The normal boiling point of bromine (Br2) is 58.8°C, and its molar enthalpy of vaporization is ΔHvap = 29.6 kJ/mol.
Calculate the value of ΔS when 1.00 mol of Br2(l) is vaporized at 58.8°C.?
Chemistry - DrBob222, Tuesday, December 4, 2012 at 9:46pm
dG = dH - TdS
At equilibrium (which you have at the boiling point), dG = 0, then
0 = dH -TdS
TdS = dH
dS = dH/T = ?
Remember T must be in kelvin; also dH in kJ gives dS in kJ or dH in J gives dS in J.
Chemistry - Anonymous, Friday, October 31, 2014 at 3:26pm