15.4  Spontaneity 
15.4.1: Predict whether a reaction or process will be spontaneous using the sign of ΔGº
Entropy is nature's arrow. Overall, everything tends towards increasing disorder, as this is overwhelmingly the most probable state for particles and associated heat energy.
Joshua Willard Gibb realised that for a process to occur the overall entropy of the universe must increase. The universe being made up of the system under study and its surroundings.
Therefore the total entropy in the universe must be equal to the entropy of the system + the surroundings, and any change in the universal entropy must be due to a change in entropy of system, surroundings or both. i.e.
ΔS(universe) = ΔS(system) + ΔS(surroundings)
The surroundings can only be affected by heat exchange with the system. For an exothermic reaction the surroundings receives energy that increases its disorder (entropy) by a factor of q/T where q (in terms of the system) must be ΔH/T
Hence universal entropy can be written:
ΔS(universe) = ΔS(system) + ΔH/T(surroundings)
But when the surroundings increases in energy so the system decreases in energy, therefore in terms of the system:
ΔS(universe) = ΔS(system)  ΔH/T(system)
Multiply through by T to remove fractions:
TΔS(universe) = TΔS(system) + ΔH(system)
and rearrange:
TΔS(universe) = ΔH(system) TΔS(system)
Gibbs lumped together the TΔSuniverse term and called it ΔG, Gibbs free energy.
So when Gibbs free energy is negative the entropy change of the universe must be positive. This can be measured by looking at the enthalpy and entropy change of the system under study:
ΔG = ΔH(system) TΔS(system)
So when the universal entropy increases (a spontaneous process) then the Free energy decreases.
ΔG is a measure of the free energy of the UNIVERSE, not the system under study. You can think of free energy as that portion of the energy of the universe that is available to do work.
The universal entropy is always increasing and this decreases the available free energy.
Gibbs free energy equation relates the free energy of the universe to the systems enthalpy and entropy change:
ΔG = ΔH  TΔS
When the value of ΔG is negative this means that the universal free energy has decreased and hence the universal entropy has increased. The process is consequently favourable (thermodynamically spontaneous)
It is impossible to increase the free energy of the universe.
So what exactly is free energy?
It has been interpreted as being the energy that could be used to do work. All of this came about with the study of engines. We know that energy tends to flow from high temperature to ambient. It cannot go in reverse. So a region of high energy can be used to do work, such as boiling water to drive a steam engine. But the same energy spread out evenly throughout the universe cannot do any work. All it succeeds in doing is increasing the average heat of the universe by a miniscule fraction.
Energy does not have to exist as heat. It could also be chemical potential energy. In that state it could be transformed into heat energy and do work, before dissipating and being useless.
So the idea of 'Free energy' is energy that is still available to do work. Energy that has been released and dissipated has increased the entropy of the universe and is no longer 'useful'.
Hence an exothermic reaction releases heat energy, this increases universal entropy and decreases the free energy of the universe.
Spontaneity
A reaction is said to be thermodynamically spontaneous (i.e. the reaction can happen) if its free energy change is negative.
There are several possibilities depending on the sign of the enthalpy change and the entropy change for the reaction, remembering that:
ΔG = ΔH(system) TΔS(system)
Sign of the enthalpy change  Sign of the entropy change  Spontaneity 
positive (+)  positive (+)  The reaction is spontaneous at high temperature 
positive (+)  negative ()  The reaction is never spontaneous 
negative ()  negative ()  The reaction is spontaneous at low temperature 
negative ()  positive (+)  The reaction is always spontaneous 
In order to find out the temperature at which the reaction just becomes spontaneous, set ΔG = 0 and fill in the values for ΔHand ΔS to find the temperature.
15.4.2: Calculate Gibb's free energy change (ΔGº) for a reaction using the equation ΔGº= ΔHº  TΔ Sºand by using values of the standard free energy change of formation, ΔGfº
Gibbs free energy change is calculated using the Gibbs free energy equation:
ΔG = ΔH(system) TΔS(system)
It is important to remember that the units of enthalpy are often given in kiloJoules whereas the units of entropy are given in Joules per Kelvin. In order to enter values into the Gibbs free energy equation all values must be in either kiloJoules or Joules.
Example: The Haber process N_{2} + 3H_{2} 2NH_{3} .... ΔH = 92 kJ The absolute entropy values are:
On the left hand side there is 1 mole of nitrogen and 3 moles of hydrogen, hence the total absolute entropy on the left hand side = 192 + 3(131) = 585 J K^{1} On the right hand side there are 2 moles of ammonia, hence the total absolute entropy on the right hand side = 2(193) = 386 J K^{1} Hence the entropy has decreased from 585 to 386 = 199 J K^{1} ΔGº = ΔHº  TΔSº At 298K ΔGº = 92 + (298 x 0.199) = 32.7 kJ 
15.4.3: Predict the effect of a change in temperature on the spontaneity of a reaction using standard entropy and enthalpy changes and the equation ΔGº= ΔHº TΔSº
To predict the effect of changing temperature refer to the table of enthalpy and entropy signs on spontaneity.
Sign of the enthalpy change  Sign of the entropy change  Spontaneity 
positive (+)  positive (+)  The reaction is spontaneous at high temperature 
positive (+)  negative ()  The reaction is never spontaneous 
negative ()  negative ()  The reaction is spontaneous at low temperature 
negative ()  positive (+)  The reaction is always spontaneous 
Clearly, if a reaction becomes spontaneous at higher temperatures, then any increase in temperature will make Gibbs free energy more negative (or less positive). The same logic can be applied to any reaction.
Example: The Haber process N_{2} + 3H_{2} 2NH_{3} .... ΔH = 92 kJ The absolute entropy values are:
On the left hand side there is 1 mole of nitrogen and 3 moles of hydrogen, hence the total absolute entropy on the left hand side = 192 + 3(131) = 585 J K^{1} On the right hand side there are 2 moles of ammonia, hence the total absolute entropy on the right hand side = 2(193) = 386 J K^{1} Hence the entropy has decreased from 585 to 386 = 199 J K^{1} ΔGº = ΔHº  TΔSº At 298K ΔGº = 92 + (298 x 0.199) = 32.7 kJ As the sign of enthalpy is negative and the sign of the entropy change is negative then the value of  TΔSº bwcomes more positive as the temperature is increased. A point will be reached at which the reaction is no longer spontaneous, as increasing temperature makes Gibbs Free energy change more positive and less negative. When the value of ΔGº = 0 the reaction reaches equilibrium. Any further increase in temperature drives the reaction backwards. 
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