Thermodynamics and Equilibrium - PowerPoint PPT Presentation

1 / 41
About This Presentation
Title:

Thermodynamics and Equilibrium

Description:

Chapter 20 Thermodynamics and Equilibrium – PowerPoint PPT presentation

Number of Views:48
Avg rating:3.0/5.0
Slides: 42
Provided by: pbw121
Category:

less

Transcript and Presenter's Notes

Title: Thermodynamics and Equilibrium


1
Chapter 20
  • Thermodynamics and Equilibrium

2
Overview
  • First Law of Thermodynamics
  • Spontaneous Processes and Entropy
  • Entropy and the Second Law of Thermodynamics
  • Standard Entropies and the Third Law of
    Thermodynamics

3
  • Free Energy Concept
  • Free Energy and Spontaneity
  • Interpretation of Free Energy
  • Free Energy and Equilibrium Constants
  • Relating DG to the Equilibrium Constant
  • Change of Free Energy with Temperature

4
Definitions
  • Spontaneous or Product-favored reaction reaction
    in which most of the reactants can eventually be
    converted to products, given sufficient time
  • Nonspontaneous or Reactant-favored reaction
    misleading - does not mean that it does not occur
    at all, rather, it means that when equilibrium is
    achieved, not many reactant molecules have been
    converted into products.

5
Definitions Continued
  • Thermodynamics the science of energy transfer,
    it helps us to predict whether a reaction can
    occur given enough time. Thermodynamics tells us
    nothing about the speed of the reactions.

6
Reaction Probability
  • After an exothermic reaction, energy is
    distributed more randomly - dispersed over a much
    larger number of atoms and molecules - than it
    was before. Energy dispersal is favored because
    it is much more probable that energy will be
    dispersed than that it will be concentrated.
  • Just as there is a tendency for highly
    concentrated energy to disperse, highly
    concentrated matter also tends to disperse.

7
There are two ways that the final state of a
system can be more probable than the initial one
  • 1. Having energy dispersed over a greater number
    of atoms and molecules and
  • 2. Having the atoms and molecules themselves
    more disordered

8
Entropy A Measure of Matter Dispersal or
Disorder
  • The dispersal or disorder in sample of matter can
    be measured with a calorimeter, the same
    instrument needed to measure the enthalpy change
    when a reaction occurs.
  • The result is a thermodynamic function called
    entropy and symbolized by S.

9
Entropy and the Third Law
  • Measurement of entropy depends on the assumption
    that in a perfect crystal at the absolute zero
    temperature all translational motion ceases and
    there is not any disorder.

10
Calculating Entropy Change
  • When energy is transferred to matter in very
    small increments, so that the temperature change
    is very small, the entropy change can be
    calculated as DS q/T, the heat absorbed divided
    by the absolute temperature at which the change
    occurs.

11
Standard Molar Entropies
  • Applies to one mole of a substance at standard
    pressure. Expressed in units of joules per
    molekelvin.
  • DS Sproducts Sreactants

12
Example 1
  • Calculate the standard molar entropy change for
    the formation of gaseous propane.
  • Soc 6 J/K
  • SoH 131 J/K
  • SoC3H8 207 J/K

13
Generalizations About Entropy
  • 1. When comparing the same or very similar
    substances, entropies of gases are much larger
    than those of liquids, which are larger than for
    solids.
  • 2. Entropies of more complex molecules are
    larger than those of simpler molecules,
    especially in a series of closely related
    compounds.
  • 3. Entropies of ionic solids become larger as
    the attractions among the ions become weaker.
  • 4. Entropy usually increases when a pure liquid
    or solid dissolves in a solution.
  • 5. Entropy increases when a dissolved gas
    escapes from a solution.

14
Example 2
  • Indicate whether the entropy increases or
    decreases in each of the following processes.
  • Moisture condenses on the outside of cold glass
  • Gasoline vaporizes in the carburetor of an
    automobile engine
  • Sugar dissolves in coffee
  • Iron rusts

15
Example 3
  • Predict the sign of DS for the following
  • AgCl(s) ? Ag(aq) Cl-(aq)
  • 2H2(g) O2(g) ? 2H2O(l)
  • H2O(l) ? H2O(g)
  • N2(g) 3H2 (g) ? 2NH3(g)

16
Entropy and the Second Law of Thermodynamics
  • In a product-favored process there is a net
    increase in the entropy of the system and the
    surroundings.
  • In other words when a product favored reaction
    occurs the entropy (disorder) of the universe
    increases. This means that even if the entropy
    of a particular system decreases in a product
    favored process, the total change in the entropy
    of the universe (the system and all its
    surroundings) must be positive.

17
Entropy of the Surroundings
  • The sign of DSsurr depends on the direction of
    heat flow (q).

18
  • A positive DSsurr occurs when heat flows out of
    the system into the surroundings and increases
    thermal motion. This should make sense to you
    because heat is a form of energy and when the
    energy increases the molecular motion increases
    which causes more randomness.

19
  • A negative DSsurr occurs when heat flows into the
    system from the surroundings, decreasing the
    thermal motion of the surroundings and therefore
    decreasing the entropy.

20
DSsurr and absolute temperature
  • If the surroundings are at a high temperature,
    the various types of molecular motion are already
    sufficiently energetic.
  • Therefore, the absorption of heat from an
    exothermic process in the system will have
    relatively little impact on molecular motions and
    the resulting increase in entropy will be small.

21
  • If the temperature of the surroundings is low,
    then the addition of the same amount of heat will
    cause a more drastic increase in molecular
    motions and hence a larger increase in entropy.
  • So, the entropy change produced when a given
    amount of heat is transferred is greater at low
    temperatures than at high temperatures.

22
  • The result is that DSsurr qsurr/T where qsurr
    is the heat flow into the surroundings at the
    absolute (Kelvin) temperature, T.
  • For a constant-pressure processes DH was defined
    as being equal to the qsystem which would be
    equal to -qsurr.
  • So this means that DSsurr -DH/T

23
DS and DH
  • For a product-favored process
  • DSsystem - DH/T gt 0 or
  • TDS - DH gt 0

24
Second Law of Thermodynamics
  • To be product favored a reaction must lead to an
    increase in the entropy of the universe.
  • For a product favored process TDS - DH gt 0 or if
    we multiply the equation throughout by -1 DH -
    TDS lt 0.
  • Now we have a criterion for a product-favored
    reaction that is expressed only in terms of the
    properties of the system and we no longer need be
    concerned with the surroundings.

25
Gibbs Free Energy
  • A new thermodynamic function can now be
    introduced, its called the Gibbs Free Energy, or
    simply Free Energy, as follows
  • G H - TS
  • G has the units of energy and, like H and S, it
    is a state function. (State Function A
    quantity whose value is determined only by the
    state of the system)

26
DG
  • The change in free energy (DG) of a system (which
    if what we're interested in) for a process at
    constant temperature is given by DG DH - TDS

27
Example 4
  • Given the values of DH and DS, which of the
    following changes will be spontaneous at constant
    T and P?
  • DH 25KJ DS 5.0 J/K T 300 K
  • DH 25 KJ DS 100J/K T 300 K
  • DH -10 KJ DS 5.0 J/K T 298 K
  • DH -10 KJ DS -40.0 J/K T 200 K

28
Standard Free Energies of Formation
  • The standard free energy of formation, Gf, of
    a substance is defined similarly to the standard
    enthalpy of formation. That is, DGf is the
    free-energy change that occurs when 1 mol of a
    substance is formed from its elements in their
    most stable states at 1 atm and at a specified
    temperature (usually 25 C)

29
  • By tabulating DGf for substances we can easily
    calculate DG for any reaction involving those
    substances using the following formula
  • DG DGf (products) - DGf (reactants)
  • NOTE If you are confused about the two
    different ways to calculate the DG - this last
    one can only be used when the temperature is that
    of the tabulated values - usually 25 C.

30
Example 5
  • For the following reaction at 298 K, determine
    the value of the Gibbs Free Energy
  • 4PH3(g) 8O2(g) ? P4O10(s) 6H2O(l)
  • DGf PH3 13 KJ/mol
  • DGf O2 0 KJ/mol
  • DGf P4O10 -2698 KJ/mol
  • DGf H2O -237 KJ/mol

31
(No Transcript)
32
Product-Favored or Reactant-Favored?
  • Reactions at constant temperature and pressure go
    in such a direction as to decrease the free
    energy of the system.
  • DG lt 0 Product-Favored Reaction
  • DG gt 0 Reactant-Favored.
  • DG 0 The system is at equilibrium. There
    is no net change.

33
Free Energy and Temperature
  • The value of G and consequently the
    directionality of the reaction change with
    temperature.

34
Free-Energy and Temperature
  • DH DS Low Temperature High Temperature
  • reactant-favored product-favored
  • - reactant-favored reactant-favored
  • - product-favored product-favored
  • - - product-favored reactant-favored

35
?G Temperature Dependence
  • Spontaneity Reaction becomes spontaneous when ?G
    goes from to ?. We use ?G 0 to tell us when
    reaction just becomes spontaneous or 0 ?H ? T?S
    or T ?H/?S.

36
Example 6
  • Determine the temperature at which the synthesis
    of HI(g) becomes spontaneous.
  • ?Ho 52.96 kJ and ?So 166.4 J/mol
  • H2(g) I2(g) ? 2HI(g)

37
EQUILIBRIUM CONSTANTS AND ?G
  • Equilibrium constant for a reaction
  • aA bB ... ? mM nN
  • ... is defined as
  • Tells how far to right reaction proceeds.
  • Large value ? mostly products.
  • Small value ? mostly reactants.
  • At equilibrium this equation must always be
    obeyed no matter what relative amount of reactant
    and was started with.

38
Thermodynamics and the Equilibrium Constant
  • DG DGo RT lnQ
  • At equilibrium Q K and DG 0
  • DGo -RT lnK where R is 8.31 J/mol K

39
Example 7
  • Chloroform, formerly used as an anesthetic, and
    now believed to be a carcinogen has a heat of
    vaporization of 31.4 KJ/mol. The entropy of
    vaporization is 94.2 J/molK. At what
    temperature would you expect chloroform to boil
    (i.e. at what temperature will the liquid and
    vapor be in equilibrium)?

40
Thermodynamics
  • First Law The total energy of the universe is
    constant
  • Second Law The total entropy of the universe is
    always increasing
  • Third Law The entropy of a pure, perfectly
    formed crystalline substance at absolute zero is
    zero

41
  • Neither of the first two laws of thermodynamics
    has ever been or can be proven. However, there
    has never been a single, concrete example showing
    otherwise.
Write a Comment
User Comments (0)
About PowerShow.com