# 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.

Dec 31, 2016

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) ${\text{FeO"(g) + "CO"(g) rightleftharpoons "Fe"(s) + "CO}}_{2} \left(g\right)$

If we assume that there is no further reaction occurring, then there are $\boldsymbol{2}$ independent components ($\text{FeO} \left(g\right)$ and $\text{CO} \left(g\right)$ should no longer be in the system if the reaction went to completion at 100% yield), so $\textcolor{b l u e}{{c}_{i}} = c - {\cancel{r}}^{0} - {\cancel{a}}^{0} = c = \textcolor{b l u e}{\boldsymbol{2}}$.

The number of phases, $p$, is easily countable. You can see that there are $\textcolor{b l u e}{p = \boldsymbol{2}}$ phases: gas and solid.

Therefore, $\textcolor{b l u e}{f} = 2 + {c}_{i} - p = 2 + 2 - 2 = \textcolor{b l u e}{\boldsymbol{2}}$. 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 ${\text{AlCl}}_{3}$ would allow water to coordinate onto the empty $p$ orbital on aluminum (${\text{AlCl}}_{3}$ is a Lewis acid!). That would create another component in solution, $\text{AlCl"_3cdot"H"_2"O}$:

CASE I

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

$\text{AlCl"_3(aq) + 3"H"_2"O"(l) -> "Al"("OH")_3(s) + 3"HCl} \left(g\right)$

We include these non-independent components:

• $\text{H"_2"O} \left(l\right)$
• ${\text{OH}}^{-} \left(a q\right)$
• ${\text{H}}^{+} \left(a q\right)$
• "Al"("OH")_3(s)
• $\text{HCl} \left(g\right)$

With the water ions in solution, we have $a = 1$ for the charge balance, and $r = 1$ to account for the autoionization process, so $\textcolor{b l u e}{{c}_{i}} = c - r - a = 5 - 1 - 1 = \textcolor{b l u e}{\boldsymbol{3}}$.

The number of phases is $\textcolor{b l u e}{p = 3}$ (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 ${\text{AlCl}}_{3}$ since it was fully reacted):

• $\text{H"_2"O} \left(l\right)$
• ${\text{OH}}^{-} \left(a q\right)$
• ${\text{H}}^{+} \left(a q\right)$
• "Al"("OH")_3(s)
• $\text{HCl} \left(g\right)$

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

This gives that $\textcolor{b l u e}{{c}_{i}} = c - r - a = 5 - 1 - 1 = \textcolor{b l u e}{\boldsymbol{3}}$, 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 $\textcolor{b l u e}{p = 3}$. Therefore, the number of degrees of freedom are:

$\textcolor{b l u e}{f} = 2 + {c}_{i} - p$

$= 2 + 3 - 3 = \textcolor{b l u e}{\boldsymbol{2}}$

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.