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Chapter 8 Spontaneous Processes & Thermodynamic Equilibrium
 ch 7, recall: energy, First Law, enthalpy, endothermic, exothermic; depends on energy?
(enthalpy not enough); similarly, ch 6, solution formation
8.1 Nature of Spontaneous Processes
 reactions are spontaneous (“product-favoured”) in one direction, not in the other; brick
analogy, etc.
 chemical example:
N 2 (g) + 3 H 2 (g)  2 NH 3 (g)
 spontaneous from starting materials at 472o C, but reaction from the 1:3:1 mol mixture is not
obvious
 but in this case (details in chapter 9):
Q =
(1) 2
(1)(3) 3
= 0.037
 compared to equilibrium constant of 0.105
 direction of spontaneous reaction to right (ie.- more NH3)

NB: this implies nothing about rate of reaction (ie.- kinetics); only about direction and
extent of spontaneous reaction (ie.- thermodynamics)
8.2 Entropy & Spontaneity
Spontaneity
 many exothermic processes occur spontaneously (ie.- heat released), but many examples of
spontaneous processes which are not exothermic
1. gas expansion from one flask to another (Figure 8.2), no enthalpy change; reverse does
not occur
2. melting of ice at room temperature, endothermic but occurs readily; reverse does not
3. dissolution of many salts in water, chills the solution
 common occurrence in the above: atoms or molecules are more dispersed or randomized (less
"specified" relative to neighbours) or disordered
 molecular statistical interpretation (Figure 8.3), due to large numbers of molecules
 randomness or disorder termed entropy (symbol, S)
 a state function like enthalpy
Chem 59-110 (’02)
2
 related to number of microstates available, loss of constraints; by Boltzmann’s
constant, R/No
 units of entropy: joules per kelvin per mole (ie.- J K-1 mol-1)
Entropy & Disorder (Generalizations)
 for the same or similar substances, entropies of gases larger than those of liquids which are
larger than those of solids (intermolecular interactions and freedom of movement)
 entropies of more complex molecules are larger than those of simpler molecules
 entropies of ionic solids become larger as attractions among the ions become weaker
 entropy usually increases when a pure liquid or solid dissolves in a solvent
 entropy increases when a dissolved gas escapes from a solution
 entropy increases if the number of molecules of gas increases during a reaction
 entropy increases if the temperature of a substance increases: molecular rotations and
vibrations increase with temperature
8.3 Entropy & Heat: Background to the Second Law


Entropy defined as (heat transferred/temp at which transfer occurs) = q/T , a state function
with dimensions J K-1
Change in entropy = q/T integrated over a reversible path
f dq
rev
S  
i
T
in general, S  q/T, since qrev > qirrev
 eg. for phase changes, convert ice to water very slowly at 273.15 K with input of energy (6.01
kJ mol-1, the heat of fusion) but without temp change:
 entropy change, S for H2O(s)  H2O(l) = q/T = (+6010 J mol-1)/273.15 K = + 22.0 J
K-1 mol-1

read 8.4 for interest
8.5 Entropy Changes & Spontaneity
 Example: enthalpy of vaporization of benzene is +30.8 kJ mol-1 at its boiling point of 80.1oC;
calculate S for liquid  vapour phase change and for reverse phase change
 for liquid  vapour, S = (+30800 J mol-1)/353.5 K = +87.5 J K-1 mol-1
 for vapour  liquid, S = - 87.5 J K-1 mol-1
 related: Example 8.4
 eg. does entropy increase or decrease for:
2 NO(g) + O2 (g)  2 NO2 (g)
 all gases but number of molecules decreases, therefore, entropy decreases (this
reaction spontaneous as written at 25oC)
Chem 59-110 (’02), ch 8, Spontaneous Processes & Thermodynamic Equilibrium
3
 eg. does entropy increase or decrease for:
4 Fe(s) + 3 O2 (g)  2 Fe2O3 (s)
 number of molecules decreases and gas incorporated into a solid, i.e.- matter is less
dispersed or random, therefore, S decreases
 Caution: may be opposing order and disorder effects, eg. dissolution of salt
Calculation of Entropy Changes in the Surroundings
 for consideration of entropy change accompanying a spontaneous process, we need entropy
created by dispersal of both matter and energy
 have done matter above for the system
 for energy, assume that enthalpy of reaction (from Tables like Appendix D) is transferred to
or from surroundings (if exothermic reaction, energy transferred to the surroundings)
 eg. for the system: CO(g) + 2 H2(g)  CH3OH(l), calculated Sosystem = - 332.2 J K-1 and
known that Hosystem = - 128.14 kJ
 Sosurroundings = qsurroundings/T = - Hosystem/T = + 430 J K-1 at 25oC (note: sign
change)
 Example 8.6
The Second Law of Thermodynamics
 In any spontaneous process there is always an increase in the entropy of the universe
(best stated in index of your text!)
Suniverse = Ssystem + Ssurroundings > 0
Calculation of Entropy Changes in the System & Surroundings
 continue from CO(g) + 2 H2(g)  CH3OH(l) example:
Souniverse = Sosystem + Sosurroundings
= - 332.2 + 430. J K-1 = 98 J K-1
 Second Law confirmed, also process is spontaneous as written (i.e.- product-favoured) under
these conditions
 foregoing is cumbersome; can we make some generalizations based on Hosystem and Sosystem
that will allow us to predict, at least qualitatively, the spontaneity of a given reaction?
 Will do this at end of chapter:
 signs opposite, prediction clear-cut
 signs the same, depends on T and relative magnitudes (see above eg for methanol
formation)
 in rust example, above, system entropy decreases, therefore, surroundings’ entropy must
increase to greater extent
Chem 59-110 (’02), ch 8, Spontaneous Processes & Thermodynamic Equilibrium
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 reaction exothermic – dispersal of energy as heat creates disorder in surrounding
molecules
8.6 Third Law of Thermodynamics

entropy of a pure crystalline substance at 0 K is zero
Standard-State Entropies
 standard entropies, So, have been estimated for many compounds at 298 K and 1 atm
pressure (in Tables, eg. Appendix D)
298.15 c p
 zero at 0 K, then integrate:
So  
dT  S phase changes
0
T

change accompanying a process or reaction has sign and magnitude
Calculation of Entropy Changes for Chemical Systems
 like enthalpy (ch 7), free energy changes (below), for the general reaction , use standard
entropies (So) from tables (Appendix D):
a A + b B + ...  p P + q Q + ...
S o = [pS o (P) + qS o (Q) + ...] - [aS o (A) + bS o (B) + ...]



Example: calculation for the NO + O2 reaction, above, So = - 147 J K-1
Example: calculate So for formation of CaCO3 from the elements
Ca(s) + C(graphite) + 1.5 O2(g)  CaCO3(s)
So = SoCaCO3 - (SoCa + SoC(graphite) + 1.5 SoO2)
So = 92.9 - (41.4 + 5.7 + 1.5 (205.1))
So = - 261.2 J K-1 (note: units)
 negative since a molecule of gas has disappeared and 3 types of atoms or molecules
have been converted to one, more complex, molecule/substance
Example 8.9
8.7 Gibbs Free Energy
 both enthalpy and entropy involved in spontaneity of a reaction
 hence, combined state function: (Gibbs) Free Energy
G = H - TS
 and for a process/system (no need to consider “universe”):
G = H - TS
Chem 59-110 (’02), ch 8, Spontaneous Processes & Thermodynamic Equilibrium
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 sign of free energy change is criterion of spontaneity:
 negative: spontaneous in forward direction (product-favoured)
 zero: reaction at equilibrium
 positive: spontaneous in reverse direction (reactant-favoured)
 boulder analogy: free energy always decreases in any spontaneous process at constant T, p
 note: approach from either direction
 for phase transitions, H and TS compensate, Figure 8.10, and G = 0
Calculating Standard Free Energy Changes, Gorxn
 either, calculated from standard enthalpies of formation (Hof) and standard entropies (So) as
in Appendix D, via intermediate Horxn and Sorxn values
 or, calculated from Table of standard free energies of formation, if available (also
Appendix D)
 in either case, based on reference points or standard states:
State
solid
liquid
gas
solution
elements

Standard State
pure solid
pure liquid
1 atm pressure
1M
(std. free energy 0 in normal state)
ie.- can be calculated for a reaction:
aA  bB  ...  pP  qQ  ...
G o  [pG of (P)  qG of (Q) ...] - [aG of (A)  bG of (B) ...]
= standard free energy change for reaction
 used to predict direction of spontaneous reaction
Calculating Gorxn from Horxn and Sorxn
 Example: using Hf and So values for the components of the reaction:
1
3
N 2 (g) +
H (g)  NH 3 (g)
2
2 2
N2(g)
H2(g)
NH3(g)
-1
Hf (kJ mol )
0
0
-46.11
So (J K-1 mol-1)
191.61
130.68
192.45
o
H rxn = - 46.11 – 0 = - 46.11 kJ
Sorxn = 192.45 - [(191.61/2) + 3(130.68)/2]
= 192.45 - 291.83
= - 99.38 J K-1 = - 99.38/1000 kJ K-1
o
 G rxn = Horxn - TSorxn
= - 46.11 - 298.15 (-99.38 x 10-3)
Chem 59-110 (’02), ch 8, Spontaneous Processes & Thermodynamic Equilibrium
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= - 16.47 kJ


reaction spontaneous to right from components at 1 atm and 25oC
if above an equilibrium, what happens to G with increasing temperature? eg., at
500C?
Calculating Gorxn from Gof of Components
 previous Exercise “reaction” is actually the formation of a molecule from the corresponding
elements in their natural state; Gorxn in such cases gives the standard free energies of
formation, Gof
 if Gof’s available, they can be used directly
 Example 8.10
Product-Favoured or Reactant-Favoured? Direction of the Spontaneous Reaction
 criteria listed above, all based on the sign of Gorxn
 both enthalpy and entropy changes could favour the spontaneous reaction, but not
necessarily; then talk of entropy-driven or enthalpy-driven reactions (two of the four
possibilities in Figure 8.13)
Free Energy & Temperature
 temperature-dependence due to entropy term (otherwise, all exothermic reactions would be
spontaneous), Figure 8.12
 4 situations; magnitudes important when enthalpy and entropy changes have same sign
 see Haber process example, above (change in spont. direction 25oC vs. 500oC)
 Example: MgO reduction with C; favoured at 25oC? If not, at what temp. does it become
so?
MgO(s) + C(graphite)  Mg(s) + CO(g)
Horxn = 0 + (-110.53) - [(-601.70) + 0] = 491.17 kJ
Sorxn = 32.68 + 197.67 - (26.94 + 5.74) = 197.67 J K-1
at 25oC: Gorxn = Horxn - TSorxn
= 491.17 - (298.15)(197.67 x 10-3)
= 432.23 kJ; not favoured
changeover from unfavoured to favoured at temp when Gorxn = 0
0 = 491.17 - (temp)(197.67 x 10-3)
temp = 2485 K or 2212oC
Suggested Problems
1, 5, 9, 13, 17 – 37, odd
Chem 59-110 (’02), ch 8, Spontaneous Processes & Thermodynamic Equilibrium