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

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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.

In reactions:

a) FeO(g)+CO(g)-> Fe(s)+(CO2)g

b) water solution of AlCl3

determine

##### 1 Answer

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).

If we assume that there is *no further reaction occurring*, then there are

The number of phases,

Therefore,

An aqueous solution of

**CASE I**

If we assume a ** sufficiently dilute solution**, we can pretend that

#"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

The number of phases is

**CASE II**

If the solution is ** NOT sufficiently dilute**, and we assume a reaction

**occur (to completion), then we must include all of these non-independent components, assuming excess water (but neglecting the coordination of water to**

*DOES*#"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

This gives that

You should see that in both cases, we get the same number for

If the reaction is in a closed system, then we again have **three** phases (liquid, solid, and gas), so *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.