# 16.13: The Free Energy

- Page ID
- 49578

In the previous section, we were careful to differentiate between the entropy change occurring in the reaction system Δ*S*_{sys}, on the one hand, and the entropy change occurring in the surroundings, Δ*S*_{surr}, given by –Δ*H*/*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

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

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

$$-\Delta H + T \Delta S > 0$$

(From now on we will abandon the subscript "sys".) If –Δ*H* + *T* Δ*S* 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

$$G=H-TS$$

When a chemical reaction occurs at constant temperature, the free energy will change from an initial value of *G*, given by

$$G_{1}=H_{1}=TS_{1}$$

to a final value

$$G_{2}=H_{2}-TS_{2}$$

The change in free energy ΔG will thus be

$$\Delta G = G_{2} - G_{1} = H_{2} - H_{1} - T(S_{2}-S_{1})$$

or

$$\Delta G = \Delta H - T\Delta S $$

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

$$\Delta G = \Delta H - T \Delta S <0$$

$$\Delta G < 0 \label{10}$$

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 Δ*G* for the reaction using the tables. If Δ*G* 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 Δ*G _{f}*°. A table of values of Δ

*G*° (298 K) for a limited number of substances is given in the following table.

_{f}**Table** \(\PageIndex{1}\)** ***Some Standard Free Energies of Formation at 298.15 K (25°C)*

Compound |
ΔG_{f}^{o}/kJ mol^{-1} |
Compound |
ΔG_{f}^{o}/kJ mol^{-1} |

AgCl(s) |
-109.789 | H_{2}O(g) |
-228.572 |

AgN_{3}(s) |
591.0 | H_{2}O(l) |
-237.129 |

Ag_{2}O(s) |
-11.2 | H_{2}O_{2}(l) |
-120.35 |

Al_{2}O_{3}(s) |
-1582.3 | H_{2}S(g) |
-33.56 |

Br_{2}(l) |
0.0 | HgO(s) |
-58.539 |

Br_{2}(g) |
3.110 | I_{2}(s) |
0.0 |

CaO(s) |
-604.03 | I_{2}(g) |
19.327 |

CaCO_{3}(s) |
-1128.79 | KCl(s) |
-409.14 |

C--graphite | 0.0 | KBr(s) |
-380.66 |

C--diamond | 2.9 | MgO(s) |
-569.43 |

CH_{4}(g) |
-50.72 | MgH_{2}(s) |
76.1 |

C_{2}H_{2}(g) |
209.2 | NH_{3}(g) |
-16.45 |

C_{2}H_{4}(g) |
68.15 | NO(g) |
86.55 |

C_{2}H_{6}(g) |
-32.82 | NO_{2}(g) |
51.31 |

C_{6}H_{6}(l) |
124.5 | N_{2}O_{4}(g) |
97.89 |

CO(g) |
-137.168 | NF_{3}(g) |
-83.2 |

CO_{2}(g) |
-394.359 | NaCl(s) |
-384.138 |

CuO(s) |
-129.7 | NaBr(s) |
-348.983 |

Fe_{2}O_{3}(s) |
-742.2 | O_{3}(g) |
163.2 |

HBr(g) |
-53.45 | SO_{2}(g) |
-300.194 |

HCl(g) |
-95.299 | SO_{3}(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 Δ*G* 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:

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

**Solution** 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 NH_{3} with O_{2} is very slow, so that when NH_{3} is released into the air, no noticeable reaction occurs. In the presence of a catalyst, though, NH_{3} burns with a yellowish flame in O_{2}. This reaction is very important industrially, since the NO produced from it can be reacted further with O_{2} and H_{2}O to form HNO_{3}:

$$\ce{2NO + \frac{3}{2}O2 + H2O→2HNO3}$$

Nitric acid, HNO_{3} 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:

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

**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 NO_{x} to O_{2} and N_{2}. Then, in the oxidation step, CO and O_{2} are converted to CO_{2}. If Δ*G _{m}*° 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 Δ*G _{m}*° 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 Δ

*G*from the relationship

$$\Delta G_{m}^{\circ}=\Delta H_{m}^{\circ}-T\Delta S_{m}^{\circ} $$

If we assume, as we did previously, that neither Δ*H _{m}*° nor Δ

*S*° varies much as the temperature changes from 298 K to the temperature in question, we can then use the values of Δ

_{m}*H*°(298 K) obtained from the Table of Some Standard Enthalpies of Formation at 25°C and Δ

_{m}*S*°(298 K) obtained from the Table of Standard Molar Entropies to calculate Δ

_{m}*G*° for the temperature in question.

_{m}

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

Using the enthalpy values and the entropy values, calculate Δ*H _{m}*° and Δ

*S*° for the reaction

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

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.

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

### Contributors

Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.