THERMODYNAMICS

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THERMODYNAMICS

• Chapter 19

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SPONTANEOUS PROCESS
? A process that occurs without ongoing outside
intervention.

• Examples
• Nails rusting outdoors

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• Ice melting at room temperature

• Expansion of gas into an evacuated space

• Formation of water from O2(g) and H2(g)

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Why are some processes spontaneous and others not?
We know that temperature has an effect on the
spontaneity of a process.
e.g. Tgt0oC ? ice melts ? spontaneous at this
temp. H2O (s) ? H2O (l)
Tlt0oC ? water freezes ? spontaneous at this
temp. H2O (l) ? H2O (s)
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Exothermic processes tend to be spontaneous.
Example
Rusting of nail - SPONTANEOUS! 4Fe(s) 3O2(g) ?
Fe2O3(s) ?H - 822.2 kJ.mol-1
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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.
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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
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Entropys effects on the mind
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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
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However, entropy does not always increase for a
spontaneous process
At room temperature
Spontaneous
Non-spontaneous
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SECOND LAW OF THERMODYNAMICS
The entropy of the universe increases in any
spontaneous process.
?Suniverse ?Ssystem ?Ssurroundings
Spontaneous process ?Suniverse gt 0 Process at
equilibrium ?Suniverse 0
Thus ?Suniv is continually increasing! ?Suniv
must increase during a spontaneous process, even
if ?Ssyst decreases.
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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.

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?Sunivgt0
For example Rusting nail spontaneous
process 4Fe(s) 3O2(g) ? 2Fe2O3(s) ?Ssystlt0
BUT reaction is exothermic, ? entropy of
surroundings increases as heat is evolved by the
system thereby increasing motion of molecules in
the surroundings.
? ?Ssurrgt0
-ve
ve
ve
Thus for ?Suniv ?Ssyst ?Ssurr gt0
??Ssurr? gt ??Ssyst?
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Special circumstance Isolated system

• Does not exchange energy nor matter with
  surroundings ??Ssurr 0

Spontaneous process ?Ssyst gt 0 Process at
equilibrium ?Ssyst 0
Spontaneous ? Thermodynamically favourable (Not
necessarily occur at observable rate.)
Thermodynamics ? direction and extent of
reaction, not speed.
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EXAMPLE

• State whether the processes below are
  spontaneous, non-spontaneous or in equilibrium
• CO2 decomposes to form diamond and O2(g)
• Water boiling at 100oC to produce steam in a
  closed container
• Sodium chloride dissolves in water

NON-SPONTANEOUS
EQUILIBRIUM
SPONTANEOUS
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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
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Molecules have 3 types of motion
Translational motion - Entire molecule moves in a
direction (gas gt liquid gt solid)
Vibrational motion within a molecule
Rotational motion spinning
Greater the number of degrees of freedom ?
greater entropy
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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.
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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

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THIRD LAW OF THERMODYNAMICS
The entropy of a pure crystalline substance at
absolute zero is zero.
S(0 K) 0 ? perfect order
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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.

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Entropy increases for s ? l ? g
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EXAMPLE

• Predict whether the entropy change of the system
  in each reaction is positive or negative.
• CaCO3(s) ? CaO(s) CO2(g)
• 2SO2(g) O2(g) ? 2SO3(g)
• N2(g) O2(g) ? 2NO(g)
• H2O(l) at 25oC ? H2O(l) at 55oC

ve
-ve
3 mol gas ? 2 mol gas
Cant predict, but it is close to zero
?
2 mol gas ? 2 mol gas
ve
Increase thermal energy
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Standard molar entropy (So)
molar entropy for substances in their standard
state

• NOTE
• So ? 0 for elements in their standard state
• So(gas) gt So(liquid) gt So(solid)
• So generally increases with increasing molar
  mass
• So generally increases with increasing number
  of atoms in the formula of the substance

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Calculation of ?S for a reaction
Stoichiometric coefficients
(So from tabulated data)
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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
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Calculation of ?S for the surroundings
For a process that occurs at constant temperature
and pressure, the entropy change of the
surroundings is
(T P constant)
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GIBBS FREE ENERGY (G)
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Defined as G H TS
- state function - extensive property
?Suniv ?Ssys ?Ssurr
At constant T and P
? - T?Suniv - T?Ssys ?Hsys
(at constant T P)
?G ?H T?S
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We know
Spontaneous process ?Suniv gt 0 Process at
equilibrium ?Suniv 0
Therefore
Spontaneous process Process at equilibrium
lt 0
-T?Suniv
0
-T?Suniv
?Gsyst -T?Suniv
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?G ?H T?S
Spontaneity involves
?S
?H
T
Spontaneity is favoured by increasing ?S and ?H
is large and negative.
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?G allows us to predict whether a process is
spontaneous or not (under constant temperature
and pressure conditions)
?G lt 0 ? spontaneous in forward direction
?G gt 0 ? non-spontaneous in forward
direction/spontaneous in reverse direction
?G 0 ? at equilibrium
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But nothing about rate
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Standard free energy (?Go)
?Go ?Ho T?So
Standard states Gas - 1 atm Solid - pure
substance Liquid - pure liquid Solution -
Concentration 1M
?Gfo 0 kJ/mol for elements in their standard
states
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Tabulated data of ?Gfo can be used to calculate
standard free energy change for a reaction as
follows
Stoichiometric coefficients
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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 H2
O(g) -228.57 H2O(l) -237.13

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C3H8(g) 5O2(g) ?3CO2(g) 4H2O(l)
?Gfo/kJ.mol-1 C3H8(g) -23.47 CO2(g) -394.4 H2
O(g) -228.57 H2O(l) -237.13
?Go 3(-394.4) 4(-237.13) (-23.47)
5(0) ?Go -2108 kJ
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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

-
-
-
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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
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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?
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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 gt 0
? ?Ho T?So will be less negative than ?Ho
i.e. ?Go will be less negative than ?Ho
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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.
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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
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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.
equilibrium
Set ?G 0
0 (851.5 kJ) T(38.5x10-3 kJ.K-1)
T 22117 K
at equilibrium
For spontaneous reaction ?G lt 0
? T gt 22117 K
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Free Energy and the equilibrium constant
Recall ?G? Change in Gibbs 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
Q reaction quotient R gas constant 8.314
J.K-1.mol-1
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Under standard conditions (1 M, 1 atm) Q 1
? ln Q 0 ? ?G ?Go
At equilibrium ?G 0 and Q Keq
?
If ?Go lt 0 ? ln Keq gt 0 ? Keq gt
1 i.e. the more negative ?Go, the larger K
etc.
?Go lt 0 ? Keq gt 1 ?Go gt 0 ? Keq
lt 1 ?Go 0 ? Keq 1
Also
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EXAMPLE
Calculate K for the following reaction at
25oC 2H2O(l) 2H2(g) O2(g)
?Gfo/kJ.mol-1 H2O(g) -228.57 H2O(l)
-237.13
?Go 2(0) (0) 2(-237.13) ?Go 474.26
kJ.mol-1
474.26x103 J.mol-1 -(8.314 J.K-1.mol-1)(298 K)
lnK
lnK -191.4 K 7.36x10-84