Volume Changes (Equation of State)

VdP

dG

T

PT

V

V

1

TP

V

V

1

Volume is related to energy changes:

Mineral volume changes as a function of T: , coefficient of thermal expansion

Mineral volume changes as a function of P: , coefficient of isothermal expansion

For Minerals:

Volume Changes (Equation of State)

• Gases and liquids undergo significant volume changes with T and P changes

• Number of empirically based EOS solns..• For metamorphic environments:

– Redlich and Kwong equation:

• V-bar denotes a molar quatity, aRw and bRK are constants

)(2/1RK

Rw

RK bVVT

a

bV

RTP

Hess’s LawKnown values of H for reactions can be used to determine H’s for other reactions.

H is a state function, and hence depends only on the amount of matter undergoing a change and on the initial state of the reactants and final state of the products.

If a reaction can be carried out in a single step or multiple steps, the H of the reaction will be the same regardless of the details of the process (single vs multi- step).

CH4(g) + O2(g) --> CO2(g) + 2H2O(l) H = -890 kJ

If the same reaction was carried out in two steps:

CH4(g) + O2(g) --> CO2(g) + 2H2O(g) H = -802 kJ

2H2O(g) --> 2H2O(l) H = -88 kJ

CH4(g) + O2(g) --> CO2(g) + 2H2O(l) H = -890 kJ

Net equation

Hess’s law : if a reaction is carried out in a series of steps, H for the reaction will be equal to the sum of the enthalpy change for the individual steps.

Reference States• We recall that we do not know absolute

energies!!!• We can describe any reaction or description

of reaction relative to another this is all we need to describe equilibrium and predict reaction direction, just need an anchor…

• Reference States:– Standard state: 1 atm pressure, 25°C– Absolute states – where can a value be defined?

entropy at 0 Kelvin

• Heat of reaction H0R

• H0R is positive exothermic

• H0R is negative endothermic

• Example: 2A + 3B A2B3

• H0R =H0

f(A2B3)-[2H0f(A) + 3H0

f(B)]

)()( 000 reactantsHnproductsHnHi

fiifii

iR

Heat of Reaction

Entropy of reaction

• Just as was done with enthalpies:

• Entropy of reaction S0R:

• When S0R is positive entropy increases as a

result of a change in state

• When S0R is negative entropy decreases as

a result of a change in state

)()( 000 reactantsSnproductsSnS ii

iii

iR

J. Willard Gibbs• Gibbs realized that for a reaction, a certain

amount of energy goes to an increase in entropy of a system.

• G = H –TS or G0R = H0

R – TS0R

• Gibbs Free Energy (G) is a state variable, measured in KJ/mol

• Tabulated values of G0R are in Appendix

)reactants()( 000i

iii

iiR GnproductsGnG

G is a measure of driving force

• G0R = H0

R – TS0R

• When G0R is negative forward reaction

has excess energy and will occur spontaneously

• When G0R is positive there is not

enough energy in the forward direction, and the BACKWARD reaction will occur

• When G0R is ZERO reaction is AT

equilibrium

Free Energy Examples

G0R = H0

R – TS0R

H2O(l)=-63.32 kcal/mol (NIST value: http://webbook.nist.gov/chemistry/)

• Fe2+ + ¼ O2 + H+ Fe3+ + ½ H2O=[-4120+(-63320*0.5)]-[-21870+(3954*0.25)]

=[-67440]-[-19893]=-47547 cal/mol

)reactants()( 000i

iii

iiR GnproductsGnG

• Now, how does free energy change with T and P?

• From G=H-TS:

2

1

2

1

2

)1(

2

1

)1(11222)( 121,1,,

T

T

P

P

T

PT

T

PTPTPTP dPVdTT

CTdTCTTSGG P

P

Phase Relations• Rule: At equilibrium, reactants and products have

the same Gibbs Energy– For 2+ things at equilibrium, can investigate the P-T

relationships different minerals change with T-P differently…

• For GR = SRdT + VRdP, at equilibrium, Grearranging:

R

R

G V

S

T

P

0

Clausius-Clapeyron equation

V for solids stays nearly constant as P, T change, V for liquids and gases DOES NOT

• Solid-solid reactions linear S and V nearly constant, S/V constant + slope in diagram

• For metamorphic reactions involving liquids or gases, volume changes are significant, V terms large and a function of T and P (and often complex functions) – slope is not linear and can change sign (change slope + to –)

R

R

G V

S

T

P

0

P

R

TR

R

T

V

V

S

T

C

T

S P

P

R

SR change with T or P?

V = Vº(1-P)

21

22

00

00

(2

1

2

1

2

PPPVS

VdPSdPP

SSS

P

PT

P

P

P

R

R

G V

S

T

P

0

Example – Diamond-graphite• To get C from

graphite to diamond at 25ºC requires 1600 MPa of pressure, let’s calculate what P it requires at 1000ºC:

graphite diamond

(K-1)

1.05E-05 7.50E-06

(MPa-1)

3.08E-05 2.27E-06

Sº

(J/mol K)

5.74 2.38

Vº

(cm3/mol)

5.2982 3.417

Clausius-Clapyron Example

Phase diagram• Need to represent how mineral reactions

at equilibrium vary with P and T

R

R

G V

S

T

P

0P

R

R

R

T

V

V

S

T

C

T

S P

P

R

Gibbs Phase Rule

• The number of variables which are required to describe the state of a system:

• p+f=c+2 f=c-p+2– Where p=# of phases, c= # of components,

f= degrees of freedom– The degrees of freedom correspond to the

number of intensive variables that can be changed without changing the number of phases in the system

Variance and f• f=c-p+2

• Consider a one component (unary) diagram

• If considering presence of 1 phase (the liquid, solid, OR gas) it is divariant

• 2 phases = univariant

• 3 phases = invariant