The Free Energy

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 15:23

In the previous section, we were careful to differentiate between the entropyA thermodynamic state function, symbol S, that equals the reversible heat energy transfer divided by temperature; higher entropy corresponds to greater dispersal of energy on the molecular scale. See also standard entropy. change occurring in the reaction system ΔSsys, on the one hand, and the entropy change occurring in the surroundings, ΔSsurr, given by –ΔH/T, on the other. By doing this we were able to get a real insight into what controls the direction of a reaction and why. In terms of calculations, though, it is a nuisance having to look up both entropy and enthalpyA thermodynamic state function, symbol H, that equals internal energy plus pressure x volume; the change in enthalpy corresponds to the energy transferred as a result of a temperature difference (heat transfer) when a reaction occurs at constant pressure. data in order to determine the direction of a reaction. For reasons of convenience, therefore, chemists usually combine the entropy and the enthalpy into a new function called the Gibbs free energyGibbs energy: a thermodynamic function corresponding to the tendency for spontaneous change in a system; represented by the symbol G., or more simply the free energy, which is given the symbol G. If free-energy tables are available, they are all that is needed to predict the direction of a reaction at the temperatureA physical property that indicates whether one object can transfer thermal energy to another object. for which the tables apply.

In order to introduce free energy, let us start with the inequality


\frac{-\Delta H}{T}+ {\Delta S}_{sys}>0


This inequality must be true if a reaction occurring at constant pressureForce per unit area; in gases arising from the force exerted by collisions of gas molecules with the wall of the container. in surroundings at constant temperature is to be spontaneousCapable of proceeding without an outside source of energy; refers to a reaction in which the products are thermodynamically favored (product-favored reaction).. It is convenient to multiply this inequality by T; it then becomes

–ΔH + T ΔS > 0


(From now on we will abandon the subscript "sys".) If –ΔH + T ΔS is greater than zero, it follows that multiplying it by –1 produces a quantity which is less than zero, that is,


ΔHT ΔS < 0      (1)


This latest inequality can be expressed very neatly in terms of the free energy G, which is defined by the equation

G = HTS      (2)


When a chemical reactionA process in which one or more substances, the reactant or reactants, change into one or more different substances, the products; chemical change involves rearrangement, combination, or separation of atoms. Also called chemical change. occurs at constant temperature, the free energy will change from an initial value of G, given by


G1 = H1TS1

to a final value


G2 = H2TS2


The change in free energy ΔG will thus be


         ΔG = G2G1 = H2H1T(S2 S1)

or      ΔG = ΔHT ΔS      (3)


Feeding this result back into inequality (1) gives the result


ΔG = ΔHT ΔS < 0

ΔG < 0      (4)


This very important and useful result tells us that when a spontaneous chemical reaction occurs (at constant temperature and pressure), the free-energy change is negative. In other words a spontaneous change corresponds to a decrease in the free energy of the system.

If we have available the necessary free-energy data in the form of tables, it is now quite easy to determine whether a reaction is spontaneous or not. We merely calculate ΔG for the reaction using the tables. If ΔG turns out to be positive, the reaction is nonspontaneous, but if it turns out to be negative, then by virtue of Eq. (4) we can conclude that it is spontaneous. Data on free energy are usually presented in the form of a table of values of standard free energies of formation. The standard free energy of formation of a substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula. is defined as the free-energy change which results when 1 mol of substance is prepared from its elements at the standard pressure of 1 atmAbbreviation for atmosphere, a unit of pressure equal to 101.325 kPa or 760 mmHg. and a given temperature, usually 298 K. It is given the symbol ΔGf°. A table of values of ΔGf° (298 K) for a limited number of substances is given in the following table.

Some Standard Free Energies of Formation at 298.15 K (25°C)

Compound ΔGfo /kJ mol-1 Compound ΔGfo /kJ mol-1
AgCl(s) -109.789 H2O(g) -228.572
AgN3(s) 591.0 H2O(l) -237.129
Ag2O(s) -11.2 H2O2(l) -120.35
Al2O3(s) -1582.3 H2S(g) -33.56
Br2(l) 0.0 HgO(s) -58.539
Br2(g) 3.110 I2(s) 0.0
CaO(s) -604.03 I2(g) 19.327
CaCO3(s) -1128.79 KCl(s) -409.14
C--graphite 0.0 KBr(s) -380.66
C--diamond 2.9 MgO(s) -569.43
CH4(g) -50.72 MgH2(s) 76.1
C2H2(g) 209.2 NH3(g) -16.45
C2H4(g) 68.15 NO(g) 86.55
C2H6(g) -32.82 NO2(g) 51.31
C6H6(l) 124.5 N2O4(g) 97.89
CO(g) -137.168 NF3(g) -83.2
CO2(g) -394.359 NaCl(s) -384.138
CuO(s) -129.7 NaBr(s) -348.983
Fe2O3(s) -742.2 O3(g) 163.2
HBr(g) -53.45 SO2(g) -300.194
HCl(g) -95.299 SO3(g) -371.06
HI(g) 1.7 ZnO(s) -318.3



This table is used in exactly the same way as a table of standard enthalpies of formation. This type of table enables us to find ΔG values for any reaction occurring at 298 K and 1 atm pressure, provided only that all the substances involved in the reaction appear in the table. The two following examples illustrate such usage.



EXAMPLE 1 Determine whether the following reaction is spontaneous or not:


4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)      1 atm, 298 K


SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. Following exactly the same rules used for standard enthalpies of formation, we have

ΔGm° = ΣΔGf°(products) – ΣΔGf°(reactants)

         = 4ΔGf°(NO) + 6ΔGf°(H2O) – 4ΔGf°(NH3) – 5ΔGf°(O2)


Inserting values from the table of free energies of formation, we then find


ΔGm° = [4 × 86.7 + 6 × (–237.3) – 4 × (–16.7) – 5 × 0.0] kJ mol–1

         = – 1010 kJ mol–1


Since ΔGm° is very negative, we conclude that this reaction is spontaneous.



The reaction of NH3 with O2 is very slow, so that when NH3 is released into the air, no noticeable reaction occurs. In the presence of a catalystA substance that increases the rate of a chemical reaction but that undergoes no net change during the reaction., though, NH3 burns with a yellowish flame in O2. This reaction is very important industrially, since the NO produced from it can be reacted further with O2 and H2O to form HNO3:


2NO + 1½ O2 + H2O → 2 HNO3


Nitric acidIn Arrhenius theory, a substance that produces hydrogen ions (hydronium ions) in aqueous solution. In Bronsted-Lowry theory, a hydrogen-ion (proton) donor. In Lewis theory, a species that accepts a pair of electrons to form a covalent bond., HNO3 is used mainly in the manufacture of nitrate fertilizers but also in the manufacture of explosives.



EXAMPLE 2 Determine whether the following reaction is spontaneous or not:


2NO(g) + 2CO(g) → 2CO2(g) + N2 (g)     1 atm, 298 K


Solution Following previous procedure we have


ΔGm° = (–2 × 394.4 + 0.0 – 2 ×86.7 + 2 × 137.3) kJ mol–1

         = –687.6 kJ mol–1


The reaction is thus spontaneous.



This example is an excellent illustration of how useful thermodynamicsThe study of the conversion of energy from one form to another and the availability of energy to do work. can be. Since both NO and CO are air pollutants produced by the internal-combustionVigorous combination of a material with oxygen gas, usually resulting in a flame. engine, this reaction provides a possible way of eliminating both of them in one reaction, killing two birds with one stone. A thee-way catalytic converter is able to perform the equivalent of this reaction. The reductionThat part of a chemical reaction in which a reactant gains electrons; simultaneous oxidation of a reactant must occur. step coverts NOx to O2 and N2. Then, in the oxidationThat part of a chemical reaction in which a reactant loses electrons; simultaneous reduction of a reactant must occur. step, CO and O2 are converted to CO2. If ΔGm° had turned out be +695 kJ mol–1, the reaction would be nonspontaneous and there would be no point at all in developing such a device.[1]

We quite often encounter situations in which we need to know the value of ΔGm° for a reaction at a temperature other than 298 K. Although extensive thermodynamic tables covering a large range of temperatures are available, we can also obtain approximate values for ΔG from the relationship


ΔGm° = ΔHm° – T ΔSm°      (3)


If we assume, as we did previously, that neither ΔHm° nor ΔSm° varies much as the temperature changes from 298 K to the temperature in question, we can then use the values of ΔHm°(298 K) obtained from the Table of Some Standard Enthalpies of Formation at 25°C and ΔSm°(298 K) obtained from the Table of Standard Molar Entropies to calculate ΔGm° for the temperature in question.



EXAMPLE 3 Using the enthalpy values and the entropy values, calculate ΔHm° and ΔSm° for the reaction


CH4(g) + H2O (g) → 3H2 (g) + CO(g)      1 atm


Calculate an approximate value for ΔGm° for this reaction at 600 and 1200 K and determine whether the reaction is spontaneous at either temperature.


Solution From the tables we find


ΔHm°(298 K) = 3ΔHf°(H2) + ΔHf°(CO) – ΔHf° (CH4) – ΔHf° (H2O)

                   = ( 3 × 0.0 – 110.5 + 74.8 + 241.8) kJ mol–1 = +206.1 kJ mol–1


and similarly


ΔSm°(298 K) = (3 × 130.6 + 197.6 – 187.9 – 188.7) J K–1 mol–1 = + 212.8 J K–1 mol–1


At 600 K we estimate


ΔGm° = ΔH°(298 K) – T ΔS°(298 K)

         = 206.1 kJ mol–1 – 600 × 212.8 J mol–1

         = (206.1 – 127.7) kJ mol–1 = + 78.4 kJ mol–1


Since ΔG is positive, the reaction is not spontaneous at this temperature

At 1200 K by contrast


ΔGm° = 206.1 kJ mol–1 – 1200 × 212.8 J mol–1

         = (206.1 – 255.4) kJ mol–1 = –49.3 kJ mol–1


At this higher temperature, therefore, the reaction is spontaneous.

From more extensive tables we find that accurate values of the free-energy change are ΔGm°(600 K) = +72.6 kJ mol–1 and ΔGm°(1200 K) = –77.7 kJ mol–1. Our approximate value at 1200 K is thus about 50 percent in error. Nevertheless it predicts the right sign for ΔG, a result which is adequate for most purposes.


  1. Baird, C., Cann, M. Environmental Chemistry. 3rd edition. 2005. W. H. Freeman and Company. 83-85.