For bromine,ΔH°vap = 30.91 kJ/mol and ΔS°vap = 93.23 JK^-1mol^-1 at 25°C.What is the normal boiling point for bromine?

Question

Quizplus · Accepted Answer

The normal boiling point of a substance is the temperature at which its vapor pressure equals atmospheric pressure. At this point, the Gibbs free energy change (ΔG) for the phase transition is zero. The relationship between ΔG, ΔH, and ΔS is given by ΔG = ΔH - TΔS. Setting ΔG to zero for the boiling point and solving for T gives T = ΔH/ΔS. Given ΔH°vap = 30.91 kJ/mol = 30910 J/mol and ΔS°vap = 93.23 J/mol·K, the boiling point T in Kelvin is T = 30910 J/mol / 93.23 J/mol·K = 331.6 K. Converting this to Celsius gives approximately 58°C.

For Bromine,ΔH°vap = 30