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16.4 Free energy  (Page 5/12)

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The free energy change for a process taking place with reactants and products present under nonstandard conditions, Δ G , is related to the standard free energy change, Δ G °, according to this equation:

Δ G = Δ G ° + R T ln Q

R is the gas constant (8.314 J/K mol), T is the kelvin or absolute temperature, and Q is the reaction quotient. We may use this equation to predict the spontaneity for a process under any given set of conditions as illustrated in [link] .

Calculating δ G Under nonstandard conditions

What is the free energy change for the process shown here under the specified conditions?

T = 25 °C, P N 2 = 0.870 atm , P H 2 = 0.250 atm , and P NH 3 = 12.9 atm

2NH 3 ( g ) ⟶ 3H 2 ( g ) + N 2 ( g ) Δ G ° = 33.0 kJ/mol

Solution

The equation relating free energy change to standard free energy change and reaction quotient may be used directly:

Δ G = Δ G ° + R T ln Q = 33.0 kJ mol + ( 8.314 J mol K × 298 K × ln ( 0.250 3 ) × 0.870 12.9 2 ) = 9680 J mol or 9.68 kJ/mol

Since the computed value for Δ G is positive, the reaction is nonspontaneous under these conditions.

Check your learning

Calculate the free energy change for this same reaction at 875 °C in a 5.00 L mixture containing 0.100 mol of each gas. Is the reaction spontaneous under these conditions?

Answer:

Δ G = −136 kJ; yes

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For a system at equilibrium, Q = K and Δ G = 0, and the previous equation may be written as

0 = Δ G ° + R T ln K ( at equilibrium )
Δ G ° = − R T ln K or K = e − Δ G ° R T

This form of the equation provides a useful link between these two essential thermodynamic properties, and it can be used to derive equilibrium constants from standard free energy changes and vice versa. The relations between standard free energy changes and equilibrium constants are summarized in [link] .

Relations between Standard Free Energy Changes and Equilibrium Constants
K Δ G ° Comments
>1 <0 Products are more abundant at equilibrium.
<1 >0 Reactants are more abundant at equilibrium.
= 1 = 0 Reactants and products are equally abundant at equilibrium.

Calculating an equilibrium constant using standard free energy change

Given that the standard free energies of formation of Ag + ( aq ), Cl − ( aq ), and AgCl( s ) are 77.1 kJ/mol, −131.2 kJ/mol, and −109.8 kJ/mol, respectively, calculate the solubility product, K sp , for AgCl.

Solution

The reaction of interest is the following:

AgCl ( s ) ⇌ Ag + ( a q ) + Cl − ( a q ) K sp = [ Ag + ] [ Cl − ]

The standard free energy change for this reaction is first computed using standard free energies of formation for its reactants and products:

Δ G ° = Δ G 298 ° = [ Δ G f ° ( Ag + ( a q ) ) + Δ G f ° ( Cl − ( a q ) ) ] − [ Δ G f ° ( AgCl ( s ) ) ] = [ 77.1 kJ/mol − 131.2 kJ/mol ] − [ − 109.8 kJ/mol ] = 55.7 kJ/mol

The equilibrium constant for the reaction may then be derived from its standard free energy change:

K sp = e − Δ G ° R T = exp ( − Δ G ° R T ) = exp ( − 55.7 × 10 3 J/mol 8.314 J/mol·K × 298.15 K ) = exp ( − 22.470 ) = e − 22.470 = 1.74 × 10 −10

This result is in reasonable agreement with the value provided in Appendix J .

Check your learning

Use the thermodynamic data provided in Appendix G to calculate the equilibrium constant for the dissociation of dinitrogen tetroxide at 25 °C.

2NO 2 ( g ) ⇌ N 2 O 4 ( g )

Answer:

K = 6.9

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To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of Q ), equilibrium is established when the system’s free energy is minimized ( [link] ). If a system is present with reactants and products present in nonequilibrium amounts ( Q ≠ K ), the reaction will proceed spontaneously in the direction necessary to establish equilibrium.
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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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