# 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 Sep 5 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? – user586228 Sep 5 at 18:29
• I have no clue, but maybe someone else will be able to answer. – Ed V Sep 5 at 18:38