How to calculate the Gibbs energy for the vaporisation of solid ammonia? [duplicate]

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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 $K_c=K_p/\left(RT\right)$?

marked as duplicate by orthocresol♦, Todd Minehardt, Jan, M.A.R., bonNov 29 '15 at 17:01

Actually for this reaction $\Delta G=\Delta G^\circ+RT\ln Q$ holds only when $Q$ is used in terms of partial pressures. This is because the activity of an ideal gas is measured by its pressure, not its concentration. This equation can be derived from the definition of the Gibbs energy.
• Your first statement is still wrong. For any substance $\Delta G=\Delta G^\circ+RT\ln Q$ holds for any $Q$. They can be interconverted. – Martin - マーチン Aug 31 '15 at 16:18