Determining the temperature at which 90% of the reactants in a equilibrium have reacted I have the following problem I am stuck on, specifically the last bit (iii). The first parts are fairly trivial and I have arrived at the following equation to determine the temperature:

$$T=\frac{\Delta H}{\Delta S - R \ln K}$$

However I am unsure on how to proceed. I do not have a value for $K_p$ and I can't think of how to deduce it. I have tried the following, by assuming that
$n_{\ce{Fe3O4}}=n_{\ce{CO}}$, going to the following expression

$$K = \frac{(0.9 x)^3 (0.9 x)}{(0.1x)(0.1x)}$$

but obviously that doesn't cancel.

Any help? I'm probably barking up the wrong tree with the above expression as I'm aware I'm looking for $K_p$ in terms of partial pressures but it would surely result in the same overall problem.

Thanks.
Assuming $\ce{FeO}$ and $\ce{Fe3O4}$ are solid throughout the range of temperatures we are interested in, they won't appear in the expression for the equilibrium constant $K$ (which in this case is equivalent to $K_p$). If $n$ is the total number of moles of $\ce{CO}$ that you start with,

$$K_p = K = \frac{p_{\ce{CO2}}}{p_{\ce{CO}}} = \frac{n_{\ce{CO2}}}{n_{\ce{CO}}} = \frac{0.9 n}{0.1 n} = 9$$

Plugging this into the formula previously derived, we find $T = 1388~\mathrm{K}$.