Question

Determine f, c and p in these reactions (Gibbs' Phase Rule)?
In reactions:
a) $ Fe{O_{(g)}} + C{O_{(g)}} \to F{e_{(s)}} + C{O_{2(g)}} $
b) water solution of $ AlC{l_3} $

Answer 1

Hint: The Gibbs phase rule is given as: $ F = {C_i} - P + 2 $ and $ {C_i} = C - r - a $
Where, F is the no. of degrees of freedom, $ {C_i} $ is the no. of chemically independent components in the system, C is the no. of components in the system (ignoring their chemical independence), P is the no. of phases, r is the no. of reactions taking place and a is the no. of additional restrictions (example: charge balance, etc.)

Complete answer:
a) $ Fe{O_{(g)}} + C{O_{(g)}} \to F{e_{(s)}} + C{O_{2(g)}} $
Let us assume that there is no further reaction taking place. There are 2 independent components (if the reaction proceeds to completion there should be no FeO and CO present in the system). So, $\require{cancel} {C_i} = C - \cancel{r} - \cancel{a} \to {C_i} = C = 2 $
The no. of phases P can be easily found out. There are two phases (gas and solid) present in the reaction. Hence P=2. The no. of degrees of freedom will be: $ F = 2 - 2 + 2 = 2 $
The system is bivariant (F=2) i.e., you can change two variables (be it T or P (pressure)) without moving away from the phase equilibrium.
b) water solution of $ AlC{l_3} $ : Let us assume that this is a sufficiently dilute solution and there is no formation of aluminium complex. The reaction that occurs is: $ AlC{l_{3(aq)}} + 3{H_2}{O_{(l)}} \to Al{(OH)_{3(s)}} + 3HC{l_{(g)}} $
The non-independent components include: $ {H_2}{O_{(l)}},OH_{(aq)}^ - ,H_{(aq)}^ + ,Al{(OH)_{3(s)}},HC{l_{(g)}} $
Therefore, a=1 (with water ions present in the solution) and for the charge balance r=1. The no. of components will be $ {C_i} = C - r - a = 5 - 1 - 1 = 3 $ and the no. of phases will be 3 (P=3) i.e. solid, liquid and gases. The no. of degrees of freedom will be $ F = 3 - 3 + 2 = 2 $
The system is bivariant (F=2) i.e., you can change two variables (be it T or P (pressure)) without moving away from the phase equilibrium.

Note:
If the $ AlC{l_3} $ is not sufficiently dilute it would have formed a complex with water forming $ AlC{l_3}.{H_2}O $ . In that case we will include all the above mentioned non independent components too. In that case too $ {C_i} = 3,P = 3 $ and F will be 2.