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Entropy, Free Energy, and Equilibrium

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Thermodynamics
Entropy, Free Energy, and Equilibrium
Chapter 19
_______________ 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
Refresh your memory: What is enthalpy?
Enthalpy is a thermodynamic property that describes heat
changes taking place at constant _____________.
Enthalpy is an ___________property—its magnitude depends
on the amount of the substance present.
Enthalpy cannot be determined, so we measure пЃ„H, the
change in enthalpy.
In earlier chapters, we examined пЃ„Hrxn, пЃ„Hsoln, and пЃ„HВєf
Enthalpy is a _____________function…
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)
H+ (aq) + OH- (aq)
H2O (l)
H2O (s)
NH4NO3 (s)
H2O (l)
H2O
NH4+(aq) + NO3- (aq)
пЃ„H0 = - 890.4 kJ
пЃ„H0 = - 56.2 kJ
пЃ„H0 = 6.01 kJ
пЃ„H0 = 25 kJ
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.
______________________________________________
…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.
order
disorder
S
S
пЃ„S = Sf - Si
If the change from initial to final results in an increase in randomness
пЃ„S > 0
Sf > Si
For any substance, the ____________ state is more ordered
than the ____________ state, and the ____________ state
is more ordered than ____________ state
Ssolid < Sliquid << Sgas
H2O (s)
H2O (l)
пЃ„S > 0
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 Вє)
for Some Substances at 25ВєC
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
Processes that
lead to an
______________
in entropy
(пЃ„S > 0)
How does the entropy of a system change
for each of the following processes?
(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 _______________
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:
пЃ„Suniv = пЃ„Ssys + пЃ„Ssurr > 0
Equilibrium process:
пЃ„Suniv = пЃ„Ssys + пЃ„Ssurr = 0
Entropy Changes in the System (пЃ„Ssys)
The ____________ _____________ of rxn
reaction (пЃ„S0 ) is the
entropy change for a reaction carried out at 1 atm and 250C.
aA + bB
cC + dD
пЃ„S0rxn = [ cS0(C) + dS0(D) ]
пЃ„S0rxn = S nS0(products)
—
—
[ aS0(A) + bS0(B) ]
S mS0(reactants)
Sn = the sum of the stoichiometric coefficients of the products
Sm = the sum of the stoichiometric coefficients of the reactants
What does the superscripted zero mean?
Entropy Changes in the System (пЃ„Ssys)
What is the standard entropy change for the following
reaction at 250C? 2CO (g) + O2 (g)
2CO2 (g)
S0(CO) = 197.9 J/K•mol
S0(CO
2)
= 213.6 J/K•mol
Where do we
get these values?
S0(O2) = 205.0 J/K•mol
S0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
S0rxn = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol
Entropy Changes in the System (пЃ„Ssys)
When gases are produced or consumed
•
If a reaction produces more gas molecules than it
consumes, пЃ„S0 > 0.
•
If the total number of gas molecules diminishes,
пЃ„S0 < 0.
•
If there is no net change in the total number of gas
molecules, then S0 may be positive or negative—
BUT пЃ„S0 will be a small number.
Entropy Changes in the System (пЃ„Ssys)
What is the sign of the entropy change for the
following reaction? 2Zn (s) + O2 (g)
2ZnO (s)
Entropy Changes in the System (пЃ„Ssys)
What is the sign of the entropy change for the
following reaction? 2Zn (s) + O2 (g)
2ZnO (s)
Since the total number of
gas molecules goes down,
пЃ„S is negative.
Entropy Changes in the Surroundings (пЃ„Ssurr)
Exothermic Process
пЃ„Ssurr > 0
Endothermic Process
пЃ„Ssurr < 0
You should keep studying this until you understand it!
Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero
at the absolute zero of temperature.
“How can we tell if a reaction is spontaneous?”
You will soon have an answer to this question!
Spontaneous process:
пЃ„Suniv = пЃ„Ssys + пЃ„Ssurr > 0
Substituting —Hsys/T for Ssurr
Multiply by T
Multiply by —1
пЃ„Hsys
Suniv = Ssys —
>0
T
TSuniv = —Hsys + TSsys > 0
—TSuniv = Hsys — TSsys < 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 that Hsys — TSsys is less than zero, then the
process must be spontaneous.
But we can predict spontaneity more directly….
Gibbs Free Energy (G)
G = H — TS
All quantities in the above equation refer to the system
For a constant-temperature process:
The change in Gibbs free energy (пЃ„G)
____________________________________
If пЃ„G is negative,
there is a release of usable energy,
and the reaction is spontaneous!
If пЃ„G is positive, the reaction is not spontaneous!
For a constant-temperature process:
G = Hsys — TSsys
пЃ„G < 0
The reaction is ___________ in the forward direction.
пЃ„G > 0
The reaction is ________________ as written.
The reaction is ___________ in the reverse direction
пЃ„G = 0
The reaction is _____________________ .
The _____________________________________
rxn
is the free-energy change for a reaction
when it occurs under standard-state conditions.
aA + bB
cC + dD
0
пЃ„Grxn
= [cпЃ„G0f (C) + dпЃ„G0f (D) ] - [aпЃ„G0f (A) + bпЃ„G0f (B) ]
0
пЃ„Grxn
= S nпЃ„G0f (products) - S mпЃ„G0f (reactants)
Sn = the sum of the stoichiometric coefficients of the products
Sm = the sum of the stoichiometric coefficients of the reactants
The superscripted zero means ______________________
Conventions for
Standard States
________________________
(пЃ„G0f ) is the free-energy change
that occurs when:
* 1 mole of the compound
* is formed from its elements
* in their standard states.
пЃ„Gf0 of anyf element in ______________________ is zero.
What is the standard free-energy change for the following
reaction at 25 0C?
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
0
пЃ„Grxn
= S nпЃ„G0f (products) - S mпЃ„G0f (reactants)
0
пЃ„Grxn
= [12пЃ„G0f (CO2) + 6пЃ„G0f (H2O)] - [ 2пЃ„G0f (C6H6)]
0
пЃ„Grxn
= [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ
Is this reaction spontaneous at 25 0C?
What is the standard free-energy change for the following
reaction at 25 0C?
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
0
пЃ„Grxn
= S nпЃ„G0f (products) - S mпЃ„G0f (reactants)
0
пЃ„Grxn
= [12пЃ„G0f (CO2) + 6пЃ„G0f (H2O)] - [ 2пЃ„G0f (C6H6)]
0
пЃ„Grxn
= [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ
Is this reaction spontaneous at 25 0C?
пЃ„G0 = -6405 kJ < 0
This reaction is spontaneous
G = H — TS
пѓјIf both пЃ„H and пЃ„S are positive, then пЃ„G will be
negative only when the TпЃ„S term is larger than пЃ„H.
This occurs only when T is large.
пѓјIf пЃ„H is positive and пЃ„S is negative, пЃ„G will always
be positive—regardless of the temperature.
пѓјIf пЃ„H is negative and пЃ„S is positive, then пЃ„G will
always be negative regardless of temperature.
пѓјIf пЃ„H is negative and пЃ„S is negative, then пЃ„G will
be negative only when TпЃ„S is smaller in magnitude
than пЃ„H. This condition is met when T is small.
G = H — TS
Factors Affecting the Sign of G in the Relationship G = H — TS
Temperature and Spontaneity of Chemical Reactions
CaCO3 (s)
CaO (s) + CO2 (g)
At what temperature will this reaction become
spontaneous? Or…at what temperature
does Gº become zero—or negative?
1. Use Eqn 6.16 to calculate пЃ„HВє
2. Use Eqn 18.4 to calculate пЃ„SВє
3. Substitute these into Gº = Hº — TSº
to calculate пЃ„GВє (130.0 kJ)
4. Since this is a large positive number, we set
пЃ„GВє = 0 and solve the equation for T
Temperature and Spontaneity of Chemical Reactions
CaCO3 (s)
CaO (s) + CO2 (g)
“Two points are worth making….”
First, we used пЃ„HВє and пЃ„SВє values at 25ВєC to calculate
changes that occur at much higher temperatures. Both пЃ„HВє
and пЃ„SВє change with temperature, so the пЃ„GВє value will not
be accurate—but it is a helpful, ballpark estimate.
Second, don’t think nothing happens below 835ºC, and then
suddenly, at 835ВєC, CaCO3 begins to decompose.
Look at the graph on the next frame to see what happens.
Temperature and Spontaneity of Chemical Reactions
CaCO3 (s)
пЃ„H0 = 177.8 kJ
пЃ„S0 = 160.5 J/K
G0 = H0 – TS0
At 25 0C, пЃ„G0 = 130.0 kJ
пЃ„G0 = 0 at 835 0C
CaO (s) + CO2 (g)
The condition пЃ„G = 0 applies to any phase transition.
At the temperature at which a phase transition occurs,
the system is at equilibrium, so пЃ„G = 0.
Substituting and rearranging,
G = H — TS
0 = H — TS
пЃ„S = пЃ„H/ T
Gibbs Free Energy and Phase Transitions
G0 = 0 = H0 – TS0
H2O (l)
пЃ„S =
H2O (g)
пЃ„H
40.79 kJ
=
T
373 K
= 109 J/K
Gibbs Free Energy and Chemical Equilibrium
пЃ„G0 describes a reaction in standard state conditions
пЃ„G must be used when the reaction is not in
standard state conditions
So we derive this equation: пЃ„G = пЃ„G0 + RT lnQ
R is the gas constant (8.314 J/K•mol)
At ________________
пЃ„G = 0
T is the absolute temperature (K)
Q is the reaction quotient
Q=K
0 = пЃ„G0 + RT lnK
пЃ„G0
= - RT lnK
Kp for gases
Kc for reactions in solution
пЃ„G0 < 0
пЃ„G0 > 0
пЃ„G0 = - RT lnK
Relation between пЃ„G0 and K as Predicted by пЃ„G0 = - RT lnK
Alanine + Glycine
пЃ„G0 = +29 kJ
ATP + H2O + Alanine + Glycine
пЃ„G0 = -2 kJ
Alanylglycine
K<1
ADP + H3PO4 + Alanylglycine
K>1
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