Question

the molar heat of vaporization of ethanol is 39.3 kj/mol and the boiling point of ethanol is 78.3 celcius calculate delta s for the vaporization of 0.60 mol of ethanol

Answer 1

heat of vaporization of ethanol = 39.3 kj/mol

boiling point = 78.3 oC = 351.45 K

heat of vapourization for 0.60 mol = 39.3 x 0.60

                                                     = 23.58 kJ

delta s = delta H / T

           = 23.58 x 10^3 / 351.45

delta s = 67.1 J / K