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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.

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  • $\begingroup$ This question might get an answer from someone if it gets some emergency CPR (and N)! How about defining C, P, R, and N? $\endgroup$
    – Ed V
    Sep 5, 2019 at 17:42
  • $\begingroup$ 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? $\endgroup$
    – user586228
    Sep 5, 2019 at 18:29
  • $\begingroup$ I have no clue, but maybe someone else will be able to answer. $\endgroup$
    – Ed V
    Sep 5, 2019 at 18:38

1 Answer 1

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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 $\ce{CO2}$.
  2. Three components.
  3. One chemical reaction.

Thus $$ F = 2 - 3 + 3 - 1 \rightarrow \boxed{F = 1} $$

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