18.3.2 Standard Free Energy Changes of Formation

Question 1

Standard Free Energy Changes of Formation

Answer 1

• The standard Gibbs free energy of formation (ΔG° f ) is the free energy change for a reaction producing one mole of a substance from the elements, with all components in standard state.
• As temperature changes, ΔG also changes.

Question 2

note

Answer 2

• The standard Gibbs free energy of reaction (ΔG° rxn ) is the difference between the standard Gibbs free energy of formation (ΔG° f ) of the products and the standard Gibbs free energy of formation of the reactants.
• ΔG° f is the free energy change for a reaction producing one mole of a substance from the elements, with all components in standard state. ΔG° f can be found by plugging values for ΔH° f and absolute values of entropy (S°) into the equation ΔG = ΔH – TΔS.
• The spontaneity of a reaction can be determined using standard Gibbs free energy of reaction calculations. If the free energy change is less than zero, the reaction is spontaneous.
• ΔG° f values for products and reactants are determined by plugging in known values of ΔH° f , S°, and T into ΔH° f – TΔS = ΔG° f .
• ΔG° of the reaction is found by subtracting ΔG° f of reactants from ΔG° f of products. A multiplier of 2 is used for reactants because there are 2 reactant molecules for every one product molecule.
• The negative sign on ΔG° rxn indicates that the reaction is spontaneous and proceeds towards products.
• Since ΔG is sensitive to temperature, ΔG at 350 K differs from ΔG at standard state (298 K, 1 atm) although the ΔH° f and the ΔS° remain the same.
• Entropy values (S°) do differ with temperature, but the difference between the products and reactants remains the same. However, since S° is multiplied by temperature, and temperature changes, ΔG° f also changes.
• ΔG° f is a positive value, indicating that, at 350 K, the reaction favors reactants over products. Because the reaction is exothermic (heat is a product), increasing the temperature will shift the reaction towards reactants.

Question 3

In the expression ΔG° f = ΔH – TΔS°, there is a value for absolute entropy, S°. Which statement about S° is not correct?

Answer 3

For some ions, the value for S° > 0. When these ions dissociate from the compound in aqueous solution, they “grab” H2O molecules and thus increase their S° value so that S° < 0, not > 0

Question 4

Look at the ΔHf diagram for the reaction of 2NO2(g) <==> N2O4(g)
Which statement about this reaction is not correct?

Answer 4

The value for ΔH for this reaction is positive.

Question 5

Which statement about the actual calculation of ΔGf for a reaction is not correct?

Answer 5

S° values are expressed in tables in units of kJ / mol • K.

Question 6

The expression for finding ΔGf for a formation reaction is ΔG°f = ΔH°f − TΔS°. Which of the following statements is not correct?

Answer 6

The value for ΔS° is very temperature sensitive. This makes sense because entropy changes with temperature.

Question 7

The expression for finding ΔGf for a formation reaction is ΔG°f = ΔH°f − TΔS°. Which statement about ΔGf is not correct?

Answer 7

Like ΔH f , ΔGf for a reaction is not very sensitive to changes in temperature.

Question 8

Which statement about standard state is not correct?

Answer 8

The table values for ΔG°f , ΔH°f , and ΔS° usually do not include standard state conditions.

Question 9

Which statement about Gibbs free energy change, ΔG, is not correct?

Answer 9

If ΔG = 0, then the reaction is at equilibrium. If ΔG > 0, the forward reaction is spontaneous. If ΔG < 0, the reverse reaction is spontaneous.

Question 10

Look at the ΔHf diagram for the reaction of 2NO2(g) <==> N2O4(g)
Which of the following statements is not correct?

Answer 10

ΔHreaction = (ΔH f for the formation of 2NO2 ) + (ΔH f for the formation of N2O4 )

Question 11

The value for ΔG° for the following reaction at 25°C is −394.4 kJ / mol.

C(s, graphite) + O2(g) <==> CO2(g)

Suppose that you raise the temperature to 108°C. What is ΔG° for the reaction at 108°C? (ΔH° = −393.5 kJ / mol and ΔS° = 2.83 J / mol • K)

Answer 11

−394.6 kJ / mol

Question 12

Calculate the standard free energy of formation of 2 mol of AgCl(s) at 25°C using the given information.

For Ag(s) at 25°C, S° = 42.55 J / mol • K and ΔH°f = 0.0 kJ / mol

For Cl2(g) at 25°C, S° = 222.96 J / mol • K and ΔH°f = 0.0 kJ / mol

For AgCl(s) at 25°C, S° = 96.2 J / mol • K and ΔH°f = −127.07 kJ / mol

Answer 12

−219.5 kJ / mol