The Gibbs' phase rule states that the number of degrees of freedom in a system $F = C - P + 2$ where $C$ are the number of components and $P$ are the number of phases. The term $C$ can be expanded to $C = s-r$, where $s$ are the number of species, and $r$ are the number of independent relations.

I'm trying to find number of components for two systems, but the two approaches doesn't add up. Where am I wrong?

System 1:

A closed, evacuated container with $NH_4Br(s)$ is heated, and the following equilibrium takes place: $NH_4Br(s) ⇌ NH_3(g) + HBr(g)$.

I believe the number of components C to be 1, $NH_4Br(s)$, since this substance alone can produce the gas mixture phase, and the solid phase. ($NH_3(g) + HBr(g)$ can also do this, but C should be the lowest number of components needed..).

Using the species approach, $s = NH_4Br, NH_3, HBr = 3$, and $r = $ 1 equilibrium reaction. Which gives that $C = s-r=3-1=$ 2 which doesn't match what I wrote above. Why?

System 2:

A saturated aqueous solution of $KCl(aq)$ with $KCl(s)$. There is also water vapor. I also assume this is an evacuated container, but the problem doesn't mention this.

I believe the number of components to be 2, $H_2O$ and $KCl$, since these substances can produce the solid phase $KCl(s)$, the gaseuous phase $H_2O(g)$, and the liquid phase $H_2O(l)+Cl^-(aq)+K^+(aq)$.

Using the species approach, $s = H_2O(aq/g), K^+(aq), Cl^-(aq), KCl(s)$, and $r =$ one phase with ion-balance + one equilibrium between water liquid and gas + one equlilibrium between $KCl(s)$ and the dissolved ions = 3. Which gives that $C = s-r = 4-3 = $ 1, which again doesn't match what I wrote above.

  • $\begingroup$ If the conclusion of a premise is wrong, compared to the observed F, check if the premise is right. $\endgroup$
    – Poutnik
    Nov 26, 2021 at 8:44
  • $\begingroup$ Your method for defining the # components seems a bit arbitrary (think how you would scale it for complex multiple reactions, or reactions with even product/reactant ratio at equilibrium - who is produced by who?). Normally you would include all species, reactions-aside. $\endgroup$ Nov 26, 2021 at 10:05
  • $\begingroup$ @Poutnik Well, I don't know the F, nor do I know what the conclusion is, as I have two answers. I've tried to think about the premise of both of them, but I can't see where I'm wrong.. That's why I ask the question. $\endgroup$
    – Quantonium
    Nov 26, 2021 at 10:50
  • $\begingroup$ @ViníciusGodim I'm going by the following description "The number of components, C, is the minimum number of substances (...) from which we could in principle prepare each individual phase of an equilibrium state of the system, (...)." Taken from the Libretexts article "The Gibbs Phase Rule for Multicomponent Systems " by Howard DeVoe. $\endgroup$
    – Quantonium
    Nov 26, 2021 at 10:51
  • 1
    $\begingroup$ @Quantonium P=2, not 3 ( solid NH4Cl and gas ), so C=1. There are 2 constrains - the reaction and the 1:1 gas ratio, so C=3-2 $\endgroup$
    – Poutnik
    Nov 26, 2021 at 11:43

1 Answer 1


The Gibbs phase rule for reacting systems is $$F = 2 - \pi + N - r - s$$ Where $F$ is number of degrees of freedom, $\pi$ is number of phases, $N$ is the number of components, $r$ the number of independent reactions and $s$ the number of extra constraints on the phase rule variables. These constraints originate by the way the system is created, for e.g. in the reaction $$\ce{NH_4Cl(s) -> NH_3(g) + HCl(g)}$$ If the only source of products is the reaction, there is the special constraint $y_\ce{NH_3} = y_\ce{HCl}$ ($= 0.5$ if these are the sole components). This counts as a constraint in this case ($s = 1$). (This example appears in the reference)

In the reaction $$\ce{NH_4Br(s) <=> NH_3(g) + HBr(g)}$$

$\pi = 2$, $C = 3$ (note that all components are considered), $r = 1$. Given the constraint $y_\ce{NH_3} = y_\ce{HBr}$, $s = 1$. So $F = 2 - 2 + 3 - 1 - 1 = 1$.

In the system $\ce{KCl(aq)}$ plus a vapor phase, I guess you model either $\ce{K^+}$ and $\ce{Cl^-}$ as independent components, or you could consider $\ce{KCl(aq)}$ as a component in equilibrium with $\ce{KCl(s)}$.

1st case: $\pi = 3$, $C = 4$, $r = 1$ (dissolution reaction), $s = 1$ (constraint given by mole balance), $F = 2 - 3 + 4 - 1 - 1 = 1$.

2nd case: $\pi = 3$, $C = 2$ ($\ce{KCl}$ treated as a single component), $s = 0$ (there is no extra constraint on $x_{\ce{KCl}}$), $r$ = 0 (no dissociation considerd, $\ce{KCl(s) -> KCl(aq)}$ is modeled as phase change). $F = 2 - 3 + 2 - 0 - 0 = 1$.

Reference: Smith, Joseph Mauk. "Introduction to chemical engineering thermodynamics." (1950): 584. (Chapter 14 - Chemical-Reaction Equilibria)

  • $\begingroup$ Thanks, but wouldn't $\pi$ = 3 in the KCl system (2)? KCl(s), an aqueous phase, and a gaseous phase. Also, shouldn't the phase change equilibria enter somewhere (water -> gas, and KCl(s) -> KCl(aq) or $K^+/Cl^-$) $\endgroup$
    – Quantonium
    Nov 26, 2021 at 12:08
  • $\begingroup$ @Quantonium Yes, you're right $\pi = 3$. The phase change equilibria is considered in the phase rule equation already (in its derivation), they should not be considered chemical reactions. $\endgroup$ Nov 26, 2021 at 12:15
  • $\begingroup$ I see. That's good, by this logic, my calculations in the OP also adds up. $\endgroup$
    – Quantonium
    Nov 26, 2021 at 12:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.