Given that $P_{\ce{NH3}}=\mathrm{10~atm}$ at $298\ \mathrm K$, calculate the $\Delta G^\circ$ of the reaction $$\ce{NH3(s) <=> NH3(g)}$$
I'll begin solving the problem...
I'm using the formula $$\Delta G=\Delta G^\circ+RT\ln Q.$$ The vaporisation of $\ce{NH3}$ is an equilibrium so $\Delta G=0$ and $Q=K_c$ . Also $K_c=\left[\ce{NH3}\right]$. So $\Delta G^\circ=-RT\ln K_c$.
How do I continue? Is it $K_c=K_p=10$ or $K_c=K_p/\left(RT\right)$?