Gibbs Free Energy

Gibbs free energy, denoted G, combines enthalpy and entropy into a single value. The change in free energy, ΔG, is equal to the sum of the enthalpy plus the product of the temperature and entropy of the system. ΔG can 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 nonspontaneous (it requires  the input of 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. This quantity is the energy associated with a chemical reaction that can be used to do work, and is the sum of its enthalpy (H) and the product of the temperature and the entropy (S) of the system. This quantity is defined as follows:

$G= H-TS$

or more completely as

$G= U+PV-TS$

where

• U = internal energy (SI unit: joule)
• P = pressure (SI unit: pascal)
• V = volume (SI unit: $$m^3$$)
• 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, a variation on this equation is often encountered:

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

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

The sign of ΔG indicates the direction of a chemical reaction:

Terminology

• If $$\Delta H^o$$ < 0 and  $$\Delta S^o$$ > 0,  then the reaction is spontaneous ($$\Delta G^o$$ < 0 ) at any temperature.
• If $$\Delta H^o$$ > 0  and entropy $$\Delta S^o$$ < 0,  then the reaction is nonspontaneous ($$\Delta G^o$$ > 0 ) at any temperature.

Standard-state free energy of reaction ( $$\Delta G^o$$) is defined as the free energy of reaction at standard state conditions:

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

Standard-State Free Energy of Formation (ΔGfo)

• The partial pressure of any gas involved in the reaction is 0.1 MPa.
• The concentrations of all aqueous solutions are 1 M.
• Measurements are generally taken at a temperature of 25° C (298 K).

The standard-state free energy of formation is the change in free energy that occurs when a compound is formed from 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 constituent elements at standard-state conditions.

The standard-state free energy of reaction can be calculated from the standard-state free energies of formation.  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:

$\Delta G^o = \sum \Delta G^o_{f_{products}} - \sum \Delta G^o_{f_{reactants}}$

Free energy and equilibrium constants

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

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

• $$\Delta G$$ = free energy at any moment
• $$\Delta G^o$$ = 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:

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

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

Relating Free energy to Cell potentials

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

$E = E^o - \dfrac {RT}{nF} \ln Q$

with

• $$E$$ = 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:

$E = E^o - \dfrac {RT}{nF} \ln Q$

multiply the entire equation by $$nF$$

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

which is similar to:

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

By juxtaposing these two equations:

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

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

it can be concluded that:

$\Delta G = -nFE$

Therefore,

$\Delta G^o = -nFE^o$

Example 1

$NH_4NO_{3(s)} \overset{H_2O}{\longrightarrow} NH_{4(aq)}^+ + NO_{3(aq)}^-$

Calculate $$\Delta{H^o}$$, $$\Delta{S^o}$$, and $$\Delta{G^o}$$ for the above reaction to determine if the above reaction is spontaneous or not.

SOLUTION

These values can be substituted into the free energy equation

$T_K = 25\;^oC + 273.15K = 298.15\;K$

Example 2

Calculate $$\Delta{G}$$ for the following reaction at $$25\; ^oC$$. Will the reaction occur spontaneously?

$NH_{3(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)}$

given for the reaction

• $$\Delta{H} = -176.0 \;kJ$$
• $$\Delta{S} = -284.8\;J/K$$

SOLUTION

calculate $$\Delta{G}$$ from the formula

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

but first we need to convert the units for $$\Delta{S}$$ into kJ/K (or convert $$\Delta{H}$$ into J) and temperature into Kelvin

• $$\Delta{S} = -284.8 \cancel{J}/K \left( \dfrac{1\, kJ}{1000\; \cancel{J}}\right) = -0.0284.8\; kJ/K$$
• $$T=273.15\; K + 25\; ^oC = 298\;K$$

The definition of Gibbs energy can then be used directly

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

$\Delta{G} = -176.0 \;kJ - (298 \cancel{K}) (-0.0284.8\; kJ/\cancel{K})$

$\Delta{G} = -176.0 \;kJ - (-84.9\; kJ)$

$\Delta{G} = -91.1 \;kJ$

Yes, this reaction is spontaneous at room temperature since $$\Delta{G}$$ is negative.

09:32, 21 Mar 2015

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