Decomposition of CaCO3 in a sealed vessel For the equilibrium

$$\ce{CaCO3(s) &lt;=&gt; CaO(s) + CO2(g)}$$

$\Delta H^\ominus = 178\ \mathrm{kJ\ mol^{-1}}$ 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 $n$?
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 $Q_\mathrm p &lt; K_\mathrm p$.

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, $n= 0.399$.