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


1 Answer 1


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.

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    $\begingroup$ It says nowhere in the definition of amount (of substance) concentration that the concept must not be used for gasses. It might not lead to the desired result in this case, but it is not necessarily wrong. Also the activity is derived from the chemical potential of a substance and for an ideal gas not measured by the pressure. However, it can be related to the pressure. $\endgroup$ Commented Aug 31, 2015 at 12:17
  • $\begingroup$ Yes,the term concentration can be used.What I meant was that the activity or fugacity of an ideal gas itself happens to be its pressure per unit atmosphere or bar,as per the definition of the standard state.The definition of activity is of course only from the chemical potential,independent of the substance $\endgroup$ Commented Aug 31, 2015 at 16:10
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    $\begingroup$ 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. $\endgroup$ Commented Aug 31, 2015 at 16:18
  • $\begingroup$ What exactly do you mean by 'they can be interconverted'?I understand that the "Q's" can be written in terms of concentration or pressure,but I don't see how both might give the same result in this case(assuming we are referring to an ideal gas). $\endgroup$ Commented Sep 2, 2015 at 12:24
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    $\begingroup$ Have a look at this question $\endgroup$ Commented Sep 3, 2015 at 2:36

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