Thermodynamics Entropy, Free Energy, and Equilibrium
Chapter 18 3/e Copyright The McGraw-Hill Companies, Inc. Permission
required for reproduction or display. _______________ Physical and
Chemical Processes
A waterfall runs downhill A lump of sugar dissolves in a cup of
coffee At 1 atm, water freezes below 00C and ice melts above 00C
Heat flows from a hotter object to a colder object A gas expands
into an evacuated bulb Iron exposed to oxygen and water forms rust
spontaneous nonspontaneous 18.2 18.2 Refresh your memory: What is
enthalpy?
Enthalpy is a thermodynamic property that describes heat changes
taking place at constant _____________. Enthalpy is an
___________propertyits magnitude depends on the amount of the
substance present. Enthalpy cannot be determined, so we measure H,
the change in enthalpy. In earlier chapters, we examined Hrxn,
Hsoln, and H Enthalpy is a _____________function f These are all
reactions:
Identify the following reactions as exothermic or endothermic.Does
a decrease in enthalpy mean a reaction proceeds spontaneously?
These are all reactions: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 =
kJ H+ (aq) + OH- (aq) H2O (l) DH0 = kJ H2O (s) H2O (l) DH0 = 6.01
kJ NH4NO3 (s) NH4+(aq) + NO3- (aq) DH0 = 25 kJ H2O 18.2 So we see
that a decrease in enthalpy does not mean a reaction proceeds
spontaneously.
What do we need to know in order to predict if a reaction is
spontaneous?Two things (p.577).
______________________________________________ AND The greater the
disorder or randomness of a system,the
________________________________________. The more ordered a system
is, the less its ___________. Entropy (S) is a measure of the
_____________________ of a system.
disorder S order S DS = Sf - Si If the change from initial to final
results in an increase in randomness Sf > Si DS > 0 For any
substance, the ____________ state is more ordered than the
____________ state, and the ____________ state is more ordered than
____________ state Ssolid < Sliquid 0 18.2 Thermodynamics
__________________ _________________ are properties that are
determined by the state of the system, regardless of how that
condition was achieved. energy, enthalpy, pressure, volume,
temperature , entropy Standard Entropy Values (S ) for Some
Substances at 25C Potential energy of hiker 1 and hiker 2 is the
same even though they took different paths. 18.7 Processes that
lead to an ______________ in entropy (DS > 0)
18.2 Look at page 580, Example 18.1. Lets work through it together.
Then, lets look at the Practice Exercise below it. (a) Condensing
water vapor
How does the entropy of a system change for each of the following
processes? Ex.18.1, p.580 (a) Condensing water vapor Randomness
_______________ Entropy _______________ (b) Forming sucrose
crystals from a supersaturated solution Randomness _______________
Entropy _______________ (c) Heating hydrogen gas from 600C to 800C
Randomness _______________ Entropy _______________ (d) Subliming
dry ice Randomness _______________ Entropy _______________ 18.2
First Law of Thermodynamics
Energy can be converted from one form to another,but energy cannot
be created or destroyed. Second Law of Thermodynamics The entropy
of the universe increases in a spontaneous process and remains
unchanged in an equilibrium process. Spontaneous process: DSuniv =
DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr
= 0 18.3 Entropy Changes in the System (DSsys)
The ____________ _____________ of reaction (DS0) is the entropy
change for a reaction carried out at 1 atm and 250C. rxn aA + bB cC
+ dD DS0 rxn dS0(D) cS0(C) = [ + ] bS0(B) aS0(A) DS0 rxn
nS0(products) = S mS0(reactants) n = the sum of the stoichiometric
coefficients of the products m = the sum of the stoichiometric
coefficients of the reactants What does the superscripted zero
mean? 18.3 Look at the bottom of page 581 and top of page 582,
Example 18.2.
Lets work through it together. Then, lets look at the Practice
Exercise below it. Entropy Changes in the System (DSsys)
What is the standard entropy change for the following reaction at
250C?2CO (g) + O2 (g) CO2 (g) S0(CO) = J/Kmol Where do weget these
values? S0(CO2) = J/Kmol S0(O2) = J/Kmol DS0 rxn = 2 x S0(CO2) [2 x
S0(CO) + S0 (O2)] DS0 rxn = [ ] = J/Kmol 18.3 You should work
through the other parts of this Practice Exercise. Entropy Changes
in the System (DSsys)
When gases are produced or consumed If a reaction produces more gas
molecules than it consumes, DS0 > 0. If the total number of gas
molecules diminishes, DS0 < 0. If there is no net change in the
total number of gas molecules, then DS0 may be positive or
negativeBUT DS0 will be a small number. 18.3 Look at page 582f,
Example 18.3. Lets work through it together. Then, lets look at the
Practice Exercise below it. Entropy Changes in the System
(DSsys)
What is the sign of the entropy change for thefollowing
reaction?2Zn (s) + O2 (g) ZnO (s) Practice Exercise 18.3, part (b),
page 583 18.3 Since the total number of gas molecules goes down, DS
is negative.
Entropy Changes in the System (DSsys) What is the sign of the
entropy change for thefollowing reaction?2Zn (s) + O2 (g) ZnO (s)
Since the total number ofgas molecules goes down, DS is negative.
Practice Exercise 18.3, part (b), page 583 18.3 You should work
through the other parts of this Practice Exercise. Entropy Changes
in the Surroundings (DSsurr)
Exothermic Process DSsurr > 0 Endothermic Process DSsurr < 0
You should keep studying this until you understand it! 18.3 Third
Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at the
absolute zero of temperature. 18.3 How can we tell if a reaction is
spontaneous?
You will soon have an answer to this question! DSuniv = DSsys +
DSsurr > 0
Spontaneous process: DSuniv = DSsys + DSsurr > 0 Substituting
DHsys/T for DSsurr DHsys DSuniv = DSsys > 0 T Multiply by T
TDSuniv = DHsys + TDSsys > 0 Multiply by 1 TDSuniv =DHsys TDSsys
< 0 This says that for a process carried out at constant P and
T, if the changes in enthalpy (H) and entropy (S) of the system are
such thatDHsys TDSsys is less than zero, then the process must be
spontaneous. But we can predict spontaneity more directly. p.586
18.4 Gibbs Free Energy (G) G = H TS For a constant-temperature
process:
All quantities in the above equation refer to the system For a
constant-temperature process: The change in Gibbs free energy (DG)
____________________________________ If DG is negative,there is a
release of usable energy,and the reaction is spontaneous! If DG is
positive, the reaction is not spontaneous! 18.4 For a
constant-temperature process:
DG = DHsys TDSsys DG < The reaction is ___________ in the
forward direction. DG > The reaction is ________________ as
written. The reaction is ___________ in the reverse direction DG =
The reaction is _____________________ . 18.4 n = the sum of the
stoichiometric coefficients of the products
The _____________________________________is the free-energy change
for a reactionwhen it occurs under standard-state conditions. rxn
aA + bB cC + dD DG0 rxn dDG0 (D) f cDG0 (C) = [ + ] - bDG0 (B) aDG0
(A) DG0 rxn nDG0 (products) f = S mDG0 (reactants) - n = the sum of
the stoichiometric coefficients of the products m = the sum of the
stoichiometric coefficients of the reactants The superscripted zero
means ______________________ 18.4 DG0 of any element in
______________________ is zero.
Conventions for Standard States ________________________ (DG0) is
the free-energy changethat occurs when:* 1 mole of the compound* is
formed from its elements* in their standard states. f DG0 of any
element in ______________________ is zero. f f 18.4 Look at page
587f, Example 18.4. Lets work through it together. Then, lets look
at the PracticeExercise below it. 2C6H6 (l) + 15O2 (g) 12CO2 (g) +
6H2O (l)
What is the standard free-energy change for the following reaction
at 25 0C? 2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l) DG0 rxn nDG0
(products) f = S mDG0 (reactants) - DG0 rxn 6DG0 (H2O) f 12DG0
(CO2) = [ + ] - 2DG0 (C6H6) DG0 rxn = [ 12x x237.2 ] [ 2x124.5 ] =
kJ Practice Exercise 18.4 (b), p.588 Is this reaction spontaneous
at 25 0C? 18.4 This reaction is spontaneous
What is the standard free-energy change for the following reaction
at 25 0C? 2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l) DG0 rxn nDG0
(products) f = S mDG0 (reactants) - DG0 rxn 6DG0 (H2O) f 12DG0
(CO2) = [ + ] - 2DG0 (C6H6) DG0 rxn = [ 12x x237.2 ] [ 2x124.5 ] =
kJ Is this reaction spontaneous at 25 0C? DG0 = kJ < 0 This
reaction is spontaneous 18.4 DG = DH TDS If both DH and DS are
positive, then DG will be negative only when the TDS term is larger
than DH.This occurs only when T is large. If DH is positive and DS
is negative, DG will always be positiveregardless of the
temperature. If DH is negative and DS is positive, then DG will
always be negative regardless of temperature. If DH is negative and
DS is negative, then DG will be negative only when TDS is smaller
in magnitude than DH.This condition is met when T is small. p.588
DG = DH TDS Factors Affecting the Sign of DG in the Relationship DG
= DH TDS p.589 18.4 Temperature and Spontaneity of Chemical
Reactions
CaCO3 (s) CaO (s) + CO2 (g) At what temperature will this reaction
become spontaneous?Orat what temperature does DG become zeroor
negative? 1. Use Eqn 6.16 to calculate DH 2. Use Eqn 18.4 to
calculate DS 3. Substitute these into DG = DH TDS to calculate
DG(130.0 kJ) 4. Since this is a large positive number, we set DG =
0 and solve the equation for T Follow along on pp.589f Temperature
and Spontaneity of Chemical Reactions
CaCO3 (s) CaO (s) + CO2 (g) Two points are worth making. [Read
quote p.590.] First, we used DH and DS values at 25C to calculate
changes that occur at much higher temperatures. Both DH and DS
change with temperature, so the DG value will not be accuratebut it
is a helpful, ballpark estimate. Second, dont think nothing happens
below 835C, and then suddenly, at 835C, CaCO3 begins to decompose.
Look at the graph on the next frame to see what happens.
Temperature and Spontaneity of Chemical Reactions
CaCO3 (s) CaO (s) + CO2 (g) DH0 = kJ DS0 = J/K DG0 = DH0 TDS0 At 25
0C, DG0 = kJ DG0 = 0 at 835 0C p.590 18.4 DG = DH TDS 0 = DH TDS DS
= DH/ T
The condition DG = 0 applies to any phase transition. At the
temperature at which a phase transition occurs, the system is at
equilibrium, so DG = 0. Substituting and rearranging, DG = DH TDS 0
= DH TDS DS = DH/ T Gibbs Free Energy and Phase Transitions
DG0 = 0 = DH0 TDS0 H2O (l) H2O (g) DS = T DH = 40.79 kJ 373 K = 109
J/K 18.4 Gibbs Free Energy and Chemical Equilibrium
DG0 describes a reaction in standard state conditions DG must be
used when the reaction is not in standard state conditions So we
derive this equation: DG = DG0 + RT lnQ R is the gas constant
(8.314 J/Kmol) T is the absolute temperature (K) At
________________ Q is the reaction quotient DG = 0 Q = K 0 = DG0 +
RT lnK Kp for gases Kc for reactions in solution DG0 = - RT lnK
pp.592ff 18.4 DG0 < 0 DG0 > 0 p.593 18.4 DG0 = - RT lnK
Relation between DG0 and K as Predicted by DG0 = - RT lnK p.593
18.4 18.5 Alanine + GlycineAlanylglycine
DG0 = +29 kJ K < 1 ATP + H2O + Alanine + Glycine ADP + H3PO4 +
Alanylglycine DG0 = -2 kJ K > 1 18.5