Part A (30) Part B (15)
Q3. For binary phase diagrams, the Market
Gibb's phase cule is reduced to
(A) F= C-B+2
(B) F=P-C+2
(0) F= 6-2
(D) F=C-P+1
Answers
Answer:
Gibbs Phase Rule: f = c – p + 2
f = Intensive Degrees of freedom = variance
Number of intensive variables that can be changed independently without
disturbing the number of phases in equilibrium
p = number of phases
gas, homogeneous liquid phases, homogeneous solid phases
c = components
Minimum number of independent constituents
Case I. No chemical reactions: c = constituents
Example 1: start with methanol and water – 2 components
Case II With chemical reactions:
Example 2: start with NaH2PO4 in water --
Ka2 Ka3
H2PO4
-
→
← HPO4
2- + H+
→
← PO4
3- + H+
Constituents: Na+
, H+
, H2PO4
-
, HPO4
2-, PO4
3-, H2O
but only 2 components -- NaH2PO4 and H2O.
Example 3: start with NaH2PO4 and Na2HPO4 in water --
Same constituents: Na+
, H+
, H2PO4
-
, HPO4
2-, PO4
3-, H2O
but now 3 components -- NaH2PO4, Na2HPO4, and H2O.
Need to know: T, P, yA, yB, xA, xB
total intensive variables = c p + 2
But yA + yB = 1
xA + xB = 1
Get p such equations, one for each phase:
Independent variables = c p + 2 – p
But, chemical potential is everywhere equal:
µA(xA) = µA(g)
µB(xB) = µB(g)
Get p–1 for each component
Get c( p–1) such equations:
Independent variables = c p + 2 – p – c( p–1)
f = c – p + 2
A & B
liquid
µA(g)
µA(xA)
=
µB(xB)
=
µB(g)
xA + xB =1
A & B
vapor
yA + yB =1
f ' = c – p + 1 P = cst
0 1
xA, yA →
liquid A & B
f ' = 2
vapor A & B
T f ' = 1
f ' = 2
T
*
bA
T
*
bB
Colby College
f' = c – p + 1 cst. P
f "= c – p cst. T&P
Binary solid-liquid Equilibrium
Melting Point Variation with Composition
c = 2
p = 3
liquid, pure solid A, pure solid B
Solid-liquid 2-phase region:
f
' = 2 – 2 + 1 = 1
Eutectic:
f' = 2 – 3 + 1 = 0
invariant at cst P
For NaCl in water:
Eutectic -21.1 oC at 23% wt/wt giving NaCl·2H2O
Add One Extensive Independent Variable for Each Phase:
Gibbs energy is extensive:
Degrees of freedom:
D = f + p
Binary Solid-Liquid at constant T & P:
Solid-liquid 2-phase region:
f " = 2 – 2 = 0
D " = f " + p = 0 + 2 = 2
dG = µA dnA + µB dnB dnA and dnB: totals for both phases
since: µA(s) = µA(l), and µB(s) = µB(l) (doesn’t matter which phase)
Step-by-step explanation: