# Gibbs 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. ΔG can also predict the direction of the chemical reaction under two conditions: (1) constant temperature and (2) constant pressure. If ΔG is positive, then the reaction is non-spontaneous (requires external energy to occur) and if it is negative, then it is spontaneous (occurs without external energy input).

### Introduction

In 1875, Josiah Gibbs introduced a thermodynamic quantity combining enthalpy and entropy into a single value called Gibbs free energy. It is the energy associated with a chemical reaction that can be used to do work and is the sum of its enthalpy (H) plus the product of the temperature (Kelvin) and the entropy (S) of the system. This quantity can be defined as:

$G= H-TS$

or more completely as

$G= U+PV-TS$

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)

### Free energy of reaction (ΔG)

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

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

$\text {change in free energy} \qquad \text {change in enthalpy} \qquad \text {(temperature) change in entropy}$

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

Terminology:

• If $$\Delta H^0$$ < 0 and  $$\Delta S^0$$ > 0  then the reaction will be SPONTANEOUS ($$\Delta G^0$$ < 0 ) at any temperature.
• If $$\Delta H^0$$ > 0  and entropy $$\Delta S^0$$ < 0  then the reaction will be NONSPONTANEOUS ($$\Delta G^0$$ > 0 ) at any temperature.

Standard-state free energy of reaction ( $$\Delta G^o$$)

• The free energy of reaction at standard state conditions:

$\Delta G^o = \Delta H^0 - T \Delta S^0$

### Standard-State Free Energy of Formation (ΔGf)

• 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:

$\Delta G^0 = \sum \Delta G^0_{f_{produts}} - \sum \Delta G^0_{f_{reactants}}$

### Free Energy and Equilibrium Constants

The following equation relates the standard-state free energy of reaction with the free energy of reaction at any moment in time during a reaction (not necessarily at standard-state conditions):

$\Delta G = \Delta G^0 + RT \ln Q$

• $$\Delta G$$ = free energy at any moment
• $$\Delta G^0$$ = standard-state free energy
• R is the ideal gas constant = 8.314 J/mol-K
• T is the absolute temperature (Kelvin)
• $$\ln Q$$ is natural logarithm of the reaction quotient

At equilibrium, ΔG = 0 and Q=K. Thus the equation can be arranged into:

$\Delta G^0 = - RT \ln K$

This equation is particuluar interesting as it relate the free energy difference under standard conditions to the properties of a system under equilibrium (which is very rarely at standard conditions).

### Free energy and Cell potentials

The Nernst equation relates the standard-state cell potential with the cell potential of the cell at any moment in time:

$E = E^0 - \frac {RT}{nF} \ln Q$

• E = cell potential in volts (Joules per Coulomb)
• n = moles of electrons
• F = Faraday's constant: 96,485 Coulombs per mole of electrons

By rearranging this equation we obtain:

$E = E^0 - \frac {RT}{nF} \ln Q \text {multiply the entire equation by nF}$

$nFE = nFE^0 - RT \ln Q$

which is similiar to:

$\Delta G = \Delta G^0 + RT \ln Q$

By showing these two equations:

$nFE = nFE^0 - RT \ln Q$

$\Delta G = \Delta G^0 + RT \ln Q$

We can conclude that:

$\Delta G = -nFE \text {and} \Delta G^0 = -nFE^0$

Example 1

Now, can plug in these values we've calculated into the free energy equation

### References

1. General Chemistry: Principles of Modern Applications 9th Edition (pages 792-795)
2. SAT Subject Test: Chemistry 6th Edition (page 66)
3. http://en.wikipedia.org/wiki/Gibbs_free_energy

08:43, 14 May 2014

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