)
  • DG0 - RT ln K

For DG0 to be negative, K must be larger than 1
and the reation is product favored.
59
Summary DG0 and K
  • The free energy at equilibrium is ________ than
    the free energy of the pure reactants and of the
    pure products.
  • DG0 rxn can be calculated from
  • DG0rxn S G0 (products) S G0 (reactants)
  • DGorxn DHorxn - DT Sorxn
  • DGorxn - RT ln K
  • DGrxn describes the direction in which a
    reaction proceeds to reach ___________, it can be
    calculated from
  • DGrxn DG0rxn RT ln Q
  • When DGrxn lt 0, Q lt K, reaction proceeds
    spontaneously to convert ______________________
    until equilibrium is reached.
  • When DGrxn gt 0, Q gt K, reaction proceeds
    spontaneously to convert ______________________
    until equilibrium is reached.
  • When DGrxn 0 , Q K, reaction is
    ___________________.

60
The formation constant for Ag(NH3)2 is 1.6
x107. Calculate DG0 for the reaction below.
  • Ag (aq) 2 NH3 (aq) ? Ag(NH3)2 (aq)
  • DG0 -RTlnK

61
The reaction below has a DG0 -16.37 kJ/mol.
Calculate the equilibrium constant.
  • 1/2 N2 (g) 3/2 H2 (g) ? NH3 (g)
  • DG0rxn DG0f NH3 (g)
  • DG0 -RTlnK

62
The value of Ksp for AgCl (s) at 25oC is 1.8 x
10-10. Determine DGo for the process Ag
(aq) Cl- (aq) ? AgCl (s) at 25oC.
63
The standard free energy change for a chemical
reaction is -18.3 kJ/mole. What is the
equilibrium constant for the reaction at 87 ?C?
(R 8.314 J/Kmol)
64
Thermodynamics
  • First Law The total energy of the universe is a
    constant.
  • Second Law The total entropy of the universe is
    always increasing.
  • Third Law The entropy of a pure, perfectly
    formed crystalline substance at 0K is zero.
  • - A local decrease in entropy (the assembly of
    large molecules) is offset by an increase in
    entropy in the rest of the universe -.

65
End of Chapter
  • Go over all the contents of your textbook.
  • Practice with examples and with problems at the
    end of the chapter.
  • Practice with OWL tutor.
  • Work on your OWL assignment for Chapter 19.
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