Title: Chapter 19 Entropy and Free Energy
1Chapter 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.
2Introduction
- 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)
3Thermodynamics
- 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
5SpontaneousProcesses
- 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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7Identify 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
8Reversible 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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10Reversible 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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12Entropy
- 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 __________.
13Entropy
- 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
14DS 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
15Calculate 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
162nd 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
18Entropy- 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
19Entropy- 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.
20Entropy and probability
Most often case is when energy is distributed
over all particles and to a large number of
states.
21Entropy 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.
22Dispersal of Energy
- Dispersal of matter often contributes to energy
dispersal.
23W 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.
24Explain why melting of ice is spontaneous process
25Explain 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.
27Predict the sign of DS each process occurs at
constant T
- Ag(aq) Cl- (aq) AgCl(s)
- H2O(l) H2O(g)
- N2(g) O2(g) 2 NO (g)
- CO(g) 3 H2(g) CH4(g) H2O(g)
283rd 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.
29Entropy Change
30Standard Molar Entropy Values
31Standard Molar Entropy Values
32Thermodynamics vs Kinetics
Diamond is thermodynamically (favored/not
favored) ___________ to convert to graphite,
and _______ (favored/not favored) kinetically.
33Standard 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.
34Entropy
- The entropy of liquid water is _______ than the
entropy of solid water (ice) at 0 C.
S(H2O sol) ____ S(H2O liq)
35Entropy 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.
36Which 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)
37Entropy 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.
38Calculate the standard entropy changes for the
evaporation of 1.0 mol of liquid ethanol to
ethanol vapor.
C2H5OH(l) ? C2H5OH(g)
39Calculate 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)
40Using standard absolute entropies at 298K,
calculate the entropy change for the system when
2.35 moles of NO(g) react at standard conditions.
41Calculate the standard entropy change for the
oxidation of ethanol vapor (CH2H5OH (g)).
42Show that DS0univ is positive (gt0) for dissolving
NaCl in water
- DS0univ DS0sys DS0surr
- 1) Determine DS0sys
- 2) Determine DS0surr
- NaCl(s) ?NaCl (aq)
43Classify 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
44Calculate 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?
45Gibbs 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
46Gibbs 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.
47Free 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 _______________.
48Calculate DGo for the reaction below at 25.0 ?C.
- P4(s) 6 H2O(l) ? 4 H3PO4(l)
- DG0rx DH0 TDS0
49Standard 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 _________.
50Gibbs 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)
51Calculating 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
52Free 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
53Changes in DG0 with T
54Consider 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
55Estimate 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)
56For 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
57DG0, 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
58DG0, 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.
59Summary 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
___________________.
60The 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
61The 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
62The 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.
63The 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)
64Thermodynamics
- 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 -.
65End 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.