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FREE ENERGY & SPONTANEOUS CHANGE
A spontaneous change is one which has a natural tendency to occur if left to itself.
For example:
 the water in a lake will freeze spontaneously if the air temperature above it is
below 0oC.
 ice cubes placed in a drink will melt spontaneously
 iron in the presence of oxygen and water will rust spontaneously
 a battery will spontaneously run down
However, the spontaneous change will only occur in one direction unless conditions
are changed. It is often difficult to reverse a spontaneous change.
 the water in the lake will not melt unless the temperature of the air above it
rises
 water needs the electrical energy used by a fridge to convert it into ice cubes
 iron oxide needs the conditions produced in a blast furnace to turn it back to
iron
 a battery can be re-charged using electrical energy from the mains
The quantity which decides whether a particular change can occur spontaneously is
the free-energy change, which is given the symbol G. For a change to occur
spontaneously:
.
G must be negative or zero
Free energy changes combine the effects, in a particular reaction, of changes in
enthalpy and changes in entropy.
G = H - TS
1. Changes in Enthalpy
The sign of an enthalpy change does not allow a prediction to be made about the
feasibility of a change.
There is a natural tendency for enthalpy to fall in a reaction, and many exothermic
changes occur spontaneously. For example, the reaction between potassium and
water occurs spontaneously and liberates a lot of heat:
2K(s) + 2H2O(l)
2KOH(aq) + H2(g)
However, spontaneous endothermic changes are also known, for example the
reaction of sodium hydrogencarbonate with dilute hydrochloric acid:
NaHCO3(s) + HCl(aq)
NaCl(aq) + H2O(l) + CO2(g)
The fact that endothermic changes can occur spontaneously clearly indicates that a
factor other than enthalpy change must also be involved. This additional factor is the
change in entropy.
TOPIC 13.14: ENTROPY 1
2. Entropy
Entropy can best be thought of as a measure of the disorder of a chemical system;
the more disordered the system, the higher the entropy. There is a natural tendency
for the entropy of a system to increase.
The particles in a solid are ordered and their movement is restricted to vibration
about a mean position. The particles in a gas are disordered and they move about
randomly. Therefore, the entropy of a gas is much higher than that of a solid.
entropy increases
solid
liquid
gas
ordered
less ordered
chaotic
When a solid is heated, the amplitude of the vibrating particles increases, and
therefore the entropy increases. When the solid reaches its melting point, the entropy
increases at constant temperature as the ordered solid becomes a less ordered
liquid.
When the liquid is heated, the kinetic energy of the particles increases, and therefore
the entropy increases. When the liquid reaches its boiling point, the entropy
increases at constant temperature as the less ordered liquid becomes a chaotic gas.
The entropy change from liquid to gas is greater than that from solid to liquid.
entropy
gas
liquid
solid
m.p
.
b.p.
temperature
TOPIC 13.14: ENTROPY 2
The units of entropy are:
J.K-1.mol-1
Entropy decreases as temperature decreases, so that at absolute zero (0K), most
substances are solids consisting of perfectly ordered particles which have ceased to
vibrate. They therefore have zero entropy. This means that there is a definite starting
point for entropy, and substances can be assigned an absolute standard entropy
vaue (So). Compare this with enthalpy, where only changes can be measured, and
where it is necessary to assign to elements the arbitrary standard enthalpy of zero.
The table below lists some standard entropies.
So / J.K-1.mol-1
CH4(g)
C(s,diamond)
C(s,graphite)
Na2CO3(s)
Cl2(g)
O2(g)
H2O(g)
C2H5OH(l)
C6H12(l)
186
2.4
5.7
136
223
205
189
161
204
H2O(l)
NaCl(s)
NaHCO3(s)
Mg(s)
H2(g)
CO2(g)
CH3OH(l)
C6H6(l)
69.9
72.4
102
32.5
131
214
127
173
2. Changes in Entropy
a) Chemical Changes
The entropy change during a chemical reaction is calculated from the expression:
So = So(products) - So(reactants)
Example: Calculate the standard entropy change for the combustion of methane:
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
So = 

o(products)
o(products)
-  So(reactants)
= 214 + 2 x 69.9 = 353.8 J.K-1.mol-1
 So(reactants) = 186 + 2 x 205 = 596 J.K-1.mol-1
So = 353.8 - 596 = -242.2 J.K-1.mol-1
There is a fall in the entropy of the system (i.e. it becomes more ordered),
because three moles of gas react to form one mole of gas and two moles of liquid.
Liquids are more ordered than gases.
TOPIC 13.14: ENTROPY 3
b) Physical Changes
When a system is in equilibrium between two physical states, for example ice
and water at 273K and 105Pa, there is no spontaneous tendency for the equilibrium
mixture to change. Unless the external conditions are changed, the mixture will not
spontaneously solidify nor liquefy. Since there is no tendency to move spontaneously
in either direction, the free energy change, G is zero.
Since G = H - TS, when G = 0
H = TS
S = H
T
Example 1: Calculate the entropy change when one mole of ethanol is vaporised.

Svap = Hvap = 43.5 x 1000 = +123.6 J.K-1.mol-1
Tb
352
Example 2: Calculate the entropy change when one mole of ethanol is melted.

Sfus = Hfus = 4.60 x 1000 = +29.5 J.K-1.mol-1
Tm
156
TOPIC 13.14: ENTROPY 4
FREE ENERGY CHANGES
1. H is negative; S is positive
+
Energy
change
Temperature /K
0
H
-
TS
G
Since G is always negative, a change of this type is feasible at any temperature.
Examples:
NH4NO3(s)
2H2O(l) + N2O(g)
C(s) + O2(g)
CO2(g)
2. H is positive; S is positive
H
+
change feasible
Temperature /K
Energy
change
0
change not feasible
-
G
TS
G becomes zero where the line crosses the x-axis (temperature). At temperatures
above this point, G is negative and the change is feasible. Below this temperature,
G is positve and the change is not feasible.
Examples:
H2O(l)
CaCO3(s)
H2O(g)
CaO(s) + CO2(g)
NaHCO3(s) + HCl(aq)
TOPIC 13.14: ENTROPY 5
boiling
NaCl(aq) + H2O(l) + CO2(g)
3. H is negative; S is negative
TS
+
Energy
change
change not feasible
0
G
Temperature /K
-
change feasible
H
G becomes zero where the line crosses the x-axis (temperature). At temperatures
below this point, G is negative and the change is feasible. Above this temperature,
G is positve and the change is not feasible.
Examples:
H2O(g)
N2(g) + 3H2(g)
H2O(l)
condensation
2NH3(g)
4. H is positive; S is negative
G
+
Energy
change
0
-
G is always positive, so a change of this type is never feasible.
TOPIC 13.14: ENTROPY 6
TS
H
Temperature /K
Calculating the temperature at which a reaction becomes
feasible.
Referring to the diagrams on the previous two pages, a change just becomes
feasible when G = 0.
Since G = H - TS, when G = 0
H = TS
T = H
S
It is important in this calculation to ensure that H and S are in compatible
units. H is usually given in kJ.mol-1, therefore, S must be converted to kJ.K-1.mol-1
by dividing by 1000. The answer will be in K.
Example: Use the data below to calculate the temperature at which the following
reaction becomes feasible.
CaCO3(s)
CaO(s) + CO2(g)
Hf /kJ.mol-1
CaCO3(s)
-1207
CaO(s)
-635
CO2(g)
-394
S /J.K-1.mol-1
88.7
40
214
Ho =  o(products) -  Ho(reactants)
= (-635-394) – (-1207)
= + 178 KJ.mol-1
So =  o(products) -  So(reactants)
= (40 + 214) – (88.7)
= + 165.3 J.K-1.mol-1
Go = Ho - TSo
when the reaction just becomes feasible, G = 0
The temperature at which this occurs is given by T = Ho
So
T = 178 x 1000 = 1077K
165.3
TOPIC 13.14: ENTROPY 7