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5.8 Free energy  (Page 2/12)

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Free energy changes may also use the standard free energy of formation ( Δ G f ° ) , for each of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpies of formation, Δ G f ° is by definition zero for elemental substances under standard state conditions. The approach to computing the free energy change for a reaction using this approach is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction

m A + n B ⟶ x C + y D ,

the standard free energy change at room temperature may be calculated as

Δ G 298 ° = Δ G ° = ∑ ν Δ G 298 ° ( products ) − ∑ ν Δ G 298 ° ( reactants ) = [ x Δ G f ° ( C ) + y Δ G f ° ( D ) ] − [ m Δ G f ° ( A ) + n Δ G f ° ( B ) ] .

Calculation of Δ G 298 °

Consider the decomposition of yellow mercury(II) oxide.

HgO ( s , yellow ) ⟶ Hg ( l ) + 1 2 O 2 ( g )

Calculate the standard free energy change at room temperature, Δ G 298 ° , using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?

Solution

The required data are available in Appendix G and are shown here.

Compound Δ G f ° (kJ/mol) Δ H f ° (kJ/mol) S 298 ° (J/K·mol)
HgO ( s , yellow) −58.43 −90.46 71.13
Hg( l ) 0 0 75.9
O 2 ( g ) 0 0 205.2

(a) Using free energies of formation:

Δ G 298 ° = ∑ ν G S 298 ° (products) − ∑ ν Δ G 298 ° (reactants)
= [ 1 Δ G 298 ° Hg ( l ) + 1 2 Δ G 298 ° O 2 ( g ) ] − 1 Δ G 298 ° HgO ( s , yellow )
= [ 1 mol (0 kJ/mol) + 1 2 mol(0 kJ/mol) ] − 1 mol(−58.43 kJ/mol) = 58.43 kJ/mol

(b) Using enthalpies and entropies of formation:

Δ H 298 ° = ∑ ν Δ H 298 ° (products) − ∑ ν Δ H 298 ° (reactants)
= [ 1 Δ H 298 ° Hg ( l ) + 1 2 Δ H 298 ° O 2 ( g ) ] − 1 Δ H 298 ° HgO ( s , yellow )
= [ 1 mol ( 0 kJ/mol ) + 1 2 mol ( 0 kJ/mol ) ] − 1 mol ( −90.46 kJ/mol ) = 90.46 kJ/mol
Δ S 298 ° = ∑ ν Δ S 298 ° (products) − ∑ ν Δ S 298 ° (reactants)
= [ 1 Δ S 298 ° Hg ( l ) + 1 2 Δ S 298 ° O 2 ( g ) ] − 1 Δ S 298 ° HgO ( s , yellow )
= [ 1 mol ( 75.9 J/mol K ) + 1 2 mol ( 205.2 J/mol K ) ] − 1 mol ( 71.13 J/mol K ) = 107.4 J/mol K
Δ G ° = Δ H ° − T Δ S ° = 90.46 kJ − 298.15 K × 107.4 J/K·mol × 1 kJ 1000 J
Δ G ° = ( 90.46 − 32.01 ) kJ/mol = 58.45 kJ/mol

Both ways to calculate the standard free energy change at 25 °C give the same numerical value (to three significant figures), and both predict that the process is nonspontaneous ( not spontaneous) at room temperature.

Check your learning

Calculate Δ G ° using (a) free energies of formation and (b) enthalpies of formation and entropies ( Appendix G ). Do the results indicate the reaction to be spontaneous or nonspontaneous at 25 °C?

C 2 H 4 ( g ) ⟶ H 2 ( g ) + C 2 H 2 ( g )

Answer:

−141.5 kJ/mol, nonspontaneous

Got questions? Get instant answers now!

Temperature dependence of spontaneity

As was previously demonstrated in this chapter’s section on entropy, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered:
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Source:  OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13
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