Question

Determine `f`, `c` and `p` in these reactions (Gibbs' Phase Rule)?

In reactions:
a) FeO(g)+CO(g)-> Fe(s)+(CO2)g
b) water solution of AlCl3

determine `f`, `c` and `p` based on Gibb's phase rule where `f` is the number of degrees of freedom, `c` is the number of components and `p` is the number of phases in thermodynamic equilibrium with each other.

Answer 1

The Gibbs' Phase rule is:

`f = 2 + c_i - p`

`c_i = c - r - a`

where:

• `f` is the number of degrees of freedom (how many independent intensive variables can be varied without affecting other thermodynamic variables).
• `c_i` is the number of chemically independent components in the system.
• `c` is the number of components in the system, ignoring their chemical independence.
• `p` is the number of phases.
• `r` is the number of reactions.
• `a` is the number of additional restrictions (e.g. charge balance).

`a)` `"FeO"(g) + "CO"(g) rightleftharpoons "Fe"(s) + "CO"_2(g)`

If we assume that there is no further reaction occurring, then there are `bb(2)` independent components (`"FeO"(g)` and `"CO"(g)` should no longer be in the system if the reaction went to completion at 100% yield), so `color(blue)(c_i) = c - cancel(r)^(0) - cancel(a)^(0) = c = color(blue)(bb(2))`.

The number of phases, `p`, is easily countable. You can see that there are `color(blue)(p = bb(2))` phases: gas and solid.

Therefore, `color(blue)(f) = 2 + c_i - p = 2 + 2 - 2 = color(blue)(bb(2))`. This says that you can change two natural variables without moving away from a phase equilibrium (`T`, temperature, `P`, pressure).

`b)` This is a very complicated system, so we'll check two scenarios.

An aqueous solution of `"AlCl"_3` would allow water to coordinate onto the empty `p` orbital on aluminum (`"AlCl"_3` is a Lewis acid!). That would create another component in solution, `"AlCl"_3cdot"H"_2"O"`:

CASE I

If we assume a sufficiently dilute solution, we can pretend that `"AlCl"_3cdot"H"_2"O"` is essentially not there, but this reaction should happen:

`"AlCl"_3(aq) + 3"H"_2"O"(l) -> "Al"("OH")_3(s) + 3"HCl"(g)`

We include these non-independent components:

• `"H"_2"O"(l)`
• `"OH"^(-)(aq)`
• `"H"^(+)(aq)`
• `"Al"("OH")_3(s)`
• `"HCl"(g)`

With the water ions in solution, we have `a = 1` for the charge balance, and `r = 1` to account for the autoionization process, so `color(blue)(c_i) = c - r - a = 5 - 1 - 1 = color(blue)(bb(3))`.

The number of phases is `color(blue)(p = 3)` (solid, liquid, gas).

CASE II

If the solution is NOT sufficiently dilute, and we assume a reaction DOES occur (to completion), then we must include all of these non-independent components, assuming excess water (but neglecting the coordination of water to `"AlCl"_3` since it was fully reacted):

• `"H"_2"O"(l)`
• `"OH"^(-)(aq)`
• `"H"^(+)(aq)`
• `"Al"("OH")_3(s)`
• `"HCl"(g)`

I will assume case II, which omits the water association reaction and omits the `"AlCl"_3cdot"H"_2"O"`, canceling out the difference between case I and case II.

This gives that `color(blue)(c_i) = c - r - a = 5 - 1 - 1 = color(blue)(bb(3))`, since it is the most realistic. Notice how the number of non-independent components we add generally equals the number of reactions we subtract.

You should see that in both cases, we get the same number for `c_i`, which is how it should be.

If the reaction is in a closed system, then we again have three phases (liquid, solid, and gas), so `color(blue)(p = 3)`. Therefore, the number of degrees of freedom are:

`color(blue)(f) = 2 + c_i - p`

`= 2 + 3 - 3 = color(blue)(bb(2))`

which means you can change the temperature and pressure (two natural variables) a little and not move off the phase equilibrium you establish, assuming a closed system.