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ChemWiki: The Dynamic Chemistry Hypertext > Physical Chemistry > Thermodynamics > State Functions > Free Energy > Gibbs Free Energy

Gibbs Free Energy

Skills to Develop

- To get an overview of Gibbs energy and its general uses in chemistry.
- Understand how Gibbs energy pertains to reactions properties
- Understand how Gibbs energy pertains to equilibria properties
- Understand how Gibbs energy pertains to electrochemical properties

Gibbs free energy, denoted \(G\), combines enthalpy and entropy into a single value. The change in free energy, \(\Delta G\), is equal to the sum of the enthalpy plus the product of the temperature and entropy of the system. \( \Delta G\) can predict the direction of the chemical reaction under two conditions:

- constant temperature and
- constant pressure.

If \(ΔG\) is positive, then the reaction is nonspontaneous (i.e., an the input of external energy is necessary for the reaction to occur) and if it is negative, then it is spontaneous (occurs without external energy input).

Gibbs energy was developed in the 1870’s by Josiah Willard Gibbs. He originally termed this energy as the “available energy” in a system. His paper published in 1873, “Graphical Methods in the Thermodynamics of Fluids,” outlined how his equation could predict the behavior of systems when they are combined. 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:

\[ G= H-TS \tag{1.1}\]

or more completely as

\[ G= U+PV-TS \tag{1.2}\]

where

- \(U\) is internal energy (SI unit: joule)
- \(P\) is pressure (SI unit: pascal)
- \(V\) is volume (SI unit: \(m^3\))
- \(T\)is temperature (SI unit: kelvin)
- \(S\) is entropy (SI unit: joule/kelvin)
- \(H\) is the enthalpy (SI unit: joule)

Spontaneous - is a reaction that is consider to be natural because it is a reaction that occurs by itself without any external action towards it. Non spontaneous - needs constant external energy applied to it in order for the process to continue and once you stop the external action the process will cease. When solving for the equation, if change of G is negative, then it's spontaneous. If change of G if positive, then it's non spontaneous. The symbol that is commonly used for FREE ENERGY is G. can be more properly consider as "standard free energy change"

In chemical reactions involving the changes in thermodynamic quantities, a variation on this equation is often encountered:

\[ \Delta G \qquad \qquad = \qquad \qquad \Delta H \qquad \qquad - \qquad \qquad T \Delta S \tag{1.3} \]

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

Example 1.1

Calculate ∆G at 290 K for the following reaction:

\[2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)}\]

Given

- ∆H = -120 kJ
- ∆S = -150 JK
^{ -1}

**SOLUTION**

now all you have to do is plug in all the given numbers into the above equation that was introduced earlier. expect you have to convert \(\Delta S\) so the same unit was \(\Delta H\).

\[\Delta S = -150 \cancel{J}/K \left( \dfrac{1\; kJ}{1000\;\cancel{J}} \right) = -0.15\; kJ/K\]

looks like:

\[∆G= -120\; kJ - (290 \;\cancel{K})(-0.150\; kJ/\cancel{K})\]

\[∆G= -120 \;kJ + 43 \;kJ\]

\[∆G= -77\; kJ\]

Exercise 1.1

What is the \(\Delta G\) for this formation of ammonia from nitrogen and hydrogen gas.

\[N_2 + 3H_2 \rightleftharpoons 2NH_3\]

The Standard free energy formations: NH_{3} =-16.45 H_{2}=0 N_{2}=0

**Answer**

\[\Delta G=-49.35\;kJ \;mol^{-1}\]

Since the changes of entropy of chemical reaction are not measured readily, thus, entropy is not typically used as a criterion. To obviate this difficulty, we can use \(G\). The sign of ΔG indicates the direction of a chemical reaction and determine if a reaction is spontaneous or not.

- \( \Delta G < 0 \): reaction is spontaneous in the direction written (i.e., the reaciton is
**exergonic**) - \( \Delta G =0 \): the system is at equilibrium and there is no net change either in forward or reverse direction.
- \( \Delta G > 0 \): reaction is not spontaneous and the process proceeds spontaneously in the reserve direction. To drive such a reaction, we need to have input of free energy (i.e., the reaction is
**endergonic**)

The factors affect \( \Delta G \)of a reaction (assume \( \Delta H \) and \( \Delta S \) are independent of temperature):

\(\Delta H\) | \(\Delta S\) | \(\Delta G\) | Example |
---|---|---|---|

+ | + | at low temperature: + , at high temperature: - | 2HgO(s) -> 2Hg (l) + O_{2} (g) |

+ | - | at all temperature: + | 3O_{2} (g) ->2O_{3} (g) |

- | + | at all temperature: - | 2H_{2}O_{2} (l) -> 2H_{2}O (l) + O_{2} (g) |

- | - | at low temperature: - , at high temperature: + | NH_{3} (g) + HCl (g) -> NH_{4}Cl (s) |

Note:

- \( \Delta G \) depends only on the difference in free energy of products and reactants (or final state and initial state). \( \Delta G \) is independent of the path of the transformation and is unaffected by the mechanism of a reaction.
- \( \Delta G \) cannot tell us anything about the rate of a reaction.

The standard Gibbs energy change \( \Delta G^o \) (at which reactants are converted to products at 1 bar) for:

\[ aA + bB \rightarrow cC + dD \tag{1.4}\]

\[ \Delta r G^o = c \Delta _fG^o (C) + d \Delta _fG^o (D) - a \Delta _fG^o (A) - b \Delta _fG^o (B) \tag{1.5}\]

\[\Delta _fG^0 = \sum v \Delta _f G^0 (\text {products}) - \sum v \Delta _f G^0 (\text {reactants}) \tag{1.6}\]

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

\[ \Delta G^o = \Delta H^o - T \Delta S^o \tag{1.7}\]

Note

- If \( \left | \Delta H \right | >> \left | T\Delta S \right |\): the reaction is
*enthalpy-driven* - If \( \Delta H \) << \( T\Delta S \): the
*reaction is entropy-driven*

- 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:

\[ \Delta G^o = \sum \Delta G^o_{f_{products}} - \sum \Delta G^o_{f_{reactants}} \tag{1.8}\]

Example 1.2

Used the below information to determine if \(NH_4NO_{3(s)}\) will dissolve in water at room temperature.

Compound | \(\Delta H_f^o\) | \(\Delta S_f^o\) |

\(NH_4NO_{3(s)}\) | -365.56 | 151.08 |

\(NH^+_{4(aq)}\) | -132.51 | 113.4 |

\(NO_{3(aq)}^-\) | 205.0 | 146.4 |

**SOLUTION**

This question is essentially asking if the following reaction is spontaneous at room temperature.

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

This would normally only require calculating \(\Delta{G^o}\) and evaluating its sign. However, the \(\Delta{G^o}\) values are not tabulated, so they must be calculated manually from calculated \(\Delta{H^o}\) and \(\Delta{S^o}\) values for the reaction.

*Calculate \(\Delta{H^o}\):*

\[ \Delta H^o = \sum n\Delta H^o_{f_{products}} - \sum m\Delta H^o_{f_{reactants}} \]

\[ \Delta H^o= \left[ \left( 1\; mol\; NH_3\right)\left(-132.51\;\dfrac{kJ}{mol} \right) + \left( 1\; mol\; NO_3^- \right) \left(-205.0\;\dfrac{kJ}{mol}\right) \right] \]

\[- \left[ \left(1\; mol\; NH_4NO_3 \right)\left(-365.56 \;\dfrac{kJ}{mol}\right) \right]\]

\[ \Delta H^o = -337.51 \;kJ + 365.56 \; kJ= 28.05 \;kJ\]

*Calculate \(\Delta{S^o}\):*

\[ \Delta S^o = \sum n\Delta S^o_{f_{products}} - \sum S\Delta H^o_{f_{reactants}} \]

\[ \Delta S^o= \left[ \left( 1\; mol\; NH_3\right)\left(113.4 \;\dfrac{J}{mol\;K} \right) + \left( 1\; mol\; NO_3^- \right) \left(146.6\;\dfrac{J}{mol\;K}\right) \right] \]

\[- \left[ \left(1\; mol\; NH_4NO_3 \right)\left(151.08 \;\dfrac{J}{mol\;K}\right) \right]\]

\[ \Delta S^o = 259.8 \;J/K - 151.08 \; J/K= 108.7 \;J/K\]

*Calculate \(\Delta{G^o}\):*

These values can be substituted into the free energy equation

\[T_K = 25\;^oC + 273.15K = 298.15\;K\]

\[\Delta{S^o} = 108.7\; \cancel{J}/K \left(\dfrac{1\; kJ}{1000\;\cancel{J}} \right) = 0.1087 \; kJ/K\]

\[\Delta{H^o} = 28.05\;kJ\]

Plug in \(\Delta H^o\), \(\Delta S^o\) and \(T\) into Equation 1.7

\[\Delta G^o = \Delta H^o + T \Delta S^o\]

\[\Delta G^o = 28.05\;kJ - (298.15\; \cancel{K})(0.1087\;kJ/ \cancel{K})\]

\[\Delta G^o= 28.05\;kJ - 32.41\; kJ\]

\[\Delta G^o = -4.4 \;kJ\]

This reaction is spontaneous at room temperature since \(\Delta G^o\) is negative. Therefore \(NH_4NO_{3(s)}\) will dissolve in water at room temperature.

Example 1.3

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

- \(\Delta{S} = -284.8 \cancel{J}/K \left( \dfrac{1\, kJ}{1000\; \cancel{J}}\right) = -0.284.8\; kJ/K\)
- \(T=273.15\; K + 25\; ^oC = 298\;K\)

The definition of Gibbs energy can then be used directly

\[\Delta{G} = \Delta{H} - T\Delta{S} \]

\[\Delta{G} = -176.0 \;kJ - (298 \cancel{K}) (-0.284.8\; kJ/\cancel{K}) \]

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

\[\Delta{G} = -91.1 \;kJ \]

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

Let's consider the following reversible reaction:

\[ A + B \leftrightharpoons C + D \tag{1.9}\]

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):

\[ \Delta G = \Delta G^o + RT \ln Q \tag{1.10}\]

- \(\Delta G\) = 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:

\[\Delta{G} = \Delta{G}^o + RT \ln \dfrac{[C][D]}{[A][B]} \tag{1.11}\]

with

- \(\Delta{G}^o\) = standard free energy change
- \(R\) = gas constant = 1.98 * 10
^{-3}kcal mol^{-1}deg^{-10} - \(T\) = is usually room temperature = 298 K
- \(K = \dfrac{[C][D]}{[A][B]}\)

The Gibbs free energy \(\Delta{G}\) depends primarily on the reactants' nature and concentrations (expressed in the \(\Delta{G}^o\) term and the logarithmic term of Equation 1.11, respectively).

At equilibrium, \(\Delta{G} = 0\): no driving force remains

\[0 = \Delta{G}^{o'} + RT \ln \dfrac{[C][D]}{[A][B]} \tag{1.12}\]

\[\Delta{G}^{o} = -RT \ln\dfrac{[C][D]}{[A][B]} \tag{1.13}\]

The equilibrium constant is defined as

\[K_{eq} = \dfrac{[C][D]}{[A][B]} \tag{1.14}\]

When \(K_{eq}\) is large, almost all reactants are converted to products. Substituting \(K_{eq}\) into Equation 1.14, we have:

\[\Delta{G}^{o} = -RT \ln K_{eq} \tag{1.15}\]

or

\[\Delta{G}^{o} = -2.303RT log_{10} K_{eq} \tag{1.16}\]

Rearrange,

\[K_{eq} = 10^{-\Delta{G}^{o}/(2.303RT)} \tag{1.17}\]

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 rarely at standard conditions).

\(K_{eq}\) | \(\Delta{G_o}\; (kcal/mole)\) |

\(10^{-5}\) | 6.82 |

\(10^{-4}\) | 5.46 |

\(10^{-3}\) | 4.09 |

\(10^{-2}\) | 2.73 |

\(10^{-1}\) | 1.36 |

1 | 0 |

\(10^{1}\) | -1.36 |

\(10^{2}\) | -2.73 |

\(10^{3}\) | -4.09 |

\(10^{4}\) | -5.46 |

\(10^{5}\) | -6.82 |

Example 1.4

What is \(\Delta{G}^{o}\) for isomerization of dihydroxyacetone phosphate to glyceraldehyde 3-phosphate?

If at equilibrium, we have \(K_{eq} = 0.0475\) at 298 K and pH 7. We can calculate:

\[\Delta{G}^{o} = -2.303\;RT log_{10} K_{eq}= (-2.303) * (1.98 * 10^{-3}) * 298 * (log_{10} 0.0475) = 1.8 \;kcal/mol\]

Given:

- The initial concentration of dihydroxyacetone phosphate = \(2 \times 10^{-4}\; M\)
- The initial concentration of glyceraldehyde 3-phosphate = \(3 \times 10^{-6}\; M\)

**SOLUTION**

From equation 2:

\(\Delta{G}\) = 1.8 kcal/mol + 2.303 RT log_{10}(3*10^{-6} M/2*10^{-4} M) = -0.7 kcal/mol

Note

Under non-standard conditions (which is essential all reactions), the spontaneity of reaction is determined by \(\Delta{G}\), not \(\Delta{G}^{o'}\).

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

\[ E = E^o - \dfrac {RT}{nF} \ln Q \tag{1.18}\]

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:

\[ E = E^o - \dfrac {RT}{nF} \ln Q \tag{1.19} \]

multiply the entire equation by \(nF\)

\[ nFE = nFE^o - RT \ln Q \tag{1.20}\]

which is similar to:

\[ \Delta G = \Delta G^o + RT \ln Q \tag{1.21}\]

By juxtaposing these two equations:

\[ nFE = nFE^o - RT \ln Q \tag{1.22}\]

\[ \Delta G = \Delta G^o + RT \ln Q \tag{1.23}\]

it can be concluded that:

\[ \Delta G = -nFE \tag{1.24}\]

Therefore,

\[ \Delta G^o = -nFE^o \tag{1.25}\]

- Free Energy is not necessarily "free": The appellation “free energy” for G has led to so much confusion that many scientists now refer to it simply as the Gibbs energy. The “free” part of the older name reflects the steam-engine origins of thermodynamics with its interest in converting heat into work: ΔG is the maximum amount of energy which can be “freed” from the system to perform useful work. By "useful", we mean work other than that which is associated with the expansion of the system. This is most commonly in the form of electrical work (moving electric charge through a potential difference), but other forms of work (osmotic work, increase in surface area) are also possible.
- Free Energy is not energy: A much more serious difficulty with the Gibbs function, particularly in the context of chemistry, is that although G has the units of energy (joules, or in its intensive form, J mol
^{–1}), it lacks one of the most important attributes of energy in that it is not conserved. Thus although the free energy always falls when a gas expands or a chemical reaction takes place spontaneously, there need be no compensating increase in energy anywhere else. Referring to G as an energy also reinforces the false but widespread notion that a fall in energy must accompany any change. But if we accept that energy is conserved, it is apparent that the only necessary condition for change (whether the dropping of a weight, expansion of a gas, or a chemical reaction) is the redistribution of energy.The quantity –ΔG associated with a process represents the quantity of energy that is “shared and spread”, which as we have already explained is the meaning of the increase in the entropy. The quotient –ΔG/T is in fact identical with ΔS_{total}, the entropy change of the world, whose increase is the primary criterion for any kind of change. - Free Energy is not even "real": G differs from the thermodynamic quantities H and S in another significant way: it has no physical reality as a property of matter, whereas H and S can be related to the quantity and distribution of energy in a collection of molecules (e.g., the third law of thermodynamics). The free energy is simply a useful construct that serves as a criterion for change and makes calculations easier.

- Chang, Raymond.
*Physical Chemistry for the Biosciences*. Sansalito, CA: University Sciences, 2005. - Atkins, Peter and de Paula, Julio.
*Physical Chemistry for the Life Sciences*. New York, NY: W. H. Freeman and Company, 2006. Page 153-163, 286. - Stryer, Lubert.
*Biochemistry (Third Edition)*. New York, NY: W.H. Freeman and Company, 1988. Page 181-184.

- Cathy Doan (UC Davis), Han Le (UC Davis)

- Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook

Last modified

04:27, 19 Jul 2015

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