Gibbs Free Energy

You learned from my previous post that processes can be spontaneous and non-spontaneous. Spontaneous processes can be endothermic and exothermic.  Usually, an spontaneous endothermic process has an increase  in disorder and entropy of the system.  On the other hand, a spontaneous exothermic process has a decrease in the entropy of the system.

Since spontaneous processes that result in the decrease in entropy are always exothermic, therefore the spontaneity of the reaction seems to involve two thermodynamic quantities the enthalpy and entropy.   An American mathematician Josiah Willard Gibbs (1839-1903) find a way to use ∆H and ∆S to predict whether a given reaction occurring at constant temperature and pressure will be spontaneous. Gibbs developed the so-called Gibbs Free Energy  (or simply free-energy) in answer to the problem above.  Gibbs Free energy is defined as

G  =   H  -     TS

where T is the absolute temperature.  For a process occurring at constant temperature, the change in free energy of the system, ∆G, is given by the equation

∆G  =  ∆H  -   T∆S

To determine how the function G relates to reaction spontaneity, we learned from my previous post that 

∆S_univ  =   ∆S_sys    +    ∆S_surr    =    ∆S_sys   +    (-∆H_sys / T)

Multiplying both sides by -T gives

-T∆S_univ  =  ∆H_sys   -  T∆S_sys

Looking at the two equations above, we can conclude that the free-energy change in the process occurring at constant temperature and pressure, ∆G, is equal to  -T∆S_univ.    We learned that the value of ∆S for spontaneous processes is positive, and so the sign of ∆G can provide us with valuable information about the spontaneity of processes that occur at constant temperature and pressure.   If both T and P are constant, the relationship between the sign  of ∆G and the spontaneity of the reaction  is as follows:

1.  If ∆G is negative, the reaction is spontaneous in the forward direction.

2.  If ∆G is zero, the reaction is at equilibrium.

3.  If ∆G is positive, the reaction in the forward direction is nonspontaneous ; work must be supplied      from the surroundings to make it occur.  However the reverse reaction will be spontaneous.

Standard Free Energy Changes

The Standard Free-Energy of reaction ( ∆G^o_rxn) is the free-energy change for a reaction when it occurs at standard state conditions, when reactants in their standard state is converted to products in their standard states.    The standard state for gaseous substances is 1 atm pressure.  For solid substances, the standard state is the pure solid; for liquids, the pure liquid.  For substances in solution, the standard state is normally a concentration of 1 M.  The temperature chosen for purposes of tabulating data is 25^oC or 298K.  For the standard heats of formation, the free energies of elements in their standard states are set to zero.  For the standard free-energy of formation values of some substances click here.

We can calculate the Standard Free-Energy of Reaction ( ∆G^o_rxn)  using the equation below:

∆G^o_rxn =    ∑nG^o_f (products) - ∑mG^o_f(reactants)

where n and m are stoichiometric coefficients, G^o_f is the standard free-energy of formation of a compound, that is, the free-energy change that occurs when 1 mole of the compound is synthesized from its elements in their standard states.

Let us see for example the standard free-energy change of the combustion of graphite.

        C(graphite)   +   O₂(g)  →    CO₂(g)

The value of ∆G^o_f of O₂(g) and C(graphite) is zero, since both are in their stable allotropic form, and so the ∆G^o_rxn is equal to the ∆G^o_f of CO₂.

Sample Problems:

Calculate the free-energy changes of the following reactions at 25^oC or 298 K.

1.  CH₄(g)  +   2O₂(g)   →   CO₂(g)   +   2H₂O(l)
2.  2MgO(s)    →   2Mg(s)   +    O₂(g)

Solution:

Before calculating be sure to find the values of free-energy of formation of the substances in the given link here.   Then substitute the values in the equation.

1.  ∆G^o_rxn =  [∆G^o_f(CO₂)  +  2∆G^o_f(H₂O)]  -  [∆G^o_f(CH₄) +   2∆G^o_f (O₂)
                 =  [ (-394.4 kJ/mol) + (2)(-237.2 kJ/mol)]  -  [(-50.8 kJ/mol) +  (2)(0 kJ/mol)]
                 =  (-394.4 kJ/mol - 474.4 kJ/mol)  - (-50.8 kJ/mol)
                 =  -868.8 kJ/mol + 50.8 kJ/mol
                 =  -818.0 kJ/mol

2.  ∆G^o_rxn  = [2∆G^o_f(Mg)  +  ∆G^o_f(O₂)]  -   [2∆G^o_f(MgO)]
                   = [(2)(0 kJ/mol)  +  (0 kJ/mol)] -  [(2)(-569.6 kJ/mol)]
                   = 0 - (-1139.2 kJ/mol)
                   = 1139 kJ/mol

Free Energy and Temperature

We calculated the Free-Energy change at 25^oC or 298K from the concept above.  However, we can also calculate Free-Energy Change at other temperatures, the formula presented above will be used.

∆G  =  ∆H  -   T∆S

In order to predict the sign of ∆G, we need to know the value of ∆H and ∆S.  Below are possible outcomes of the ∆H and ∆S relationships:

 1.  If both ∆H and ∆S are positive, then ∆G will be negative when the T∆S term is greater in                     magnitude than ∆H.   This condition is met when T is large.

2.  If ∆H is positive and ∆S is negative, ∆G will always be positive, regardless of temperature.

3.  If ∆H is negative and ∆S is positive, then ∆G will be negative regardless of temperature.

4.  If ∆H is negative and ∆S is negative, then ∆G will be negative only when T∆S is smaller in                  magnitude than ∆H.  This condition is met when T is small.

The temperatures that will cause the ∆G to be negative will depend on the actual values of ∆H and ∆S of the system.  The table below summarizes the effect of temperature  on the spontaneity of reaction:

 Let us have an example, the melting of ice to liquid water at 1 atm pressure:

H₂O(s)   →   H₂O(l)       ∆H  > 0,  ∆S >  0

The process above is endothermic, which means the ∆H is positive, and since their is an increase in disorder of molecules the ∆S is also positive, and -T∆S is negative.  At temperatures below 0^oC the magnitude of ∆H is greater than that of -T∆S, and therefore the positive ∆H dominates, leading to positive value of ∆G.  The positive value of ∆G means that the melting of ice is not spontaneous at T < 0^oC; rather the reverse process is the spontaneous, the freezing of liquid water to ice.

Before we proceed to calculation, let us first differentiate ∆G and ∆G^o.  ∆G is used to predict the direction of the reaction for nonstandard condition, while ∆G^o is used to predict the direction of the reaction for standard condition ( 1 M concentration) at 25^oC  and 1 atm pressure. The sign of ∆G^o
tell us whether the products or the reactants are favored when the reacting system reaches equilibrium.  Therefore, a negative value of  ∆G^o indicates that the reaction favors products formation whereas a positive value of ∆G^o indicates that there will be more reactants than products at equilibrium.

Sample Problem 1.

a) Using the standard enthalpies of formation and standard entropies ( click the following : Standard Enthalpy; Standard Entropies), calculate ∆H^o and ∆S^o at 298 K for the following reactions:
CaCO₃(s)    →  CaO(s)   +   CO₂(g)
b)  Using the values obtained in a) estimate ∆G^o at 298 K

Solution:

a)  ∆H^o = [∆H^o_f(CaO)  +  ∆H^o_f(CO₂)] -  [∆H^o_f(CaCO₃)]
             = [(-635.6 kJ/mol) + (-393.5 kJ/mol)] - (-1206.9 kJ/mol)
             = -1029.1 kJ/mol + 1206.9 kJ/mol
             = +177.8 kJ/mol

     ∆S^o = [S^o(CaO)  +  S^o(CO₂]  - [S^o(CaCO₃)]
             = [(39.8 J/K.mol)  +  (213.6 J/K.mol)]  -  (92.9 J/K.mol)
             = (253.4 J/K.mol) - (92.9 J/K.mol)
             = 160.5 J/K.mol

b)   ∆G^o  = ∆H^o  -  T∆S^o
               = (177.8 kJ/mol)  - [(298 K)(160.5 J/K.mol)( 1 kJ / 1000 J)
               = 177.8 kJ/mol -  47.83 kJ/mol
               =  130.0 kJ/mol
 
Sample Problem 2

a) Using standard enthalpies of formation and standard entropies, calculate ∆H^o and ∆S^o at 298 K for     the reaction:  2SO₂(g)  +  O₂(g)  →   2SO₃(g),
b) using the values obtained in a) estimate the ∆G^o at 400 K.

Solution:

a) ∆H^o = 2∆H^o_f(SO₃) - [2∆H^o_f(SO₂) +  ∆H^o_f(O₂)]
            = [(2)(-395.2 kJ/mol)] - [(2)(-296.9 kJ/mol) +  (0 kJ/mol)
            = (-790.4 kJ/mol) - (-593.8 kJ/mol)
            = -790.4 kJ/mol + 593.8 kJ/mol
            = -196.6 kJ/mol

     ∆S^o =  2S^o(SO₃)  -  [2S^o(SO₂)  +  S^o(O₂)
             =  2(256.2 J/K.mol) - [2(248.5 J/K.mol) +  205.0 J/K.mol]
             =  512.4 J/K.mol - (497 J/K.mol + 205.0 J/K.mol)
             =  512.4 J/K.mol -702 J/K.mol
             =  -189.6 J/K.mol

b)  ∆G^o  = ∆H^o  -  T∆S^o
              = -196.6 kJ/mol - [(400 K)(-186.6 J/K.mol)( 1 kJ /1000J)
              = -196.6 kJ/mol + 74.64 kJ/mol
              = - 122.0 kJ/mol