12
$\begingroup$

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 $n$?

$\endgroup$
  • $\begingroup$ Is your system Isobaric or Isochoric? $\endgroup$ – A.K. Jun 8 '16 at 14:55
  • $\begingroup$ Isochoric, I assume, since the vessel is described as "closed". $\endgroup$ – donkey Jun 8 '16 at 15:05
  • 1
    $\begingroup$ 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 '16 at 18:09
  • $\begingroup$ Hmm, you're right. If the pressure is constant, then the volume must change, though? $\endgroup$ – donkey Jun 8 '16 at 18:17
  • 3
    $\begingroup$ The system is isobaric, the question specifies "under a pressure of 1 bar". $\endgroup$ – orthocresol Jan 17 '17 at 12:33
1
$\begingroup$

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, $n= 0.399$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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