CHM1 12 Free Energy Collection

Free Energy

Gibbs free energy combines enthalpy and entropy into a single value. The change of free energy is equal to the sum of its enthalpy plus the product of the temperature and entropy of the system. `DeltaG` can also predict the direction of the chemical reaction. If `DeltaG` is positive, then the reaction is non-spontaneous. If it is negative, then it is spontaneous.

Introduction

In 1875, Josiah Gibbs introduced a thermodynamic quantity combining enthalpy and entropy into a single value called Gibbs free energy. This quantity can be defined as:

`G = H - TS`

`G = U + PV - TS`

where

• `U` = internal energy (SI unit: joule)
• `P` = pressure (SI unit: pascal)
• `V` = volume (SI unit: m3)
• `T` = temperature (SI unit: kelvin)
• `S` = entropy (SI unit: joule/kelvin)
• `H` = enthalpy (SI unit: joule)

Gibbs Free Energy (`G`) - The energy associated with a chemical reaction that can be used to do work.  The free energy of a system is the sum of its enthalpy (H) plus the product of the temperature (Kelvin) and the entropy (S) of the system:

`G=H-TS`

Free energy of reaction (`DeltaG`)

In chemical reactions involving the changes in thermodynamic quantities we often use another variation of this equation:

`DeltaG = DeltaH - TDeltaS`

The sign of delta G can allow us to predict the direction of a chemical reaction.

Hence, there are two factors affect the change in free energy `DeltaG`:

• `DeltaU` = the change in internal energy
• `DeltaH` = the change in entropy of the system

These factors are contrast to each other, which is a important criterion to determine a reaction is spontaneous or not.

• `DeltaG<0`: reaction is spontaneous in the direction written. (reactant-favored)
• `DeltaG=0`: the system is at equilibrium and there is no net change either in forward or reverse direction.
• `DeltaG>0`: reaction is not spontaneous and the process proceeds spontaneously in the reserse direction. To drive such a reaction, we need to have input of free energy. (product-favored)

Terminology:

• If `DeltaH_0 < 0` and  `DeltaS_0 > 0`  then the reaction will be SPONTANEOUS (`DeltaG_0 < 0` ) at any temperature.
• If `DeltaH_0 > 0`  and entropy `DeltaS_0 < 0`  then the reaction will be NONSPONTANEOUS (`DeltaG_0 > 0` ) at any temperature.

Standard-state free energy of reaction (`DeltaG`)

• The free energy of reaction at standard state conditions:

`DeltaG_0=DeltaH_0DeltaTDeltaS_0`

Standard-State Free Energy of Formation (`DeltaG_f`)

• The partial pressures of any gases involved in the reaction is 0.1 MPa.
• The concentrations of all aqueous solutions are 1 M.

Measurements are also generally taken at a temperature of 25C (298 K). The change in free energy that occurs when a compound is formed form its elements in their most thermodynamically stable states at standard-state conditions.  In other words, it is the difference between the free energy of a substance and the free energies of its elements in their most thermodynamically stable states at standard-state conditions.

The standard-state free energy of reaction can be calculated from the standard-state free energies of formation as well.  It is the sum of the free energies of formation of the products minus the sum of the free energies of formation of the reactants:

`DeltaG° = sum DeltaG_"f products"^° - sum DeltaG_"f reactants"^°`

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 `DeltaG` for the reaction using the tables. If `DeltaG` 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 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 `DeltaG_f^?`. A table of values of `DeltaG_f^?` (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	`DeltaG_f^° "/"kJ mol^(-1)`	Compound	`DeltaG_f^° "/"kJ mol^(-1)`
`AgCl(s)`	-109.789	`H_2O(g)`	-228.572
`AgN_3(s)`	591.0	`H_2O(l)`	-237.129
`Ag_2O(s)`	-11.2	`H_2O_2(l)`	-120.35
`Al_2O_3(s)`	-1582.3	`H_2S(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_2H_2(g)`	209.2	`NH_3(g)`	-16.45
`C_2H_4(g)`	68.15	`NO(g)`	86.55
`C_2H_6(g)`	-32.82	`NO_2(g)`	51.31
`C_6H_6(l)`	124.5	`N_2O_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_2O_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 `DeltaG` 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 13 Determine whether the following reaction is spontaneous or not:

`4NH_3(g) + 5O_2(g) -> 4NO(g) + 6H_2O(l)`      1 atm, 298 K 

Solution:

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

`DeltaG_m^° = sum DeltaG_"f products"^° - sum DeltaG_"f reactants"^°`

`= 4 DeltaG_f^? (NO) + 6 DeltaG_f^? (H_2O) - 4DeltaG_f^?(NH_3) - 5DeltaG_f^?(O_2)`

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

`DeltaG_m^? = [4*86.7 + 6 * (-273.3) - 4 * (-16.7) - 5 * 0.0] kJ"/"mol`

`= -1010 kJ"/"mol`

Since `DeltaG_m^?` 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_2O` to form `HNO_3`:

`2NO + 1 1"/"2 O_2 + H_2O -> 2 HNO_3`

Nitric acid, `HNO_3` is used mainly in the manufacture of nitrate fertilizers but also in the manufacture of explosives.

EXAMPLE 14 Determine whether the following reaction is spontaneous or not:

`2NO(g) + 2CO(g) -> 2CO_2(g) + N_2 (g)`     1 atm, 298 K 

Solution:

Following the previous procedure, we have

`DeltaG_f^? = (-2 * 394.4 + 0.0 - 2 * 86.7 + 2 * 137.3) kJ"/"mol`

`= 687.6 kJ"/"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 `DeltaG_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 `DeltaG_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 `DeltaG` from the relationship

`DeltaG_m^? = DeltaH_m^? – T DeltaS_M^?`     (3)

If we assume, as we did previously, that neither `DeltaH_m^?` nor `DeltaS_M^?` varies much as the temperature changes from 298 K to the temperature in question, we can then use the values of `DeltaH_M^?`(298 K) obtained from a Table of Some Standard Enthalpies of Formation at 25°C and `DeltaS_m^?`(298 K) obtained from a Table of Standard Molar Entropies to calculate `DeltaG_m^?` for the temperature in question.

EXAMPLE 15 Using the enthalpy values and the entropy values given previously in this text, calculate `DeltaH_M^?` and `DeltaS_M^?` for the reaction

`CH_4(g) + H_2O (g) -> 3H_2 (g) + CO(g)`      1 atm

Calculate an approximate value for `DeltaG_m^?` for this reaction at 600 and 1200 K and determine whether the reaction is spontaneous at either temperature. 

Solution:

From the tables we can find,

`DeltaH_m^? (298K) = 3DeltaH_f^? (H_2) + DeltaH_f^? (CO) - DeltaH_f^?(CH_4) - DeltaH_f^? (H_2O)`

`= (3 * 0.0 - 110.6 + 74.8 + 241.8) kJ"/"mol = +206.1 kJ"/"mol`

and similarly

`DeltaS_m^? (298K) = (3 * 130.6 + 197.6 - 187.9 - 188.7) J"/"mol K = +212.8 J"/"mol K`

At 600K we estimate,

`Delta G_m^? = DeltaH? *298K) - TDeltaS?(298K)`

`= 206.1 kJ"/"mol - 600 * 212.8 J"/"mol`

`= (206.1 - 127.7) kJ"/"mol = +78.4 kJ"/"mol`

Since `DeltaG` is positive, reaciton is not spontaneous at this temperature

At 1200K by contrast,

`DeltaG_M^? = 206.1 kJ"/"mol - 1200 * 212.8 J"/"mol`

`= (206.1 - 255.4) kJ"/"mol = -49.3 kJ"/"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 `DeltaG_m^?(600K) = +72.6 kJ"/"mol` and `DeltaG_m^? (1200K) = -77.7 kJ"/"mol`. Our most approximate value at 1200K is thus about 50 percent in error. Nevertheless it predicts the right sing for `DeltaG`, a result which is adequate for most purposes.