Transcript THERMODYNAMICS - University of the Witwatersrand

THERMODYNAMICS
Chapter 19
SPONTANEOUS PROCESS
 A process that occurs without ongoing
outside intervention.
Examples
• Nails rusting outdoors
• Ice melting at room temperature
• Expansion of gas into an evacuated space
• Formation of water from O2(g) and H2(g):
2H2(g) + O2(g)
2H2O(g)
Why are some processes spontaneous and
others not?
We know that temperature has an effect on the
spontaneity of a process.
e.g. T>0oC  ice melts
 spontaneous at this temp.
H2O (s)  H2O (l)
T<0oC  water freezes
 spontaneous at this temp.
H2O (l)  H2O (s)
T=0oC  water and ice in equilibrium
H2O (l)
H2O (s)
Exothermic processes tend to be spontaneous.
Example
Rusting of nail - SPONTANEOUS!
4Fe(s) + 3O2(g)  Fe2O3(s)
H = - 822.2 kJ.mol-1
Formation of water - SPONTANEOUS!
2H2(g) + O2(g) Pt cat 2H2O(l)
H = - 285.8 kJ.mol-1
However, the dissolution of ammonium
nitrate is also spontaneous, but it is also
endothermic.
NH4NO3(s)  NH4+(aq) + NO3-(aq)
H = +25.7 kJ.mol-1
So is:
2N2O5(s)  4NO2(g) + 2O2(g)
H = +109.5 kJ.mol-1
 a process does not have to be exothermic
to be spontaneous.
 something else besides sign of H must
contribute to determining whether a process is
spontaneous or not.
That something else is:
ENTROPY (S)
 extent of disorder!
More disordered  larger entropy
Entropy is a state function
S = Sfinal - Sinitial
Units: J K-1mol-1
Entropy’s effects on the mind
Examples of spontaneous processes where
entropy increases:
Dissolution of ammonium nitrate:
NH4NO3(s)  NH4+(aq) + NO3-(aq)
H = +25.7 kJ.mol-1
Decomposition of dinitrogen pentoxide:
2N2O5(s)  4NO2(g) + 2O2(g)
H = +109.5 kJ.mol-1
However, entropy does not always increase for
a spontaneous process
At room temperature:
Spontaneous
Non-spontaneous
SECOND LAW OF THERMODYNAMICS
The entropy of the universe increases in any
spontaneous process.
Suniverse = Ssystem + Ssurroundings
Spontaneous process: Suniverse > 0
Process at equilibrium: Suniverse = 0
Thus Suniv is continually increasing!
Suniv must increase during a spontaneous
process, even if Ssyst decreases.
Q: What is the connection between sausages and the
second law of thermo?
A: Because of the 2nd law, you can put a
pig into a machine and get sausage, but
you can't put sausage into the machine
and get the pig back.
Suniv>0
For example:
Rusting nail = spontaneous process
4Fe(s) + 3O2(g)  2Fe2O3(s)
Ssyst<0
BUT reaction is exothermic,
 entropy of surroundings increases as heat is
evolved by the system thereby increasing
motion of molecules in the surroundings.
 Ssurr>0
+ve
-ve
+ve
Thus for Suniv = Ssyst + Ssurr >0
Ssurr > Ssyst
Special circumstance = Isolated system:
Does not exchange energy nor matter with
surroundings
Ssurr = 0
Spontaneous process: Ssyst > 0
Process at equilibrium: Ssyst = 0
Spontaneous  Thermodynamically favourable
(Not necessarily occur at observable rate.)
Thermodynamics  direction and extent of
reaction, not speed.
EXAMPLE
State whether the processes below are
spontaneous, non-spontaneous or in equilibrium:
•CO2 decomposes to form diamond and O2(g)
NON-SPONTANEOUS
•Water boiling at 100oC to produce steam in a
closed container
EQUILIBRIUM
•Sodium chloride dissolves in water
SPONTANEOUS
MOLECULAR INTEPRETATION OF S
Decrease in number of gaseous molecules
 decrease in S
e.g. 2NO(g) + O2(g)  2NO2(g)
3 moles gas
2 moles gas
Molecules have 3 types of motion:
Translational motion - Entire molecule moves in
a direction (gas > liquid > solid)
Vibrational motion – within a molecule
Rotational motion – “spinning”
Greater the number of degrees of freedom
 greater entropy
Decrease in temperature
 decrease in thermal energy
 decrease in translational, vibrational
and rotational motion
 decrease in entropy
As the temperature keeps decreasing, these
motions “shut down”  reaches a point of
perfect order.
EXAMPLE
Which substance has the great entropy in each
pair? Explain.
• C2H5OH(l) or C2H5OH(g)
• 2 moles of NO(g) or 1.5 moles of NO(g)
• 1 mole O2(g) at STP or 1 mole NO2(g) at STP
THIRD LAW OF THERMODYNAMICS
The entropy of a pure crystalline substance at
absolute zero is zero.
S(0 K) = 0  perfect order
Ginsberg's Theorem
(The modern statement of the three laws of
thermodynamics)



1. You can't win.
2. You can't even break even.
3. You can't get out of the game.
Entropy increases for s  l  g
EXAMPLE
Predict whether the entropy change of the
system in each reaction is positive or negative.
• CaCO3(s)  CaO(s) + CO2(g)
+ve
• 2SO2(g) + O2(g)  2SO3(g) -ve
3 mol gas  2 mol gas
• N2(g) + O2(g)  2NO(g) ?
2 mol gas  2 mol gas
Can’t predict, but it
is close to zero
•H2O(l) at 25oC  H2O(l) at 55oC +ve
Increase thermal energy
Standard molar entropy (So)
= molar entropy for substances in their
standard state
NOTE
• So  0 for elements in their standard state
• So(gas) > So(liquid) > So(solid)
• So generally increases with increasing molar
mass
• So generally increases with increasing
number of atoms in the formula of the
substance
Calculation of S for a reaction
S  nS (products) mS (reactants)
o
o
Stoichiometric coefficients
(So from tabulated data)
o
EXAMPLE
Calculate So for the synthesis of ammonia
from N2(g) and H2(g):
N2(g) + 3H2(g)  2NH3(g)
So/J.K-1.mol-1
N2(g)
191.5
H2(g)
130.6
NH3(g)
192.5
Calculation of S for the surroundings
For a process that occurs at constant
temperature and pressure, the entropy change
of the surroundings is:
Ssurr 
- Hsys
T
(T &P constant)
GIBB’S FREE ENERGY (G)
Defined as: G = H – TS
- state function
- extensive property
Suniv = Ssys + Ssurr
At constant T and P:
Hsys
Suniv = Ssys T
 - TSuniv = - TSsys + Hsys
G = H – TS
(at constant T & P)
We know:
Spontaneous process: Suniv > 0
Process at equilibrium: Suniv = 0
Therefore:
Spontaneous process:
-TSuniv < 0
Process at equilibrium:
-TSuniv = 0
Gsyst = -TSuniv
G = H – TS
Spontaneity involves
S
T
H
Spontaneity is favoured by increasing S and
H is large and negative.
G allows us to predict whether a process is
spontaneous or not (under constant
temperature and pressure conditions):
G < 0  spontaneous in forward direction
G > 0  non-spontaneous in forward
direction/spontaneous in reverse
direction
G = 0  at equilibrium
But nothing about rate
Standard free energy (Go)
Go = Ho – TSo
Standard states:
Gas
Solid
Liquid
Solution -
1 atm
pure substance
pure liquid
Concentration = 1M
Gfo = 0 kJ/mol for elements in their standard
states
Tabulated data of Gfo can be used to calculate
standard free energy change for a reaction as
follows:
Go   nGof (products)   mGof (reactants )
Stoichiometric coefficients
Substance
AgS
AgCl(s)
Al(s)
H
(kJ mol-1)
0
127.1
0
AlCl3(s)
-704.2
Al2O3(s)
-1669.8
Br2()
0
S
(J K-1 mol-1)
+42.6
+96.2
28.32
G
(kJ mol-1)
0
-109.8
Substance
H
(kJ mol-1)
S
(J K-1 mol-1)
G
(kJ mol-1)
I2(s)
0
+116.1
0
I2(g)
+62.4
+260.6
+19.4
0
MgO(s)
-601.5
+27.0
-569.2
+110.7
-628.8
MnO2(s)
-520.0
+53.1
-465.2
+51.0
-1576.5
+152.2
0
N2(g)
N2O4(g)
0
+191.5
0
+9.3
+304.2
+97.8
0
+51.3
0
BrF3(g)
-255.6
+292.4
-229.5
Na(s)
C(g)
+716.7
+158.0
+671.3
NaF(s)
-569.0
+51.3
-546.3
C(graphite)
0
+5.8
0
NaCl(s)
-411.1
+72.4
-384.3
C(diamond)
+1.9
+2.4
+2.9
NaBr(s)
-361.1
+87.2
-349.1
CO(g)
-110.5
+197.6
-137.2
NaI(s)
-287.8
+98.5
-282.4
CO2(g)
-393.5
+213.7
-394.4
NaOH(s)
-425.6
+64.5
-379.5
CH4(g)
-74.5
+186.1
-50.8
NH3(g)
-46.2
+192.7
-16.4
C3H8(g)
-103.8
+269.9
-23.4
N2H4()
+50.6
+121.2
+149.2
NO(g)
+90.3
+210.6
+86.6
+33.2
+240.0
+51.3
-174.1
+155.6
-80.8
+205.0
0
Ca(s)
CaO(s)
CaCO3(s)(calcite)
0
41.4
0
-635.1
+38.1
-603.5
-1206.9
+92.9
-1128.8
NO2(g)
HNO3()
Cl2(g)
0
+223.0
0
O2(g)
0
Cu(s)
0
+33.2
0
O3(g)
F2(g)
0
+202.7
0
P(s)(white)
Fe(s)
0
+27.3
0
P4O10(s)
-3010.0
+231.0
-2724.0
+142.7
0
+238.8
+41.1
+163.2
0
Fe2O3(s)(hematite)
-824.2
+87.4
-742.2
PCl3(g)
-287.0
+311.7
-267.8
H(g)
+218.0
+114.6
+203.3
PCl5(g)
-374.9
+364.5
-305.0
PbO2(s)
-277.4
+68.6
-217.4
H2(g)
0
+130.6
0
HCl(g)
-92.3
+186.8
-95.3
HF(g)
-271.1
+173.8
-273.2
HI(g)
+26.4
+206.5
+1.6
HBr(g)
-36.4
+198.6
-53.5
HCN(g)
+135.1
+201.7
H2O(g)
-241.8
H2O()
H2O2()
S(s)(orthorhombic)
0
+32.0
0
-20.6
+205.6
-33.4
SiO2(s)(quartz)
-910.7
+41.5
-856.3
SiCl4()
-687.0
+239.7
-619.9
+124.7
SO2(g)
-296.8
+248.1
-300.2
+188.7
-228.6
SO3(g)
-395.7
+256.6
-371.1
-285.8
+70.0
-237.2
Zn(s)
-187.8
+109.6
-120.4
ZnO(s)
H2S(g)
0
-350.5
+41.6
+43.6
0
-320.5
EXAMPLE
The combustion of propane gas occurs as
follows:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Using thermodynamic data for Go, calculate
the standard free energy change for the
reaction at 298 K.
Gfo/kJ.mol-1
C3H8(g)
-23.47
CO2(g)
-394.4
H2O(g)
-228.57
H2O(l)
-237.13
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
C3H8(g)
CO2(g)
H2O(g)
H2O(l)
Gfo/kJ.mol-1
-23.47
-394.4
-228.57
-237.13
Go  nGof (products) mGof (reactants)
Go = [3(-394.4) + 4(-237.13)] – [(-23.47) – 5(0)]
Go = -2108 kJ
Free Energy and Temperature
How is change in free energy affected by
change in temperature?
G = H – TS
G = H - TS
H
S
-TS
+
-
+
+ at all temp
-
+
-
+
+
-
- at all temp
- at high temp
+ at low temp
-
-
+
+ at high temp
- at low temp
Note:
For a spontaneous process the maximum
useful work that can be done by the system:
wmax = G
“free energy” = energy available to do work
EXAMPLE
The combustion of propane gas occurs as
follows at 298K:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Ho = -2220 kJ.mol-1
a) Without using thermodynamic data tables,
predict whether Go, for this reaction is
more or less negative than Ho.
b) Given that So = -374.46 J.K-1.mol-1 at 298 K
for the above reaction, calculate Go. Was
your prediction correct?
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Ho = -2220 kJ.mol-1
a)
Without using thermodynamic data tables,
predict whether Go, for this reaction is more or
less negative than Ho.
Go = Ho – TSo
-ve
-ve
6 moles gas  3 moles gas
 – TSo > 0
 Ho – TSo will be less negative than Ho
i.e. Go will be less negative than Ho
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Ho = -2220 kJ.mol-1
b) Given that So = -374.46 J.K-1.mol-1 at 298 K for the
above reaction, calculate Go.
Was your prediction correct?
Go = Ho – TSo
Go = (-2220 kJ) – (298 K)(-374.46x10-3 kJ.K-1.mol-1)
Go = -2108 kJ.mol-1
 Prediction was correct.
EXAMPLE (TUT no. 5a)
At what temperature is the reaction below
spontaneous?
AI2O3(s) + 2Fe(s)  2AI(s) + Fe2O3(s)
Ho = 851.5 kJ; So = 38.5 J K-1
At what temperature is the reaction below
spontaneous?
AI2O3(s) + 2Fe(s)  2AI(s) + Fe2O3(s)
Ho = 851.5 kJ; So = 38.5 J K-1
Go = Ho – TSo
Assume H and S do not vary that
much with temperature.
Set G = 0
equilibrium
0 = (851.5 kJ) – T(38.5x10-3 kJ.K-1)
T = 22117 K at equilibrium
For spontaneous reaction: G < 0
 T > 22117 K
Free Energy and the equilibrium constant
Recall:
G = Change in Gibb’s free energy under standard
conditions.
G can be calculated from tabulated values.
BUT most reactions do not occur under standard
conditions.
Calculate G under non-standard conditions:
G  G  RT ln Q
Q = reaction quotient
R = gas constant = 8.314 J.K-1.mol-1
G  G  RT ln Q
Under standard conditions:
Q = 1  ln Q = 0
At equilibrium:
G = 0 and
0  G  RT lnK
(1 M, 1 atm)
 G = Go
Q = Keq

G  RT lnK
If Go < 0
 ln Keq > 0
 Keq > 1
i.e. the more negative Go, the larger K etc.
Go < 0
Go > 0
Go = 0
Also



Keq > 1
Keq < 1
Keq = 1
K  e  G / RT
EXAMPLE
Calculate K for the following reaction at 25oC:
2H2O(l)
2H2(g) + O2(g)
H2O(g)
H2O(l)
Gfo/kJ.mol-1
-228.57
-237.13
Go   nGof (products)   mGof (reactants)
Go = [2(0) + (0)] – [2(-237.13)]
Go = 474.26 kJ.mol-1
G  RT lnK
474.26x103 J.mol-1 = -(8.314 J.K-1.mol-1)(298 K) lnK
lnK = -191.4
2
K

[
H
]
[O2 ]
-84
2
K = 7.36x10