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

Question

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.

Answer 1

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.

Answer 2

Standard Gibbs energy at normal boiling point

At the normal boiling point, all species are at standard state (pure liquid, gas at normal pressure). The system is at equilibrium, so the Gibbs energy of the process is zero. Because all species are at standard state, the standard Gibbs energy is zero as well.

Standard Gibbs energy at room temperature

The OP used the Clausius Clapeyron equation to get the enthalpy of the process from two Gibbs energies at different temperatures. Now, we are asking for the Gibbs energy at a third temperature, which can be determined by using Clausius Clapeyron in the other direction.