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posted by Josh G. on Tuesday, December 4, 2012 at 5:26pm.

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.?

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.

89.17

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