If DH is not equal to TDS for any given reaction relationship, these quantities could be made equal by simply changing the temperature, since the term TDS is a function of temperature. This would be true if the relative heat capacity Cp of the reactants and products is the same, that is DCp = 0.
In the case of a reaction
A = B
At equilibrium DGT = DHT - TDST
At the standard state: DH0 = DH0B - DH0A
If we increase the temperature by DT,
DHT = (H0B + DT.CpB) - (H0A + DT.CpA)
= H0B - H0A +DT.(CpB - CpA)
= DH0+ DT.DCp
If DCp = 0, that is all the mineral phases have the same heat capacity, then :
DHT = DH0
Similarly, DST = (S0B + CpB.ln(T/T0) - (S0A + CpA.ln(T/T0))
DST = DS0 + DCp.ln(T/T0)
and if DCp = 0 DST=DS0
and therefore, at equilibrium, DGT = DHT - TDST = DH0 - TDS0 = 0
and the temperature at 1 bar would be:
T = DH0/DS0
If DCp is not zero, but does not vary with temperature:
DGT = (DH0
+ DT.DCp) - T(DS00
+ DCp.ln(T/298)) = 0
and DH0 + (T-298).DCp = T(DS0 + DCp.ln(T/298))
from which T can be solved by iteration.
However, if Cp varies with temperature, then:
DCp = Da + DbT
and DHT = DH0 + fDCp.dT = DH0 +[DaT + DbT^2/2 - Dc/T]298T
and DST = DS0 +DfCp/T.dT = DS0 + f(Da/T +Db + Dc/T^3).dT
= DS0 + [Da.ln(T) +Db.T - Dc/(2.T^2)]298T
and DGT = DH0 + fDCp.dT - T(DS0 + fDCp/T.dT) = 0
REACTION ALBITE TO JADEITE + QUARTZ (NalSi3O8 = NaAlSi2O6 + SiO2)
In the case of Albite -> Jadeite and Quartz at some temperature T:
DGT = DHT + T.DST
If the temperature of the system is decreased to T (point A in the above diagram), DGT = 0, and Albite, Jadeite, and Quartz will be in equilibrium.
How could we attain equilibrium by changing pressure?
G = H-TS and H = E + PV,
then G = E + PV-TS
and for a general change in thermodynamic parameters:
dG = dE + PdV + VdP -TdS -SdT
and since dE + PdV and TdS are equal to dq,
then dG = VdP -SdT
At constant temperature therefore, dG/dP = V and dG/dT = -S and dP/dT = -S/V = -H/T.V
Since, dG/dP = V and dG = V.dP
and relative to the standard state at 1 bar DG = f1P V.dP = V.(P-1),
The free energy of a phase due to a change in pressure will therefore be:
GP = Gp=1 + DG
= Gp=1 + V.(P-1)
For a reaction relationship such albite -> jadeite + quartz, the difference in the free energy of the reaction assemblages:
DGP = DGp=1 + DV.(P-1)
If, as can be assumed in most cases, the volume of solid material is independent of temperature and pressure, then DV is the volume difference of the products and reactants. In the reaction of Albite to Jadeite and Quartz (NaAlSi3O8 -> NaAlSi2O6 + SiO2), there is a decrease in molar volume. ?V is therefore negative, and an increase in pressure (a positive term) will cause a decrease in free energy, and at some limiting pressure Albite will convert to Jadeite. If the value of ?G for the reaction at some arbitrary temperature and 1 bar pressure is known, the equilibrium pressure at that temperature can be calculated:
DGp=1 = - DV.(P-1) and P = (-DGp=1 / DV) +1
Note: since under standard conditons albite is the stable phase and not jadeite and quartz, DGp = 1 is +ve, and therefore P is positive.
If, as in the case of gases, the volume of
any of the phases is sensitive to changes in pressure,
then, since PV = RT,
G - G0 = fVdp = RTfdp/P = RT ln (P/P0 )
= RT ln((DP + P0)/P0) = RT ln(DP/P0 + 1)
If P0 = 1 bar (the standard state) then
G - G0 = RT ln (P)
and if the gas is PURE, then G = G0 where G0 is the free energy at the standard state.
If the pressure is less than 1 bar, that is a fraction of a bar, the free energy of the gas will be less than the standard free energy, because the term ‘ln P’ will be negative.
G - G0 is a measure of the change in free energy when the pressure of the gas is changed. In the case of a gas phase, given that PV = nRT, we can increase the pressure by decreasing the volume, increasing the temperature or increasing the quantity of the gas.
If the gas is made up of a mixture of two gas phases, then PV = (n1+n2)RT. In this case, the total pressure is the sum of the partial pressures, P= p1 + p2, and
p1 = n1RT and p2 = n2RT. The partial free energy of each gas components is then proportional to the log of the molar proportion of the component, G - G = RT ln (n), and the molar free energy of the gas phase will be the sum of the partial molar free energies.
By analogy we can use this relationship in
the case of solid phases that are solid solutions, such as olivine, pyroxene
and plagioclase, where ‘n’ represents the mole fraction of the component
of interest in the phase, e.g. forsterite in olivine, i.e. G - G
= RT ln a where a is the molar fraction of the component.
In non-ideal systems 'a' is referred to as the activity or effective concentration of the component in the phase. If the component is PURE its activity is unity (1).
If diopside is added to the system Albite -
Jadeite - Quartz at the standard state, then:
DG = G0jadeite + RT.ln ajadeite + G0qtz - G0albite
At equilibrium DG = 0, and :
ln ajadeite = (Galbite - Gjadeite - Gqtz )/RT = -DG0/RT
In the case of a reaction involving a gas phase such as
calcite + quartz = wollastonite + carbon
CaCO3 + SiO2 = CaSiO3 + CO2
if the CO2 behaves like a perfect gas such that its
activity is equal to its pressure, then at a given temperature T the free
energy difference will be:
DG = DH0 - T DS0
The free energy of the system will be decreased by an amount ?V .(P-1) if the pressure on the solid phases is increased to a value P, that is
DG = DH0 - T DS0 +DV .(P-1)
The free energy of the system will also be change by an amount RT ln(P) as a result of the change in activity (= pressure) of the CO2, that is,
DG = DH0 - T DS0 + DV .(P-1) + RT ln(P).
Since at equilibrium DG = 0, then
DH0 - T DS0 = -DV .(P-1) - RT ln(P)
If the standard state enthalpy difference between
products and reactants is 19,343 cal, the entropy difference is is 35.07
e.u., and the molar difference between the solid phases is -16.69 cm3.
19343 - T*35.07 = -16.69/41.84*(5000-1) - 1.987*T.ln5000
Therefore T = 936
On a P-T plot the phase boundary for a reaction involving a gas phase is a curved rather than a straight line, with a low angle dP/dT slope at low pressures and a steep dP/dT slope at high pressures.
Since an increase in pressure promotes the consumption of CO2, once the pressure increase takes the assemblage into the stability field of calcite and quartz, the pressure can only be held constant by maintaining the source of CO2 from an external source at constant pressure (open system), or by decreasing the volume of the gas phase until all the CO2 is reacted out (closed system; concentration moles/v is constant). If the system is not open to CO2 at constant pressure, and the volume of the system is also held constant (concentration moles/v decreases), then the pressure of the CO2 will drop until the reaction boundary condition pressure is arrived at. At this point the reaction will stop.
If the gas phase is a mixture of two or more gas phases, the pressure of the CO2 will be its partial pressure (concentration moles/v decreases) and the reaction equilibrium boundary will move towards the point representing the total pressure. Note that RT.lnP free energy term for a gas represents both the ?V(P-1) free energy change with pressure of solids, and the RT.lnK free energy term of solid solutions. Only in the case of partial pressures of less than 1 bar will the free energy change be a negative term. As in the case of solid solutions therefore, the existence of a mixed gas phase implies the existence of a reaction boundary at all conditions of P and T.
A phase occupies a specific volume and has uniform physical and chemical characteristics. (W&F - a restricted part of the system with distinctive physical and chemical properties. K&W - a restricted portion of a system, homogeneous with respect to all its properties.)
The total free energy of a phase = its molar free energy x the number of mols.
Minimum number of chemical (atomic and molecular) species required to specify the compositions of all the phases in the system. The composition of olivine can be fully represented in terms of the components Mg2SiO4 and Fe2SiO4 (binary) or FeO MgO SiO2 CaO Al2O3 if olivine and plagioclase are the phases.
The partial molar free energy of a components of a solid solution is referred to as its chemical potential, symbolised as m; an intensive property analogous to molecular weight (neither can be added). The free energy of n moles of forsterite (a component) in olivine (a phase) is therefore G = n.m, where n is the number of mols of forsterite. (The gram weight of n moles of Mg2SiO4 = n x gram molecular weight of Mg2SiO4)
In the case of a pure phase, m = G the molar free energy (free energy per mole of a pure phase). In other words, for a pure phase the chemical potential of its constituent component is equal to the molar free energy of the phase.
In the case of a partial molar solid solution G= m.n where n is the mole fraction.
In the case of a reaction: aA +
bB = cC + dD
where a, b , c, and d are mols of components A, B, C, and D,
the net change in free energy will be DG = cmC + dmD - amA - bmB
However, since c.mC
= cG0C + c.RTln [C] where [C] is the mole fraction of
C in solution, and d.mD = etc, etc, etc. etc.
that is, the total free energy of C, c.mC, equals the total free energy of the pure phase, cG0C, ,less a quantity of free energy represented by the c.RTln [C] term, which is negative in view of fractional nature of [C].
then DG = DG0 + c.RT ln [C] + d.RTln [D] - a.RTln [A] - b.RTln [B]
= DG0 + RT ln (([C]c.[D]d)/([A]a.[B]b))
where DG0 is the free energy difference between products and reactants as PURE phases in the standard state, and DG is the free energy difference of the phases as components in solution.
If the system is at equilibrium, then DG = 0
and DG0 = -RT ln (([C]c[D]d)/([A]a.[B]b))
Where the terms in square brackets represent
the molar fraction of the phase in solution.
If all the phases are pure phases, then the term -RT ln (([C]c.[D]d)/([A]a.[B]b)) = 0, in which case DG = DG0 and the standard state must also be the equilibrium state.
It is important to note that whereas the pure phases are incapable of coexisting other than under the restricted P-T conditions represented by the equilibrium boundary of the phase diagram, the values of [A] and [B] can vary relative to [C] and [D] within the limits of the value of DG0. If DG0 is a large negative value then DG0/-RT could be a large positive value. Consequently the product [A].[B] must be small relative to the product [C].[D] - which we might expect since the large negative value for DG0 indicates that pure phases C and D are stable relative to A and B. If DG0 is a large positive value then DG0/-RT could be a large negative value. Consequently the product [A].[B] must be large relative to the product [C].[D] - which we might expect since the large positive value for ?G0 indicates that pure phases A and B are stable relative to C and D.
Mass, volume, energy, entropy, and position are extensive properties, that is they are additive. Temperature, pressure, density, chemical potential are intensive parameters, that is for example the temperature of a body at 100° cannot be added to the temperature of a body at 200° to get a body at 300°; both bodies would change to a temperature of 150°, which is the average temperature. Heat energy from the higher temperature body can however be added to the lower temperature body. Energy is the product of an extensive and intensive parameter:
e.g. dW = F dl; dQ = C dT; dQ = T dS; dE = chem. pot. dn
GIBBS PHASE RULE
Degrees of freedom (Variance) = components -phases +2
What determines the mineralogical complexity of a rock; how many phases can exist at the same time?
F = C(omponents) - P(hases) + 2
The variance (or degrees of freedom) F dictates the variation that can take place in temperature and pressure while still maintaining the same number of phases in the system.
F =1 - 3 + 2 = 0 (invariant)
F = 1 - 2 + 2 = 1 (univariant)
F = 1 - 1 + 2 = 2 (divariant)
Albite = Jadeite + Quartz
F = 2 - 3 + 2 = 1 (equilibrium boundary)
F = 2 - 1 +2 = 3 (field of Albite; three degrees of freedom because the existence of only albite requires that the system be composed of equal stoichiometric proportions of Jadeite and Quartz. Otherwise the system would be composed of Albite and Quartz or Albite and Jadeite.)
F = 2 -2 +2 = 2 (field of Jadeite and Quartz)
dG = dG/dh.
dh + dG/dP. dP =
since dG/dh = Mg = work done in increasing potential energy
and dG/dP = VT from dG = VdP - SdT
then Mgdh = - VdP
and dP/dh = - Mg/V = - Dg
= (dP/dT). (dT/dh)
dT/dh = g. (M/V). (dT/dP) = g.(M/V). T.(dV/dH) = g. M .T.(dV/V )/dH = g. T. (dV/dT/V) / (dH/dT/M)
= g. T. x coefficient of thermal expansion x molar thermal capacity
= g. T. a.CP