Review: Expressions of the thermodynamic equilibrium constant K

• K, (a dimensionless quantity) can be expressed in terms of fugacities for gas phase reactions or activities for aqueous phase reactions.

• Gas phase: Fugacity ( a dimensionless quantity) is equal to the numerical value of partial pressure, i.e. pj/pθ where pθ = 1 bar).

• Aqueous phase: 1. Neutral solution: the activity, a, is equal to the numerical value

of the molality, i.e. bj/bθ where bθ = 1 mol kg-1.

2. Electrolyte solution: The activity shall now be calculated as αj = γj*bj/bθ , where the activity coefficient, γ, denotes distance from the ideal system where there is no ion-interactions among constituents.

The activities of solids and pure liquid are equal to 1

• α(solid) = 1 and α(pure liquid) = 1 (!!!)

• Illustration: Express the equilibrium constant for the heterogeneous

reaction

NH4Cl(s) ↔ NH3(g) + HCl(g)

• Solution:

In term of fugacity (i.e. thermodynamic equilibrium constant):

Kp =

In term of molar fraction: Kx =

Estimate reaction compositions at equilibrium

• Example 1: Given the standard Gibbs energy of reaction H2O(g) → H2(g) + 1/2O2(g) at 2000K is + 135.2 kJ mol-1, suppose that steam at 200k pa is passed through a furnace tube at that temperature. Calculate the mole fraction of O2 present in the output gas stream.

• Solution: (details will be discussed in class)

lnK = - (135.2 x 103 J mol-1)/(8.3145 JK-1mol-1 x 2000K)

= - 8.13037

K = 2.9446x10-4

K =

Ptotal = 200Kpa

assuming the mole fraction of O2 equals x

PO2 = x* Ptotal,

PH2 = 2(x*Ptotal)

PH2O = Ptotal – PO2 – PH2 = (1-3x)Ptotal

PP

PPPP

OH

HO

/

)/()/( /

2

22

21

Equilibria in biological systems: Standard reaction Gibbs energy for biochemical systems

• Biological standard state: pH = 7.• For a reaction: A + vH+

(aq) ↔ P

ΔrG = ΔrGθ + RT

= ΔrGθ + RT

the first two terms of the above eq. form ΔrG‡

ΔrG‡ = ΔrGθ + 7vRTln10, A better practice is to transfer the above eq into ΔrG‡ = ΔrGθ - 7vRTln10,

and then recognizes v is the stoichiometric number of H+.

)ln(vHA

P

bb

b

A

pv b

bRT

Hln)

][ln(

1

Example: For a particular reaction of the form A → B + 2H+ in aqueous solution, it was found that ΔrGθ = 20kJ mol-1 at 28oC. Estimate

the value of ΔrG‡. • Solution: ΔrG‡ = ΔrGθ - 7vRTln10 here the stoichiometric number of H+ is 2, i.e. v = 2

ΔrG‡ = 20 kJ mol-1 - 7(2)(8.3145x10-3 kJ K-1mol-1) x(273+ 28K)ln10 = 20 kJ mol-1 – 80.676 kJ mol-1 = -61 kJ mol-1 (Notably, when measured with the biological standard, the

standard Gibbs energy of reaction becomes negative!A transition from endergonic to exergonic process )

Molecular Interpretation of equilibrium

Two factors affect the thermodynamic equilibrium constant: (1) Enthalpy, and (2) Entropy.

Boltzmann distribution is independent of the nature of the particle.

The response of equilibria to reaction conditions

• Equilibria may respond to changes in pressure, temperature, and concentrations of reactants and products.

• The equilibrium constant is not affected by the presence of a catalyst.

How equilibria respond to pressure

• The thermodynamic equilibrium constant K is a function of the standard reaction Gibbs energy, ΔrGθ .

• Standard reaction Gibbs energy ΔrGθ is defined at a single standard pressure and thus is independent of the pressure used in a specific reaction.

• The equilibrium constant is therefore independent of reaction pressure. Such a relationship can be expressed as:

0

Tp

K)(

• Although the thermodynamics equilibrium constant K is independent of pressure, it does not mean that the equilibrium composition is independent of the pressure!!!

• Example: Consider a gas reaction 2A(g) ↔ B(g)

assuming that the mole fraction of A equals xA at quilibrium, then xB = 1.0 – xA,

because K does not change, xA must change in response to any variation in Ptotal!!!

totalA

A

totalA

totalA

Px

Px

PPx

PPxK 22

)0.1(

)/(

/)0.1(

Le Chatelier’s Principle

• A system at equilibrium, when subject to a disturbance, responds in a way that tends to minimize the effect of the disturbance.

• The above statement suggests that if the total pressure of a system is increased, the system will shift to the direction that will have smaller number of molecules, i.e. smaller pressure.

• 3H2(g) + N2(g) ↔ 2NH3(g).

Example: Predict the effect of an increase in pressure on the Haber reaction, 3H2(g) + N2(g) ↔ 2NH3(g).

• Solution: According to Le Chatelier’s Principle, an increase in pressure will

favor the product.

prove:

Therefore, to keep the thermodynamic equilibrium constant K unchanged, the equilibrium mole fractions Kx must change by a factor of 4 if the pressure ptotal is doubled.

2

23

2

33

22

3

2

22

3

22

3

22

3

total

x

totalHN

NH

totalHtotalN

totalNH

HN

NH

p

K

pxx

x

pxpx

px

pp

pK

The response of equilibria to temperature

• According to Le Chatelier’s Principle: Exothermic reactions: increased temperature favors the reactants.

Endothermic reactions: increased temperature favors the products.

• The van’t Hoff equation:

(a) (7.23a)

(b) (7.23b)

2

ln

RT

H

dT

Kd r

R

H

Td

Kd r

)

1(

ln

Derivation of the van’t Hoff equation:

• Differentiate lnK with respect to temperature

• Using Gibbs-Helmholtz equation (eqn 3.53 8th edition)

thus

• Because d(1/T)/dT = -1/T2:

dT

TGd

RdT

Kd r )/(ln 1

RT

GK r

ln

2T

H

dT

TGd rr

)/(

2

ln

RT

H

dT

Kd r

R

H

Td

Kd r

)

1(

ln

• For an exothermic reaction, ΔrHθ < 0, thus , suggesting that increasing the reaction temperature will reduce the equilibrium constant.

0dT

Kd ln