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what is the entropy change for the vaporization of 3 mol H2O at 100 degrees Celcius...

what is the entropy change for the vaporization of 3 mol H2O at 100 degrees Celcius and 1 atm? H=40700 J/mol Answer in J/K

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Answer #1

We know that the equilibrium boiling temperature for water at 1 atm pressure is 100°C .So under these conditions, liquid water and water vapor are in equilibrium.

For an equilibrium reaction Gibbs free energy change is zero.

H2O(l) H2O(g)

ΔG = ΔHvap - TbΔSvap = 0

So ΔHvap = TbΔSvap

      ΔHvap /Tb = ΔSvap

Given ΔHvap = 40700 J/mol

                 = 3x40700 J/ (3 mol)          Given that it is 3 mole

                = 122100 J/(3mol)

         Tb = 100 oC = 100+273 = 373 K

ΔSvap = ΔHvap / Tb

          = 122100 / 373

          = 327.3 J/(3mol-K)

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