Transcript Thermodynamics: Entropy, Free energy, and Equilibrium

Chapter 16
•
Thermodynamics:
Entropy,
Free Energy,
and Equilibrium
spontaneous
nonspontaneous
In this chapter we will determine the direction of a chemical
reaction and calculate equilibrium constant using
thermodynamic values: Entropy and Enthalpy.
Spontaneous Processes
Spontaneous Process: A process that, once
started, proceeds on its own without a continuous
external influence.
Universe: System + Surroundings
The system is what you observe; surroundings are everything else.
Thermodynamics
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
energy, pressure, volume, temperature
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
Laws of Thermodynamics
1) Energy is neither created nor destroyed.
2) In any spontaneous process the total Entropy of
a system and its surrounding always
increases.
We will prove later:
∆G = ∆Hsys – T∆Ssys
∆G < 0 Reaction is spontaneous in forward
direction
3) The entropy of a perfect crystalline substance is
zero at the absolute zero ( K=0)
Enthalpy, Entropy, and Spontaneous
Processes
State Function: A function or property whose value
depends only on the present state, or condition, of the
system, not on the path used to arrive at that state.
Enthalpy Change (DH): The heat change in a reaction
or process at constant pressure.
Entropy (DS): The amount of Molecular randomness
change in a reaction or process at constant pressure.
Enthalpy Change
Exothermic:
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
DH° = -890.3 kJ
Endothermic:
H2O(s)
H2O(l)
H2O(l)
H2O(g)
N2O4(g)
NaCl(s)
H2O
DHfusion = +6.01 kJ
DHvap = +40.7 kJ
2NO2(g)
DH° = +57.1 kJ
Na1+(aq) + Cl1-(aq)
DH° = +3.88 kJ
Entropy Change
DS = Sfinal - Sinitial
Entropy
Entropy
The sign of entropy change, DS,
associated with the boiling of water is
_______.
1. Positive
2. Negative
3. Zero
Correct Answer:
1. Positive
2. Negative
3. Zero
Vaporization of a liquid
to a gas leads to a large
increase in volume and
hence entropy; DS
must be positive.
Entropy and Temperature
02
Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at the
absolute zero ( K=0)
Ssolid < Sliquid < Sgas
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 negative .
What is the sign of the entropy change for the following
reaction? 2Zn (s) + O2 (g)
2ZnO (s)
The total number of gas molecules goes down, DS is negative.
Standard Molar Entropies and Standard
Entropies of Reaction
Standard Molar Entropy (S°): The entropy of 1 mole
of a pure substance at 1 atm pressure and a specified
temperature.
Calculated by using S = k ln W , W = Accessible microstates
of translational, vibrational and rotational motions
Review of Ch 8:
•∆G = ∆H – T∆S see page 301 of your book
We will show the proof of this formula later in this chapter
•
Calculating DH° for a reaction:
DH° = DH°f (Products) – DH°f (Reactants)
See Appendix B, end of your book
•
For a balanced equation, each heat of formation must
be multiplied by the stoichiometric coefficient.
aA + bB
cC + dD
DH° = [cDH°f (C) + dDH°f (D)] – [aDH°f (A) + bDH°f (B)]
Entropy Changes in the System (DSsys)
0 ) is the
The standard entropy change of reaction
(DS
rxn
entropy change for a reaction carried out at 1 atm and 250C.
aA + bB
DS0rxn =
cC + dD
[ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ]
DS0rxn = S nS0 (products)
- S mS0 (reactants)
What is the standard entropy change for the following
reaction at 250C? 2CO (g) + O2 (g)
2CO2 (g)
S0(CO) = 197.6 J/K•mol
S0(O2) = 205.0 J/K•mol
S0(CO2) = 213.6 J/K•mol
DS0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
DS0rxn = 427.2 – [395.2 + 205.0] = -173.0J/K•mol
Calculating ∆S for a Reaction
∆So = S So (products) - S So (reactants)
Consider 2 H2(g) + O2(g) ---> 2 H2O(liq)
∆So = 2 So (H2O) - [2 So (H2) + So (O2)]
∆So = 2 mol (69.9 J/K•mol) [2 mol (130.6 J/K•mol) +
1 mol (205.0 J/K•mol)]
∆So = -326.4 J/K
Note that there is a decrease in Entropy because 3
mol of gas give 2 mol of liquid.
Calculate DS° for the equation below using the
standard entropy data given:
2 NO(g) + O2(g)  2 NO2(g)
DS° values (J/mol-K): NO2(g) = 240, NO(g) = 211,
O2(g) = 205.
1. +176 J/mol
2. +147 J/mol
3. 147 J/mol
4. 76 J/mol
Correct Answer:
1. +176 J/K
2. +147 J/K
3. 147 J/K
4. 76 J/K
DS = S DS  (products)  S DS  (reactants)
DS° = 2(240)  [2(211) + 205]
DS° = 480  [627]
DS° =  147
Thermite Reaction
Spontaneous Processes
and Entropy
•
07
Consider the gas phase reaction of A2 molecules (red) with
B atoms (blue).
(a) Write a balanced equation for the reaction.
(b) Predict the sign of ∆S for the reaction.
Second Law Of Thermodynamic:
In any spontaneous process the total Entropy of a system
and its surrounding always increases.
Entropy Changes in the Surroundings (DSsurr)
Exothermic Process
DSsurr > 0
Endothermic Process
DSsurr < 0
Entropy and the Second Law of
Thermodynamics
Dssurr a ( - DH)
Dssurr a (
Dssurr =
1
)
T
- DH
T
See example of tossing a rock into a calm waters vs. rough waters
Page 661 of your book
2nd Law of Thermodynamics
•
01
The total entropy increases in a spontaneous
process and remains unchanged in an equilibrium
process.
Spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0
Equilibrium: ∆Stotal = ∆Ssys + ∆Ssur = 0
•
The system is what you observe; surroundings are
everything else.
2nd Law of Thermodynamics
Calculate change of entropy of surrounding in the following reaction:
2 H2(g) + O2(g) ---> 2 H2O(liq) ∆H ° = -571.7 kJ
DSsurr =
DS
D H sys
o
surr oundings
T
=
- (-571.7 kJ)(1000 J/kJ)
298.15 K
∆Sosurroundings = +1917 J/K
Spontaneous Reactions
•
01
The 2nd law tells us a process will be spontaneous
if ∆Stotal > 0 which requires a knowledge of ∆Ssurr.
spontaneous: ∆Stotal = ∆Ssys + ∆Ssur > 0
DS sur =
•∆Stotal
= ∆Ssys + (
D H sys
T
D H sys ) > 0
T

D H sys
-T (∆Stotal = ∆Ssys + (
T
)
)
<0
Gibbs Free Energy
•
02
The expression –T∆Stotal is equated as Gibbs free
energy change (∆G)*, or simply free energy
change:
•
-T
•
• ∆G
•
. ∆S
total
= -T. ∆Ssys + ΔHsys < 0
–T∆Stotal = ΔG
= ∆Hsys – T∆Ssys
∆G < 0 Reaction is spontaneous in forward direction.
∆G = 0 Reaction is at equilibrium.
∆G > 0 Reaction is spontaneous in reverse direction.
* Driving force of a reaction, Maximum work you can get from a system.
Calculate ∆Go rxn for the following:
C2H2(g) + 5/2 O2(g) --> 2 CO2(g) + H2O(g)
Use enthalpies of formation to calculate
∆Horxn = -1256 kJ
Use standard molar entropies to calculate ∆Sorxn ( see page
658, appendix A-10)
∆Sorxn = -97.4 J/K or -0.0974 kJ/K
∆Gorxn = -1256 kJ - (298 K)(-0.0974 kJ/K)
= -1227 kJ
Reaction is product-favored in spite of negative ∆Sorxn.
Reaction is “enthalpy driven”
Calculate DG° for the equation below using
the thermodynamic data given:
N2(g) + 3 H2(g)  2 NH3(g)
DHf° (NH3) = 46 kJ;
1. +33 kJ
2. +66 kJ
3. 66 kJ
4. 33 kJ
Correct Answer:
1. +33 kJ
2. +66 kJ
3. 66 kJ
4. 33 kJ
DH° = 92 kJ
DS° = 198.3J/mol . K
DG° = DH°  TDS°
DG° = (92)  (298)(0.1983)
DG° = (92) + (59.2) = 32.9 KJ
Gibbs Free Energy
03
Using ∆G = ∆H – T∆S, we can predict the sign of ∆G
from the sign of ∆H and ∆S.
1)
2)
If both ∆H and ∆S are positive,
∆G will be negative only when the temperature
value is large.
Therefore, the reaction is spontaneous only at high
temperature.
If ∆H is positive and ∆S is negative,
∆G will always be positive.
Therefore, the reaction is not spontaneous
Gibbs Free Energy:
∆G = ∆H – T∆S
04
3) If ∆H is negative and ∆S is positive,
∆G will always be negative.
Therefore, the reaction is spontaneous
4) If both ∆H and ∆S are negative,
∆G will be negative only when the temperature value is
small.
Therefore, the reaction is spontaneous only at Low
temperatures.
Gibbs Free Energy
04
Gibbs Free Energy
•
06
What are the signs (+, –, or 0) of ∆H, ∆S, and ∆G
for the following spontaneous reaction of A atoms
(red) and B atoms (blue)?
Gibbs Free Energy
•
02
The expression –T∆Stotal is equated as Gibbs free
energy change (∆G)*, or simply free energy
change:
•
-T
•
• ∆G
•
. ∆S
total
= -T. ∆Ssys + ΔHsys < 0
–T∆Stotal = ΔG
= ∆Hsys – T∆Ssys
∆G < 0 Reaction is spontaneous in forward direction.
∆G = 0 Reaction is at equilibrium.
∆G > 0 Reaction is spontaneous in reverse direction.
* Driving force of a reaction, Maximum work you can get from a system.
Standard Free-Energy Changes for
Reactions
Calculate the standard free-energy change at 25 °C for
the Haber synthesis of ammonia using the given values
for the standard enthalpy and standard entropy changes:
N2(g) + 3H2(g)
2NH3(g)
DH° = -92.2 kJ
DS° = -198.7 J/K
DG° = DH° - TDS°
=
-92.2 kJ - 298 K
x
-198.7 J
K
x
1 kJ
1000 J
= -33.0 kJ
Gibbs Free Energy
•
05
Iron metal can be produced by reducing iron(III)
oxide with hydrogen:
Fe2O3(s) + 3 H2(g)  2 Fe(s) + 3 H2O(g)
∆H° = +98.8 kJ; ∆S° = +141.5 J/K
1.
Is this reaction spontaneous at 25°C?
2.
At what temperature will the reaction become
spontaneous?
Decomposition of CaCO3 has a DH° =
178.3 kJ/mol and DS° = 159 J/mol  K.
At what temperature does this become
spontaneous?
1.
2.
3.
4.
121°C
395°C
848°C
1121°C
Correct Answer:
1.
2.
3.
4.
121°C
395°C
848°C
1121°C
T = DH°/DS°
T = 178.3 kJ/mol/0.159 kJ/mol  K
T = 1121 K
T (°C) = 1121 – 273 = 848
Standard Free Energies of Formation
DG° = DG°f (products) - DG°f (reactants)
aA + bB
cC + dD
DG° = [cDG°f (C) + dDG°f (D)] - [aDG°f (A) + bDG°f (B)]
Products
Reactants
Standard free energy of formation (DG0f) is the free-energy
change that occurs when1 mole of the compound is formed
from its elements in their standard ( 1 atm) states.
The standard free-energy of reaction (DG0rxn) is the freeenergy change for a reaction when it occurs under standardstate conditions.
aA + bB
cC + dD
0
DGrxn
= [cDG0f (C) + dDG0f (D) ] - [aDG0f (A) + bDG0f (B) ]
0
DGrxn
= S nDG0f (products) - S mDG0f (reactants)
DG0f of any element in its stable
form is zero.
--
Calculate DG° for the equation below using
the Gibbs free energy data given:
2 SO2(g) + O2(g)  2 SO3(g)
DGf° values (kJ): SO2(g) = 300.4, SO3(g) = 370.4
1. +70 kJ
2. +140 kJ
3. 140 kJ
4. 70 kJ
Correct Answer:
1. +70 kJ
2. +140 kJ
3. 140 kJ
4. 70 kJ
DG ° = DGf (products)   DGf (reactants)
°
°
DG° = (2  370.4)  [(2  300.4) + 0]
DG° = 740.8  [600.8] = 140
Gibbs Free Energy
07
Calculation of Nonstandard ∆G
•
•
The sign of ∆G° tells the direction of spontaneous reaction
when both reactants and products are present at standard
state conditions.
Under nonstandard( condition where pressure is not 1atm or
concentrations of solutions are not 1M) conditions, ∆G˚
becomes ∆G.
∆G = ∆G˚ + RT lnQ
•
The reaction quotient is obtained in the same way as an
equilibrium expression.
Free Energy Changes and the Reaction
Mixture
DG = DG° + RT ln Q
Calculate DG for the formation of ethylene (C2H4) from
carbon and hydrogen at 25 °C when the partial
pressures are 100 atm H2 and 0.10 atm C2H4.
2C(s) + 2H2(g)
C2H4(g)
Qp =
PC H
2 4
2
PH
Calculate ln Qp:
0.10
ln
1002
= -11.51
2
Free Energy Changes and the Reaction
Mixture
Calculate DG:
DG = DG° + RT ln Q
= 68.1 kJ/mol + 8.314
( see page A-13 of your book, page 667)
DG = 39.6 kJ/mol
J
K mol
1 kJ
1000 J
(298 K)(-11.51)
Free Energy and Chemical
Equilibrium
•
At equilibrium ∆G = 0 and Q = K
• ∆G
= ∆G˚ + RT lnQ
∆G˚rxn = –RT ln K
05
Free Energy and Chemical
Equilibrium
Calculate Kp at 25 °C for the following reaction:
CaCO3(s)
CaO(s) + CO2(g)
Calculate DG°:
DG° = [DG°f (CaO(s)) + DG°f (CO2(g))] - [DG°f (CaCO3(s))]
Please see appendix B:
= [(1 mol)(-604.0 kJ/mol) + (1 mol)(-394.4 kJ/mol)]
- [(1 mol)(-1128.8 kJ/mol)]
DG° = +130.4 kJ/mol
Free Energy and Chemical
Equilibrium
Calculate ln K:
DG° = -RT ln K
-130.4 kJ/mol
-DG°
ln K =
=
RT
8.314
J
1 kJ
K mol
1000 J
ln K = -52.63
Calculate K:
-52.63
K=e
= 1.4 x 10-23
(298 K)
Free Energy and Chemical
Equilibrium
∆G˚rxn = –RT ln K
04
Suppose DG° is a large, positive value.
What then will be the value of the
equilibrium constant, K?
1.
2.
3.
4.
K=0
K=1
0<K<1
K>1
Correct Answer:
DG° = RTlnK
1.
2.
3.
4.
K=0
K=1
0<K<1
K>1
Thus, large positive values of DG°
lead to large negative values of lnK.
The value of K itself, then, is very
small.
More thermo?
You betcha!
Thermodynamics and Keq
∆Gorxn = - RT lnK
Calculate K for the reaction
N2O4 --->2 NO2
∆Gorxn = +4.8 kJ/mole ( see page A-11)
ΔG° = RT lnK
∆Gorxn = +4800 J = - (8.31 J/mol.K)(298 K) ln K
ln K = -
4800 J/mole
= - 1.94
(8.31 J/m.K)(298 K)
K = 0.14
When ∆Gorxn > 0, then K < 1