# How to calculate the standard Gibbs free energy of formation of bromine vapor at room temperature?

The normal boiling point of liquid bromine is $$\pu{58.2 ^\circ C}$$. At $$\pu{9.3 ^\circ C}$$ the equilibrium vapor pressure of liquid bromine is $$\pu{100 torr}$$.
From this data, calculate the standard state Gibbs energy of formation of bromine vapor at room temperature, $$\Delta G ^\circ_\mathrm{f}$$, $$\pu{298 K}$$.

I tried using Clausius Clapeyron equation and get $$\Delta H$$ but I can't manage to solve this problem.

I just used Clapeyron equation to get around $$\pu{32 kJ/mol}$$ but I can't really manage to get Gibbs free energy at standard state.

• @Buck Thanks for taking the time to brush up this question. It would be great if you could have a look at the titles, too, and make them more descriptive when you see buzzwords. – Martin - マーチン Sep 2 '19 at 9:22

If you calculate the equilibrium vapor pressure at 298K (25 C), you can calculate the change in Gibbs free energy of the vapor in going from this pressure to the hypothetical state of 1 bar and 298K using $$dG=VdP=RTd\ln P$$. Neglecting the Poynting correction, the free energy of the liquid at 298 and at the equilibrium vapor pressure is 0. And this is also the free energy of the vapor at 298 and at the equilibrium vapor pressure.