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Three graphs, labeled, “a,” “b,” and “c” are shown where the y-axis is labeled, “Gibbs free energy ( G ),” and, “G superscript degree sign ( reactants ),” while the x-axis is labeled, “Reaction progress,” and “Reactants,” on the left and, “Products,” on the right. In graph a, a line begins at the upper left side and goes steadily down to a point about halfway up the y-axis and two thirds of the way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is slightly higher than halfway up the y-axis. The distance between the beginning and ending points of the graph is labeled as, “delta G less than 0,” while the lowest point on the graph is labeled, “Q equals K greater than 1.” In graph b, a line begins at the middle left side and goes steadily down to a point about two fifths up the y-axis and one third of the way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is near the top of the y-axis. The distance between the beginning and ending points of the graph is labeled as, “delta G greater than 0,” while the lowest point on the graph is labeled, “Q equals K less than 1.” In graph c, a line begins at the upper left side and goes steadily down to a point near the bottom of the y-axis and half way on the x-axis, then rises again to a point labeled, “G superscript degree sign ( products ),” that is equal to the starting point on the y-axis which is labeled, “G superscript degree sign ( reactants ).” The lowest point on the graph is labeled, “Q equals K equals 1.” At the top of the graph is the label, “Delta G superscript degree sign equals 0.”
These plots show the free energy versus reaction progress for systems whose standard free changes are (a) negative, (b) positive, and (c) zero. Nonequilibrium systems will proceed spontaneously in whatever direction is necessary to minimize free energy and establish equilibrium.

Key concepts and summary

Gibbs free energy ( G ) is a state function defined with regard to system quantities only and may be used to predict the spontaneity of a process. A negative value for Δ G indicates a spontaneous process; a positive Δ G indicates a nonspontaneous process; and a Δ G of zero indicates that the system is at equilibrium. A number of approaches to the computation of free energy changes are possible.

Key equations

  • Δ G = Δ H T Δ S
  • Δ G = Δ G ° + RT ln Q
  • Δ G ° = − RT ln K

Chemistry end of chapter exercises

What is the difference between Δ G , Δ G °, and Δ G 298 ° for a chemical change?

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A reactions has Δ H 298 ° = 100 kJ/mol and Δ S 298 ° = 250 J/mol·K. Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?

The reaction is nonspontaneous at room temperature.
Above 400 K, Δ G will become negative, and the reaction will become spontaneous.

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Explain what happens as a reaction starts with Δ G <0 (negative) and reaches the point where Δ G = 0.

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Use the standard free energy of formation data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.

(a) MnO 2 ( s ) Mn ( s ) + O 2 ( g )

(b) H 2 ( g ) + Br 2 ( l ) 2HBr ( g )

(c) Cu ( s ) + S ( g ) CuS ( s )

(d) 2LiOH ( s ) + CO 2 ( g ) Li 2 CO 3 ( s ) + H 2 O ( g )

(e) CH 4 ( g ) + O 2 ( g ) C ( s , graphite ) + 2H 2 O ( g )

(f) CS 2 ( g ) + 3Cl 2 ( g ) CCl 4 ( g ) + S 2 Cl 2 ( g )

(a) 465.1 kJ nonspontaneous; (b) −106.86 kJ spontaneous; (c) −53.6 kJ spontaneous; (d) −83.4 kJ spontaneous; (e) −406.7 kJ spontaneous; (f) −30.0 kJ spontaneous

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Use the standard free energy data in Appendix G to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions.

(a) C ( s , graphite ) + O 2 ( g ) CO 2 ( g )

(b) O 2 ( g ) + N 2 ( g ) 2NO ( g )

(c) 2Cu ( s ) + S ( g ) Cu 2 S ( s )

(d) CaO ( s ) + H 2 O ( l ) Ca ( OH ) 2 ( s )

(e) Fe 2 O 3 ( s ) + 3CO ( g ) 2Fe ( s ) + 3CO 2 ( g )

(f) CaSO 4 · 2H 2 O ( s ) CaSO 4 ( s ) + 2H 2 O ( g )

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Given:
P 4 ( s ) + 5O 2 ( g ) P 4 O 10 ( s ) Δ G 298 ° = −2697.0 kJ/mol
2H 2 ( g ) + O 2 ( g ) 2H 2 O ( g ) Δ G 298 ° = −457.18 kJ/mol
6H 2 O ( g ) + P 4 O 10 ( g ) 4H 3 PO 4 ( l ) Δ G 298 ° = −428.66 kJ/mol

(a) Determine the standard free energy of formation, Δ G f ° , for phosphoric acid.

(b) How does your calculated result compare to the value in Appendix G ? Explain.

(a) −1124.3 kJ/mol for the standard free energy change. (b) The calculation agrees with the value in Appendix G because free energy is a state function (just like the enthalpy and entropy), so its change depends only on the initial and final states, not the path between them.

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Is the formation of ozone (O 3 ( g )) from oxygen (O 2 ( g )) spontaneous at room temperature under standard state conditions?

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Consider the decomposition of red mercury(II) oxide under standard state conditions.
2HgO ( s , red ) 2Hg ( l ) + O 2 ( g )

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

(a) The reaction is nonspontaneous; (b) Above 566 °C the process is spontaneous.

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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