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

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.