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16.14: The Free Energy

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    In the previous section, we were careful to differentiate between the 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 enthalpy 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 energy, 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 temperature for which the tables apply. 

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

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

    This inequality must be true if a reaction occurring at constant pressure in surroundings at constant temperature is to be spontaneous. It is convenient to multiply this inequality by T; it then becomes 

    \[-\Delta H + T \Delta S > 0 \nonumber \]

    (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,

    \[\Delta H - T \Delta S < 0 \label{3} \]

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

    \[G=H-TS \nonumber \]

    When a chemical reaction occurs at constant temperature, the free energy will change from an initial value of G, given by \[G_{1}=H_{1}=TS_{1} \nonumber \]

    to a final value

    \[G_{2}=H_{2}-TS_{2} \nonumber \]

    The change in free energy ΔG will thus be \[\Delta G = G_{2} - G_{1} = H_{2} - H_{1} - T(S_{2}-S_{1}) \nonumber \]


    \[\Delta G = \Delta H - T\Delta S \nonumber \]

    Feeding this result back into inequality \(\ref{3}\) gives the result

    \[\Delta G = \Delta H - T \Delta S <0 \nonumber \]

    \[\Delta G < 0 \label{10} \]

    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. \(\ref{10}\) 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 substance 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 atm 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.

    Table \(\PageIndex{1}\): 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 \(\PageIndex{1}\): Spontaneous Reactions

    Determine whether the following reaction is spontaneous or not:

    \[\ce{4NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(l)}\qquad 1 \text{ atm, 298K} \nonumber \]


    Following exactly the same rules used for standard enthalpies of formation, we have

    \(\Delta G_{m}^{\circ} = \Sigma \Delta G_{f}^{\circ}\text{ (products)} - \Sigma \Delta G_{f}^{\circ} \text{ (reactants)}\)

    \(\qquad = \ce{4 \Delta G_{f}^{\circ}(NO) + 6 \Delta G_{f}^{\circ}(H2O) - 4\Delta G_{f}^{\circ}(NH3) - 5\Delta G_{f}^{\circ}(O2)}\)

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

    \(\Delta G_{m}^{\circ} = [4 \times 86.7 + 6 \times (-273.3) - 4 \times (-16.7) - 5 \times 0.0]\frac{ \text{kJ }}{\text{mol}}\)

    \(\qquad = -1010\frac{\text{ kJ}}{\text{ mol}}\)

    Since \(\Delta G_{m}^{\circ} \) 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 catalyst, 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:

    \[\ce{2NO + \frac{3}{2}O2 + H2O→2HNO3} \nonumber \]

    Nitric acid, HNO3 is used mainly in the manufacture of nitrate fertilizers but also in the manufacture of explosives.

    Example \(\PageIndex{2}\): Spontaneous Reactions

    Determine whether the following reaction is spontaneous or not:

    \[\ce{2NO(g) + 2CO(g) → 2CO2(g) + N2 (g)}\qquad 1 \text{ atm, 298K} \nonumber \]

    Solution Following previous procedure we have

    \((\Delta G_{m}^{\circ} = (-2\times 394.4 + 0.0 - 2 \times 86.7 + 2 \times 137.3)\frac{\text{ kJ}}{\text{ mol}}\)

    \(\qquad = -687.6\frac{ \text{ kJ}}{\text{ mol}}\)

    The reaction is thus spontaneous.

    This example is an excellent illustration of how useful thermodynamics can be. Since both NO and CO are air pollutants produced by the internal-combustion 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 reduction step coverts NOx to O2 and N2. Then, in the oxidation 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

    \[\Delta G_{m}^{\circ}=\Delta H_{m}^{\circ}-T\Delta S_{m}^{\circ} \nonumber \]

    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 \(\PageIndex{3}\): Spontaneous at Different Temperatures

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

    \[\ce{CH4(g) + H2O(g) → 3 H2(g) + CO(g)} \qquad 1 \text{ atm} \nonumber \]

    Calculate an approximate value for \(\Delta G_{m}^{º}\) for this reaction at 600 and 1200 K and determine whether the reaction is spontaneous at either temperature.

    Solution From the tables we find

    \(\Delta H_{m}^{\circ}(298 \text{ K})= 3\Delta H_{f}^{\circ}(\text{H}_{2})+\Delta H_{f}^{\circ}\text{(CO)} - \Delta H_{f}^{º}( \text{CH}_{4}) - \Delta H_{f}^{\circ}(\text{H}_{2}\text{O})\)

    \(\qquad = (3 \times 0.0 - 110.6 + 74.8 + 241.8) \frac{\text{ kJ}}{\text{ mol}} = +206.1\frac{\text{kJ}}{\text{ mol}}\)

    and similarly

    \(\Delta S_{m}^{\circ}(298K) = (3 \times 130.6 + 197.6 - 187.9 - 188.7)\frac{\text{ J}}{\text{ mol K}} = +212.8\frac{\text{ J}}{\text{ mol K}}\)

    At 600 K we estimate

    \(\Delta G_{m}^{\circ}=\Delta H^{\circ}(298 \text{ K}) - T\Delta S^{\circ}(298 \text{ K})\)

    \(\qquad =206.1\frac{ \text{ kJ}}{\text{ mol}} - 600 \times 212.8\frac{\text{ J}}{\text{ mol}}\)

    \(\qquad = (206.1 - 127.7)\frac{\text{kJ}}{\text{ mol}} = +78.4\frac{\text{ kJ}}{\text{ mol}}\)

    Since \(\Delta G\) is positive, the reaction is not spontaneous at this temperature 

    At 1200 K by contrast 

    \(\Delta G_{m}^{\circ}=206.1\frac{\text{ kJ}}{\text{ mol}} - 1200 \times 212.8\frac{\text{ J}}{\text{ mol}}\)

    \(\qquad = (206.1 - 255.4)\frac{\text{ kJ}}{\text{ mol}} = -49.3\frac{\text{ kJ}}{\text{ mol}}\)

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

    From more extensive tables we find that accurate values of the free-energy change are \(\Delta G_{m}^{\circ}(600 \text{ K}) = +72.6\frac{\text{ kJ}}{\text{ mol}}\) and \(\Delta G_{m}^{\circ}(1200 \text{ K})=-77.7\frac{\text{ kJ} }{\text{ mol}}\). Our approximate value at 1200 K is thus about 50 percent in error. Nevertheless it predicts the right sign for \(\Delta G\), 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.

    This page titled 16.14: The Free Energy is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn.