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 $n$?