Chemistry and the material world123.102

Unit 4, Lecture 4Matthias Lein

Gibbs free energy

Gibbs free energy to predict the direction of a chemical process.

Exergonic and endergonic reactions.

Temperature dependence of the Gibbs free energy.

The chemical equilibrium.

The law of mass action.

Reaction quotients.

Gibbs free energy and equilibrium constants.

Equilibrium temperature.

Consider the reaction:

N2(g) + 3 H

2(g) 2 NH→

3(g)

Let us calculate H° and S° for the following reaction to decide in which Δ Δdirection each of these factors will drive the reaction.

HΔf°(N

2) = 0 kJ/mol, HΔ

f°(H

2) = 0 kJ/mol, HΔ

f°(NH

3) = -46.11 kJ/mol

SΔf°(N

2) = 191.61 J/molK, SΔ

f°(H

2) = 130.68 J/molK, SΔ

f°(NH

3) = 192.54 J/molK

H° = -92.22 kJΔ S° = -198.78 J/KΔ

Enthalpy favours the reaction. It is exothermic, heat is released.Entropy dis-favours the reaction. Order is increased, the number of particles decreases.

The questions remains:Will the reaction occur spontaneously?

Josiah Willard GibbsUnited States,1839 – 1903

The Gibbs free energy G is defined as:

G = H -TS

a measure that connects the two previously treated thermodynamic functions H, the enthalpy and S, the entropy.

G < 0Δ favored reaction (Spontaneous) G = 0Δ Neither the forward nor the reverse reaction prevails (Equilibrium) G > 0Δ disfavored reaction (Non-spontaneous)

Just like reactions are called exothermic and endothermic depending on the heat produced ( H), reactions are also called:Δ

G < 0Δ exergonic (Spontaneous)

G > 0Δ endergonic (Non-spontaneous)

Wether a reaction is exergonic or endergonic depends on the reaction enthalpy, the reaction entropy and the temperature at which the reaction is carried out.

Consider the reaction (at 25° C):

N2(g) + 3 H

2(g) 2 NH→

3(g)

Let us calculate H° and S° for the following reaction to decide in which Δ Δdirection each of these factors will drive the reaction.

HΔf°(N

2) = 0 kJ/mol, HΔ

f°(H

2) = 0 kJ/mol, HΔ

f°(NH

3) = -46.11 kJ/mol

SΔf°(N

2) = 191.61 J/molK, SΔ

f°(H

2) = 130.68 J/molK, SΔ

f°(NH

3) = 192.54 J/molK

H° = -92.22 kJΔ S° = -198.78 J/KΔ

G = H° -T S°Δ Δ ΔG = -92,220 J - (298.15 K × -198.75 J/K)ΔG = -32.96kJΔ

Answer: at 25° C the reaction occurs spontaneously (it is exergonic)

How about the same reaction at 500° C?

N2(g) + 3 H

2(g) 2 NH→

3(g)

H° = -92.22 kJΔ S° = -198.78 J/KΔ

G = H° -T S°Δ Δ ΔG = -92,220 J - (773.15 K × -198.75 J/K)ΔG = 61.4kJΔ

Answer: at 500° C the reaction does not occur spontaneously(it is endergonic)

The entropic term becomes more and more important as the temperature rises.

G = H° -T S°Δ Δ Δ

The sign of G tells us the direction in which a reaction will proceed to reach Δan equilibrium and the magnitude of G tells us how far from equilibrium the Δreaction still is ( G = 0 means equilibrium).Δ

Thus, as the reaction progresses, the magnitude of G will become smaller Δand smaller until an equilibrium is reached. The chemical reaction is then

finished.

G = H° -T S°Δ Δ Δ

When we calculate G values for reactions, we are comparing G values for Δthe reactants and products. Therefore, the calculated G value for the Δreaction only represents the true G value for the very start of the reaction. ΔOver the course of the reaction the magnitude of G will go to zero.Δ

Equilibrium:

N2O

4(g) 2 NO→

2(g)

brown

colourless

Equilibrium:

N2O

4(g) 2 NO→

2(g)

Equilibrium:

N2O

4(g) 2 NO→

2(g)

Equilibrium:

N2O

4(g) 2 NO→

2(g)

Law of mass action (Kc is the equilibrium constant)

This equation only describes the equilibrium.

Equilibrium:

N2O

4(g) 2 NO→

2(g)

Qc is called the reaction quotient

Remember that Qc and K

c depend on temperature and pressure.

Qc

Systems that are not at equilibrium

can be described by a similar

expression.

Equilibrium:

N2O

4(g) 2 NO→

2(g)

The connection between the equilibrium constant

and the Gibbs free energy:

G° = -RT Δ ln K

So, to calculate G for a given reaction quotient:Δ

G = G° +RT Δ Δ ln Q

(note that the expression becomes zero for Q = K)

What does the magnitude of Kc mean?

K = 1 means that the reaction quotient is balanced. There is approximately the same amount of educt and product in the reaction mixture.

K > 1 means that that the numerator in the reaction quotient is larger. There is more product in the reaction mixture.

K < 1 means that the denominator in the reaction quotient is larger. There is more educt in the reaction mixture.

For example:

2 SO2(g) + O

2(g) 2 SO→

3(g) G° = -1.40 10Δ ⋄ 2 kJ/mol

What is the value of the equilibrium constant K?

We know:

G°Δ = -RT ln K (R = 8.314 J/molK, T = 298K)

So:

ln K = - G° / RTΔ

= -(-1.40 10⋄ 2 kJ/mol) / (8.314 J/molK × 298K)

= 56.5

K = e56.5 = 3 10⋄ 24

The equilibrium temperature

We know that at equilibrium:

G = 0Δ

which means that

H – T S = 0Δ Δ

That means that if we know two quantities out of H, S and T, we can Δ Δcalculate the value of the third at the equilibrium.

For example:

Br2(l) Br→

2(g) H° = 30.9 kJ/mol, S° = 93.2 J/molKΔ Δ

At what temperature is the two phases of liquid and gaseous Br2in

equilibrium?

At equilibrium: G = H – T S = 0Δ Δ Δ

soH = T SΔ Δ

and T = H / SΔ Δ

T = 30900 J/mol / 93.2J/molK = 332 K

332 K = 59° C

●Today we covered:●

Gibbs free energy to predict the direction of a chemical process.

Exergonic ( G < 0) and endergonic ( G > 0) reactions.Δ ΔTemperature dependence of the Gibbs free energy.

The chemical equilibrium.

The law of mass action.

Reaction quotients.

Gibbs free energy and equilibrium constants ( G° = -RT Δ ln K).

Equilibrium temperature ( H = T S)Δ Δ .●