thermo_BM_2

Energetics of crystalline solutions

B. Mishra

PART – 2

ENERGETICS OF CRYSTALLINE SOLUTIONS

In the previous section, dealing with pure crystalline phases, we derived an expression for G = G (P, T). However, most of the rock forming minerals (silicates) and those constitute ore deposits (sulfides) occur as solid solutions of two or more end member components.

Hence, it is appropriate to bring in the compositional variable for deducing the free energy of formation of a phase, i.e., G = G (P, T, X). Accordingly, in this section, we deal with energetics (total change in Gibbs free energy), pertaining to formation binary crystalline solution phases.

The activity of a pure phase component (i) is related to free energy through the relation

Gi

P ,T

= Gi

where Gi

Gi

0 , P ,T

φ , P ,T

(1)

is the free energy of component ‘i’ in phase φ at P, T.

P ,T

0 , P ,T

+ RT ln ai

is the free energy of component ‘i’ if it were a pure phase at P, T

(as derived in eqn. (24) in Part –1) ai

φ , P ,T

is the activity of the component ‘i’ in phase ‘ φ ’ at P, T.

Also,

ai

φ , P ,T

φ

= X i .γ i

φ , P ,T

(2)

φ

where X i = mole fraction of component ‘i’ on phase φ and, γ i

φ , P ,T

= activity coefficient of ‘i’ in phase φ

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Energetics of crystalline solutions

B. Mishra

In pure phases φ

Q X i = 1.0, γ i = 1.0

∴ ai = 1 and Gi

P ,T

= Gi

0 , P ,T

[from eqn. (1)]

In solid solution phases with ideal mixing (Fig. 1)

γ i = 1.0, hence ai = X i and Gi

P ,T

= Gi

0 , P ,T

+ RT ln X i

(3) Fig.1 Activity-composition relationship of a binary solid solution showing ideal (ai = Xi) and non-ideal (ai >Xi and ai < Xi) behaviors.

Gi − Gi = RT ln X i id

0

In solid solutions phases with non-ideal mixing (Fig. 1)

γ i ≠ 1.0 ∴ ai ≠ X i if γ i > 1.0, positive deviation from ideality. if γ i < 1.0, negative deviation from ideality.

At any known P and T, the free energy of a mechanical mixture (G MM ) of A and B should 0

0

lie on a straight line joining G A and G B , in the G-X space as shown below.

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B. Mishra

Energetics of crystalline solutions Hence, G MM = X A G A + X B G B 0

0

(4)

where G MM =free energy of mechanical mixture of A and B at P, T 0

0

G A , G B = free energy of pure phases A and B at P and T X A , X B = mole fractions of A an B in the mechanical mixture

But for solid solution of components A and B forming a homogenous crystalline solution AB, the G of Thus for a point (X) in the G-X plot, G SS can be expressed in terms of G MM and g mix as

G SS = X A G A + X B G B + g mix 0

0

(5)

Fig.2 Gâ&#x20AC;&#x201C;X plot for a model binary solid solution AB showing the variation in Gibbs free energy of the mechanical mixture (Gmm) and solid solution (Gss).

Note the shift in free energy of mixing (gmix) as consequence of change in the composition of the solid solution.

X1

X2

Alternatively, a tangent can be drawn to the GSS -X curve at (X), as shown in Fig.2 and extrapolated to intersect the G-X curve at X A = 1.0 and X B = 1.0 .The points of intersections are labeled as G A and G B that are free energies of components A and B for a specific composition (X1 or X2). But it is important to note that if (X) is shifted along the G SS curve (say from X1 to X2), the shape of the tangent changes and so does its intersections at X A = 1.0 and X B = 1.0 . Hence, G A and G B are functions of the composition of solid solution (AB) at the given P-T. 3

B. Mishra

Energetics of crystalline solutions Now, GSS at (any X value) can be represented as

GSS = X A G A + X B G B

(6)

From eqns. (5) and (6) we may write

X A G A + X B G B = X A G A + X B G B + g mix 0

0

⇒ g mix = X A (G A − G A ) + X B (G B − G B ) 0

0

(7)

Eqn. (7) can be written as

g mix = X A Gm , A + X B Gm , B where Gm, A = G A − G A = RT ln a A (see eqn. 1) 0

Gm , B = G B − G B = RT ln a B (see eqn. 1) 0

For a general case Gi − Gi = RT ln ai 0

(8)

In eqn. (8), by subtracting and adding Gi ,id one obtains

(

)

(Gi − Gi ,id ) + Gi ,id − Gi = RT ln ai = RT ln X i + RT ln γ i RTln γ i

RTln X i

From eqn. (3),

Q Gi ,id − Gi = RT ln X i 0

∴ G i − G i , id = RT ln γ

i

(9) Since (Gi ,id − Gi ) is the change in free energy of a solid solution due to ideal mixing.

0

Hence, (Gi − Gi ,id ) in eqn. (9) refers to an excess free energy of mixing over and above ideal mixing in a crystalline solid solution. If the excess free energy of mixing is represented by g mex,i , then we can write

g mex,i = RT ln γ i

(10)

Hence, the total free energy of excess mixing in a solid solution g mex, ss can be expressed as

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B. Mishra

Energetics of crystalline solutions

g mex,ss = X A g mex, A + X B g mex, B or , g mex,ss = X A ( RT ln γ A ) + X B ( RT ln γ B ) ∴ X A ln γ A + X B ln γ B =

g mex,ss RT (11)

Eqn. (11) can be written in the general form as Y = (1 − X 2 ) y1 + X 2 y 2

where Y =

g mex,ss RT

(12)

, y1 = ln γ A , y 2 = ln γ B , x1 = X A , x2 = xX B

From eqn. (12) we can write

dY = − y1 + y 2 dx 2 ⇒ y1 = y 2 −

dY dx 2

Substituting y1 in eqn. (12)

Y = (1 − x2 )( y 2 −

dY dY ) + x2 y 2 = (1 − x2 ) y 2 − (1 − x2 ) + x2 y 2 dx2 dx2

⎛ dY ⎞ ⎛ dY ⎞ ⎟⎟ ⇒ y 2 = Y + (1 − x2 )⎜⎜ ⎟⎟ ∴Y = y 2 − (1 − x2 )⎜⎜ ⎝ dx2 ⎠ ⎝ dx2 ⎠ (13a) ⎛ g mex,ss ⎜ ∂ ⎜ RT g mex,ss ln γ B = + (1 − X B ) ⎝ RT ∂X B Q y1 = y 2 −

dY dx2

⎞ ⎟ ⎟ ⎠

⎛ dY ⎞ dY ⎟⎟ − = Y + (1 − x2 )⎜⎜ ⎝ dx2 ⎠ dx2

⎛ dY ⎞ ⎟⎟ ∴ y1 = Y − x2 ⎜⎜ ⎝ dx2 ⎠

(13b)

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B. Mishra

Energetics of crystalline solutions

⎛ g mex,ss ⎜ ∂ ⎜ RT g mex,ss − xB ⎝ ln γ A = RT ∂X B

⎞ ⎟ ⎟ ⎠

From eqn. (13), we obtain ⎛ g mex,ss ∂⎜ ⎜ RT g mex,ss ln γ B = + (1 − X B ) ⎝ ∂X B RT

⎞ ⎟ ⎟ ⎠

(14a) ⎛ g mex,ss ∂⎜ ⎜ RT g mex,ss ln γ A = − XB ⎝ RT ∂X B

⎞ ⎟ ⎟ ⎠

(14b)

Symmetric mixing In case of symmetric mixing g mex, ss can be expressed as

g mex,ss = W AB . X A X B = W AB (1 − X B ) X B or , g mex,ss = W AB ( X B − X B2 ) (15)

Where WAB is the interaction or interchange free energy that arises as a result of molecular interaction between A and B, and has the same unit as gmix (joules or calories) Significantly from eqn. (15) both at X B = 1 or X A = (1 − x B ) = 1.0, g mex, ss → 0 Now, from eqn. (15), we can write ⎛ g mex, ss ⎞ ⎟ d ⎜⎜ ⎟ RT ⎠= ⎝ dxB

(

)

⎡W X − X B2 ⎤ d ⎢ AB B ⎥ RT ⎦ ⎣ dxB

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Energetics of crystalline solutions

⎛ g mex,ss d⎜ ⎜ RT ∴ ⎝ dx B

B. Mishra

⎞ ⎟ ⎟ W ⎠ = AB (1 − 2 x ) B RT

Substituting in eqn. (14a) ln γ B =

W AB ( X B − X B2 ) ⎛W ⎞ + (1 − x B )⎜ AB ⎟(1 − 2 x B ) RT ⎝ RT ⎠

[

W AB x B − x B2 + 1 − 2 x B − x B + 2 x B2 RT W W 2 ⇒ AB x B2 − 2 x B + 1 = AB (1 − x B ) RT RT ∴ RT ln γ B = W AB (1 − X B ) 2 ⇒

[

]

]

(16a)

Again from eqn. (14b), we get W AB ( X B − X B2 ) ⎛W ⎞ − X B ⎜ AB ⎟(1 − 2 x B ) RT ⎝ RT ⎠ W .X = AB B [1 − X B − 1 + 2 X B ] RT W X2 = AB B RT ∴ RT ln γ A = W AB X B2 ln γ A =

(16b)

Thus for a binary symmetrical solid solution (AB) G AP ,T = G A0, P ,T + RT ln X A + RT ln γ A or , G AP ,T = G A0, P ,T + RT ln X B + W AB X B2

(17)

Similarly, GBP ,T = GB0, P ,T + RT ln X B + W AB (1 − X B ) 2

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Energetics of crystalline solutions

B. Mishra

In eqn. (17) G AP ,T and G BP ,T can be determined at any temperature if (i) G AP ,T and G BP ,T are estimated from known constants (eqn.24, in the first part). (ii) X A and X B being intensive thermodynamic variables, can be estimated from mineral-chemical data. (iii) W AB is known, determined from solution chemistry.

Configurational Entropy of mixing

Smix = k ⋅ ln ω

(18)

where k is the Boltzman constant (= R/NA) and ω is permutability, which is given by

ω=

(N1 + N 2 )!

(19)

N 1! N 2 !

If we evaluate ω in terms of n sites in the mineral formula, then the configurational entropy of mixing will be conf S mix =

nR ω NA

If the number of mixing cations are N1 and N2 (strictly binary solid solution), then evaluation of ω is a simple problem in permutation, i.e., how many ways can N1 and N2 cations be distributed over N1+N2 positions Since the number of sites and atoms involved in mixing are too large (multiples of Avogadro’s number NA = 6.02 x 1023 per mole), Sterling’s approximation (ln N! = N ln N – N, for large numbers) can be applied to eqn. (18) and we obtain k[ln( N1 + N 2 )!−k ln N1!−k ln N 2 !] conf = k [N ln N − N − N1 ln N1 + N1 − N 2 ln N 2 + N 2 ] S mix

Now since N = N1 + N2, we get conf S mix = k [N ln N − N1 ln N1 − N 2 ln N 2 ]

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