Chapter 9: Chemical Equilibrium

9.1 The Gibbs energy minimum1. Extent of the reaction (ξ):

The amount of reactants being converted to products. Its unit is mole.Consider: A ↔ B

Assume an infinitesimal amount dξ of A turns into B,dnA = -dξ

On the other hand, dnB = dξ

Example 1: N2(g) + 3H2(g) ↔ 2NH3(g)when the extent of the reaction changes from ξ = 0 to ξ = 1.0 mole, what are the changes of each reagent?

Solution: identify vj:v(N2) = -1; v(H2) = -3; v(NH3) = 2.since dξ = 1.0 mole,

dn(N2) = -1x1.0 mole = -1.0 mole,dn(H2) = -3x1.0 mole = -3.0 moles,dn(NH3) = 2x1.0 mole = 2.0 moles,

Example 2: CH4(g) + Cl2(g) ↔ CHCl3(l) + HCl(g)Solution: (chalkboard + discussion).

2. The Reaction Gibbs energy: ∆rG,the slope of the Gibbs energy plotted against the extent of

reaction:

∆rG = here ∆r signifies a derivative

• A reaction for which ∆rG < 0 is called exergonic.

• A reaction for which ∆rG > 0 is called endergonic.

• ∆rG < 0, the forward reaction is spontaneous.

• ∆rG > 0, the reverse reaction is spontaneous.

• ∆rG = 0, the reaction is at equilibrium!!!

Tp

G

,

∂∂

ξξξξ

An useful description of the reaction Gibbs energy (∆rG)

• Consider the reaction A ↔ Binitial amount nA0 nB0final amount nAf nBf

Ginitial = uBnB0 + uAnA0 Gfinal = uBnBf + uAnAf∆G = Gfinal - Ginitial = (uBnBf + uAnAf ) – (uBnB0 + uAnA0 )

= uB(nBf- nB0) + uA (nAf - nA0) = uB∆ξ + uA(-∆ξ) = (uB - uA )∆ξ

∆rG = = uB - uA (9.2)

When uA > uB, the reaction A → B is spontaneous.When uB > uA, the reverse reaction (B → A) is spontaneous.When uB = uA, the reaction is spontaneous in neither direction (equilibrium

condition).

Tp

G

,

∂∂

ξξξξ

Molecular interpretation of the minimum in the reaction Gibbs energy

9.2 The description of equilibrium

1. Perfect gas equilibrium:A(g) ↔ B(g)

∆rG = uB – uA

quotientreactionaisQ

PPQQRTGG

PPRTG

PRTuPRTu

A

Brr

A

Br

AABB

=+∆=∆

+∆=

+−+=

ln

ln

)ln()ln(

θθθθ

θθθθ

θθθθθθθθ

• At equilibrium, ∆rG = 0, therefore

Note: The difference in standard molar Gibbs energies of the products and reactants is equal to the difference in their standard Gibbs energies of formation, thus,

∆rGθ = ∆fGθ(B) - ∆fGθ(A)

θθθθ

θθθθ

GKRT

thusmequilibriuatpressurespartialofratiothedenotetousedisK

KRTG

r

r

∆−=

+∆=

ln

:)(

ln0

Equilibrium for a general reaction

For example: 2A + B ↔ C + 3D

• The reaction Gibbs energy, ∆rG, is defined in the same way as discussed earlier:

where the reaction quotient, Q, has the form:Q = activities of products/activities of reactants

in a compact expression Q = Π αjνj

vj are the corresponding stoichiometric numbers; positive for products and negative for reactants.

vA = -2; vB = -1; vC = 1; vD = 3

).(ln 119QRTGG rr +∆=∆ θθθθ

• Justification of the equation 9.11dG = ∑ujdnj

Assuming that the extent of reaction equals dξ, one gets dnj = d(vjξ) then dnj = vjdξ (because vj are constants)since dG = ∑ujdnj

= ∑ujvjdξ(dG/dξ) = ∑ujvj � ∆rG = ∑ujvj

because uj = ujθ + RTln(aj)

∆rG = ∑{vj(ujθ + RTln(aj))}

= ∑(vjujθ) + ∑vj(RTln(aj))

= ∆rGθ + RT ∑ln(aj)vj

= ∆rGθ + RT ln (∏(aj)vj) = ∆rGθ + RT ln (Q)

θθθθθθθθθθθθ GvGvGtsac

foducts

fr ∑∑ ∆−∆=∆tanRePr

• Again, we use K to denote the reaction quotient at the equilibrium point,

K = Qequilibrium = (∏ajvj)equilibrium (9.12)

K is called thermodynamic equilibrium constant. Note that until now K is expressed in terms of activities)

Example: calculate the quotient for the reaction: A + 2B ↔ 3C + 4DSolution: first, identify the stoichiometric number of each reactant:

vA = -1, vB = -2, vC = 3, and vD = 4.

At the equilibrium condition:

2

43

BA

DC

aa

aaQ =

2

43

BA

DC

aa

aaK =

Examples of calculating equilibrium constants

1. Consider a hypothetical equilibrium reactionA(g) + B(g) ↔ C(g) + D(g)

While all gases may be considered ideal. The following data are available for this reaction:

Compound µθ(kJ mol-1)A(g) -55.00B(g) -44.00C(g) -54.00D(g) -47.00

Calculate the value of the equilibrium constant Kp for the reaction at 298.15K.

Solution:

Example 2: Using the data provided in the data section, calculate the standard Gibbs energy and the equilibrium constant at 25oC for the following reaction

CH4(g) + Cl2(g) ↔ CHCl3(g) + HCl(g)

Solution: (chalkboard)∆fGθ(CHCl3, g) = -73.66 kJ mol-1

∆fGθ(HCl, g) = - 95.30 kJ mol-1

∆fGθ(CH4, g) = - 50.72 kJ mol-1