Thermodynamics - Aqueous and Environmental Geochemistry

Geochemistry 

DM Sherman, University of Bristol 

2009/2010 

Thermodynamics 

Part 3: Multicomponent Systems and 

Chemical Equilibria 


D.M. Sherman, University of Bristol 

The story so far.. 

€ 

€ 

We’ve invented a new state-function G (Gibbs free 

energy): 

For a spontaneous process, 

dG < 0 

At equilibrium, 

dG = 0 

ΔG is the non-PV work a system can do on the 

surroundings at constant T and P.



2009/2010 

Free Energy of a Gas 

Let’s determine G of a gas as a function of pressure 

At constant T, 

dG = −SdT +VdP 

€ 

€ 

dG = VdP 

which we can integrate to get 

G(P) − G(P 0 ) = 

P 

∫VdP 

P 0 

€ 

For an ideal gas, 

Free Energy of a Gas (cont.) 

PV = nRT or V = nRT 

P 

So the integral becomes 

€ 

G(P) − G(P 0 ) = 

P 

∫ VdP = nRT ∫ 

1 P dP 

P 0 

P 

P 0 

€ 

⎛ 

G(P) = G(P 0 ) + nRT ln P ⎞ 

⎜ ⎟ 

⎝ P 0 ⎠ 

€



2009/2010 

Free Energy of Mixtures 

Define the chemical potential of component i as 

€ 

⎛ 

µ i = ∂G ⎞ 

⎜ ⎟ 

⎝ ∂n i ⎠ 

T,P,n j≠i 

Using the chemical potential, we can define the free 

energy of a mixture: 

G = G(P,T,n 1 ,…,n i ) 

(n i is the number of moles of i) 

dG = VdP − SdT + ∑ 

i 

µ i 

dn i 

€ 

Free Energy of Mixtures (cont.) 

In a gas which consists of a mixture, we can define the 

partial pressure p i of each component. For an ideal gas, 

we have 

p i V = n i RT 

∑ 

i 

p i 

= P tot 

€ 

A useful result for an ideal gas is that partial pressure 

relates to mole fraction: 

p i 

P tot 

= X i 

€



2009/2010 


By analogy with the G(P) equation we derived, the 

chemical potential of a component in a gas will depend 

on its partial pressure p: 

⎛ 

µ i (p) = µ(p 0 ) + RT ln p ⎞ 

⎜ ⎟ 

⎝ p 0 ⎠ 

From now on, we’ll let p 0 be 1 bar and write 

€ 

µ i = µ 0 + RT ln(p / bar) 

€ 


Now, let’s calculate the change in free energy when we 

mix two gases at constant total pressure: 

G start = n A µ A + n B µ B 

= n A (µ A0 + RT ln(p)) + n B (µ B0 + RT ln(p)) 

After mixing we have 

€ 

G final = n A µ A + n B µ B 

= n A (µ A0 + RT lnp A ) + n B (µ B0 + RT lnp B ) 

€



2009/2010 


Since, 

P = p A + p B and X A = p A 

P , 

X B = p B 

P 

we get 

€ 

⎛ 

ΔG = n A RT ln p ⎞ ⎛ 

A 

⎜ ⎟ + n 

⎝ P B RT ln p ⎞ 

B 

⎜ ⎟ 

⎠ ⎝ P ⎠ 

= nRT( X A lnX A + X B lnX B ) < 0 

€ 

Entropy of Mixing 

Since, 

ΔG = ΔH − TΔS 

It follows that, 

€ 

ΔS = −nR( X A lnX A + X B lnX B ) 

€ 

ΔS is the entropy of mixing the two gases. The gases 

will mix spontaneously because of the positive entropy of 

mixing.



2009/2010 

Activities and Standard State 

We’d like to generalize the concept of chemical potential 

beyond the ideal gas. To do this, we introduce the 

concept of activity: 

€ 

µ i = µ 0 + RT ln(a i ) 

The activity must be defined relative to some standard 

state where a i = 1. 

Examples of Standard States 

• For solid solutions, we can define a standard state of 

a component as being the solid phase made of the pure 

component. 

• For ions in aqueous solution, we usually define the 

standard state as being a 1 m solution with the 

properties of infinite dilution (!). The activity of an ion in a 

dilute aqueous solution will be its molal concentration. 

• For a component in a gas, the activity will be its partial 

pressure divided by 1 bar.



2009/2010 

Law of Mass Action: Chemical Equilibria 

Consider the chemical reaction where A and B reversibly 

transform to C and D: 

αA + βB = γC + δD 

If the reaction occurs at constant T and P, then 

€ 

dG = VdP − SdT + ∑ µ i dn i = ∑ µ i dn i 

i 

i 

= µ A dn A + µ B dn B + µ C dn C + µ D dn D 

Define the reaction progress variable ξ so that 

€ 

dn A = −αdξ, dn B = −βdξ, dn C = γdξ dn D = δdξ 

€ 


Then we can write: 

dG 

dξ = −αµ A 0 − αRT ln(a A ) − βµ B0 − βRT ln(a B ) 

+ γµ c0 + γRT ln(a C ) + δµ D0 + δRT ln(a D ) 

( ) + RT ln (a C) γ (a D ) δ 

⎜ 

= γµ c0 + δµ D0 − αµ A0 − βµ B 

0 

⎛ ⎞ 

⎝ (a A ) α (a B ) β ⎟ 

⎠ 

⎛ 

= ΔG 0 + RT ln (a C )γ (a D ) δ ⎞ 

⎜ 

⎝ (a A ) α (a B ) β ⎟ 

⎠ 

€



2009/2010 


The system will be in chemical equilibrium when: 

dG 

dξ = 0 

⎛ 

ΔG 0 = −RT ln (a C )γ (a D ) δ ⎞ 

⎜ 

⎝ (a A ) α (a B ) β ⎟ 

⎠ 

We define the equilibrium constant K as: 

€ 

⎛ 

K = 

(a C )γ (a D ) δ ⎞ 

⎜ 

⎝ (a A ) α (a B ) β ⎟ 

⎠ 

then lnK = −ΔG 0 / RT 

€ 

The Phase Rule 

f = c - p + 2 

• A phase is something that can be mechanically 

separated from the other phases (e.g., ice, water). 

• A component is one of the smallest number of 

chemical species needed to define the compositions 

of the phases in the system (e.g., H 2 O). 

• A degree of freedom is a property that can be 

independently varied (e.g., P, T).



2009/2010 


For each of the p phases we have 

Since 

dG = VdP −SdT + ∑µ i 

dn i 

i 

€ 

€ 

G = ∑ n i 

µ i 

i 

dG = ∑µ i 

dn i 

+ ∑n i 

dµ i 

= 0 

i 

i 

dG = VdP −SdT −∑ n i 

dµ i 

= 0 

c 

i =1 

€ 

€ 


For each of the p phases we have an equation 

VdP −SdT −∑ n i 

dµ i 

= 0 

c 

i =1 

with 2 (i.e., P and T) + c variables. Hence the variance 

of the system is 

f = c - p + 2



2009/2010 


Example: the Al 2 SiO 5 

system. 

c = 1 (Al 2 SiO 5 ) 

f = c - p + 2 


f = c - p + 2 

Example: SiO 2 (qtz) + H 4 SiO 4 (aq) + Aqueous solution 

c = 2 (e.g., SiO 2 + H 2 O) 

p = 2 (quartz + aqueous solution) 

f = 2 (P,T, [H 2 SiO 4 ])



2009/2010 


Example: the system calcite + water + CO 2 

There are over 9 species within this system defined by 

the following equilibria: 

CaCO 3 = Ca +2 + CO 3 

-2 

CO 3 

-2 

+ H+ = HCO 3 

- 

HCO 3 

- 

+ H + = H 2 CO 3 

H 2 CO 3 = H 2 O + CO 2 

H + + OH - = H 2 O 


Example: the system calcite + water + CO 2 

However, all of the species can be expressed in terms 

of four components, c = 4: 

(e.g., Ca 2+ + CO 3 

-2 

+ H 2 O + H + ) 

The number of phases p is 3: 

Calcite (CaCO 3 ) + Aqueous solution + CO 2 (g) 

The number independent degrees of freedom f is: 

c-p + 2 = 3. (e.g., P,T, pCO 2 ; or P, T, pH)



2009/2010 

For a reaction 

Summary 

αA + βB = γC + δD 

⎛ 

K = (a C )γ (a D 

) δ ⎞ 

⎜ ⎟ 

⎝ (a A 

) α (a B 

) β 

⎠ 

€ 

with lnK = −ΔG 0 / RT 

€ 

Where a i is the activity of species i. 

ΔG 0 is the change in free energy when all components 

are in their standard state.

Thermodynamics - Aqueous and Environmental Geochemistry

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