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DM Sherman
Practical 2: Thermodynamic Calculations
(SOLUTIONS)
Background: Gibbs Free Energy of a Chemical Reaction
As you will learn in the next few lectures, the Gibbs free energy (G) of a
substance at a temperature T (in K) given by
G = H - TS
where H is the enthalpy and S is the entropy. In practice, we take H to be that
required for the formation of the substance from the elements (designated ΔH f
(This means that the ΔH f for any pure element is simply zero.). For a chemical
reaction from A to B, the change in G, ΔG, will be
Δ G = ( Δ H
f
B
- Δ H f
A ) – T(S B
- S A ) = Δ H - T Δ S
The problem is that H and S are functions of pressure (P) and temperature (T).
A pretty good approximation (valid when the change in heat capacity, Δ C p = 0) is
that
Δ G ( P , T ) = Δ H ( P
0
, T
0
) + T Δ S ( P
0
, T
0 ) + Δ VdP
1
P
where Δ H(P 0
,T
0 ) means Δ H when P=P 0 and T=T 0
. P
0 is usually 1 bar and T 0
298 K.
Part I: Equilibria Between Pure Solid Phases
Most minerals are not very compressible; over the pressure range of the Earth’s
crust, the molar volume of a mineral is nearly constant. If the reaction involves
only mineral (solid) phases, then to a good approximation Δ V is constant. With
P
0 = 1 bar, we have
Δ G = Δ H
1bar
- T Δ S + (P-1) Δ V
where Δ V is the molar volume change of the reaction. If two substances A and
B are in chemical equilibrium , then the difference in free energy between them
will be zero:
Δ G = Δ H
1bar
- T Δ S + (P-1) Δ V = 0
DM Sherman
we can use this to calculate the boundary in P,T space between two phases.
Along this boundary, both phases will coexist.
1. Using the approximation that the P,T phase boundary is defined by
Δ H
1bar, 298K
- T Δ S
**298K
- (P-1)** Δ V = 0, calculate the phase diagram (plot P vs
T) for calcite-aragonite given the following thermodynamic data: (Be careful!
Use dimensional analysis to convert (P-1) Δ V into kJ/mole.)
Species Δ H f, 1bar
(kJ/mol)
S
298K
(kJ/mol-K)
V
(cm 3 /mole)
Calcite, CaCO 3
Aragonite, CaCO 3
Solution: For the reaction Calcite = Aragonite:
ΔH = - 1297.43-(-1207.37) = - 0.06 kJ/mole
ΔS = 0.08799-0.09171 = - 0.00372 kJ/mol-K
ΔV = 49.9 – 44.09 = - 2.78 cm 3 /mole
Thus
ΔG = - 0.06 (kJ/mole) – T(K)(-0.00372) kJ/mol-K)
- (P-1)(bar)(-2.78 cm 3 /mole)(
= - 0.06 + 0.00372T - (2.78 x 10
or P (bar) = - 214.8 + 13.4T
DM Sherman
2. Calculate the phase diagram for the dehydration of brucite Mg(OH) 2 to
periclase MgO and H 2 O(gas) given the following thermodynamic data.
(Assume steam is an ideal gas; do we need to include the volume change of the
solids?)
Species Δ H f, 1bar
(kJ/mol)
S
(kJ/mol-K)
V
(cm 3 /mole)
Brucite (Mg(OH) 2
Periclase (MgO) - 601.49 0.02694 11.
Steam (H 2
O) - 241.81 0.18872 24789.
Solution:
For the reaction Mg(OH) 2 = MgO + H 2 O(g)
ΔH = - 241.81-601.49 - (-924.54) = 81.24 kJ/mole
ΔS = 0.18872 + 0.02694 – 0.06318 = 0.1525 kJ/mol-K
Δ VdP
1
P
∫ ≈^ VgasdP
1
P
∫ =^
RT
P
1
P
∫ dP^ =^ RT^ ln^
P bar
1 bar
kJ
mol − K
T^ ( K )ln^
P bars
1 bar
Thus, along the reaction boundary:
ΔG = 81.24 – T (0.1525) + 0.00831T(ln(P bar/1bar)) = 0
or P (bar)/(1bar) = Exp(-9771/T + 18.2)
DM Sherman
Pressure (kbar)
Temperature (K)
Mg(OH) 2
MgO + H 2
O
Part III: Equilibria Involving Mixtures
In the two previous problems, the products and reactants were all separate pure
phases. Many problems in geochemistry involve phases that are mixtures of
several components. An example is an aqueous solution (e.g., 1 mole of NaCl in
water) or a solid solution between two endmembers (e.g., forsterite, Mg 2 SiO 4
and fayalite, Fe 2 SiO 4 in olivine). As we will derive in the next few lectures, the
free energy of a solution is defined in terms of the activities of the components.
For now, accept the following rules:
- the activity of an ion in aqueous solution is numerically equal to its molal
concentration (= moles/kg of water). The standard state is 1 molal with
the properties of infinite dilution (!).
- The activity of a gas phase component is numerically equal to its partial
pressure. The standard state is pure gas component at 1 bar pressure.
The activity of a pure solid substance is 1.
The activity of a component in a solid solution is a bit more
complicated…
For a chemical equilibrium aA + bB = cC + dD, we can write an equilibrium
constant
DM Sherman
3. Now calculate the effect of temperature on the log(K) for the dissolution
of galena at pH 2. Plot log[Pb] vs T. Is PbS very soluble at any T? How
do we form ore deposits of PbS?
Solution:
€
2.303log( K ) = ln( K ) =
−Δ H
RT
Δ S
R
log( K ) =
1
− 79.
0.00831 T
−0.
=^
−4128.
T
− 0.
€
log( K ) = 2 log[ Pb ] + pH
log[ Pb ] =
1
2
(log( K )^ −^ pH ) =^
−2064.
T
− 1.
200 300 400 500 600 700
log[Pb
2+
]
Temperature (K)
pH =
