# What is the degree of freedom of the calcium carbonate dissociation reaction?

### Question

$$\ce{CaCO3}$$ dissociated in a closed system according to the reaction:

$$\ce{CaCO3(s) -> CaO(s) + CO2(g)}$$

Assuming the reaction is in thermodynamic equilibrium, what is/are the degree(s) of freedom?

### Doubt

I tried applying the condensed phase rule. All that I could arrive at was

$$C - P + 1 = 3 - 2 + 1 = 2.$$

However, the answer is slightly modified in the solution. The solution reads that $$C = N - R$$

where $$N = 3$$ and $$R = 1.$$ These are the factors I do not understand. All that I ask myself is where did I go wrong? Where was the exception?Please note here N stands for the number of reactants and R for the number of reactions.

• This question might get an answer from someone if it gets some emergency CPR (and N)! How about defining C, P, R, and N?
– Ed V
Commented Sep 5, 2019 at 17:42
• C are the number of components.P stands for phases .Now C=N-R from the solution.However I am yet to figure out what are they.Secondly,why do I need to use the condensed phase rule and not the simple phase rule? Commented Sep 5, 2019 at 18:29
• I have no clue, but maybe someone else will be able to answer.
– Ed V
Commented Sep 5, 2019 at 18:38

The degrees of freedom for a closed reacting chemical system are $$F = 2 - \text{Num. of phases} + \text{Num. of components} - \text{Num. of chemical reactions}$$ We have:
1. Three phases: $$\ce{CaCO3}$$ solid phase, $$\ce{CaO}$$ solid phase, and a gas phase with $$\ce{CO2}$$.
Thus $$F = 2 - 3 + 3 - 1 \rightarrow \boxed{F = 1}$$