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For the equilibrium

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

$\Delta H^\ominus = 178\ \mathrm{kJ\ mol^{-1}}$ and $\Delta S^\ominus = 160\ \mathrm{J\ mol^{-1}}$.

I'm asked to suppose that $1\ \mathrm{mol}$ of $\ce{CaCO3}$ and $1\ \mathrm{mol}$ of $\ce{CO2}$ are heated reversibly in a closed vessel under a pressure of one bar.

How much $\ce{CaCO3}$ is present in the container when the temperature reaches $1100\ \mathrm K$?

I know that at this temperature $\Delta G^\ominus = 2\ \mathrm{kJ\ mol^{-1}}$ and $K_p = 0.80$. I'm thinking that since the equilibrium constant $K = p(\ce{CO2})/p^\ominus$ is less than $1$, then the $\ce{CO2}$ would combine with any $\ce{CaO}$ around to make $\ce{CaCO3}$, but since there isn't any the composition must remain unchanged at $1\ \mathrm{mol}$ $\ce{CaCO3}$ and $1\ \mathrm{mol}$ $\ce{CO2}$. Is this right?

Now suppose the temperature reaches $1120\ \mathrm K$, at which $K_p = 1.14$. Some of the $\ce{CaCO3}$ will start to decompose, right? Suppose $n$ mol do, so that I have $1-n$ mol of CaCO3, $n$ mol of $\ce{CaO}$ and $1+n$ mol of $\ce{CO2}$.

How do I find $n$?

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  • $\begingroup$ Is your system Isobaric or Isochoric? $\endgroup$
    – A.K.
    Jun 8, 2016 at 14:55
  • $\begingroup$ Isochoric, I assume, since the vessel is described as "closed". $\endgroup$
    – donkey
    Jun 8, 2016 at 15:05
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    $\begingroup$ In that case Gibb's free energy equations are not appropriate. You should use Hemholtz energy for an isochoric system. $\endgroup$
    – A.K.
    Jun 8, 2016 at 18:09
  • $\begingroup$ Hmm, you're right. If the pressure is constant, then the volume must change, though? $\endgroup$
    – donkey
    Jun 8, 2016 at 18:17
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    $\begingroup$ The system is isobaric, the question specifies "under a pressure of 1 bar". $\endgroup$
    – orthocresol ♦
    Jan 17, 2017 at 12:33

1 Answer 1

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You are right that at $1100~\mathrm{K}$ there will not be any decomposition of $\ce{CaCO3}$ in the presence of $\ce{CO2}$.

But, as the temperature increases further to $1120~\mathrm{K}$, $\ce{CaCO3}$ starts decomposing because at this temperature $Q_\mathrm p < K_\mathrm p$.

As the reaction is taking place in a closed container, by viewing the reaction mixture as an ideal gas mixture, we can calculate the amount of $\ce{CaCO3}$ decomposed at $1120~\mathrm{K}$:

$$\mathrm P(\ce{CO2}) \times V = 1 \times R \times 1100 \tag1$$ $$\mathrm P(\ce{CO2}) = 0.80$$ $$\mathrm P(\ce{CO2}) \times V = (1+n) \times R \times 1120 \tag2$$ $$\mathrm P(\ce{CO2}) = 1.14$$

On solving, $n= 0.399$.

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    $\begingroup$ The system is isobaric, the question specifies "under a pressure of 1 bar". – orthocresol♦ Jan 17 at 12:33 $\endgroup$
    – Mrigank
    Jan 23, 2017 at 14:57

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