Question

The normal boiling point of ethanol (c2h5oh) is 78.3 °c and its molar enthalpy of vaporization is 38.56 kj/mol. what is the change in entropy in the system in j/k when 42.2 grams of ethanol at 1 atm condenses to a liquid at the normal boiling point?

Answer 1

When the entropy change of vaporization
       
 = enthalpy of vaporization/boiling point temperature 

when we have the enthalpy of vaporization = 38560 J/mol

and the boiling point temperature in Kelvin = 78.3 + 273 = 351.3 K

by substitution:

∴the entropy change of vaporization = 38560J/mol/351.3K

                                                               = 109.76 J/K/mol

and when the liquid has lesser entropy than the gas and we here convert from

gas to liquid so, the change in entropy = -109.76 J/K/mol

now, we need the moles of C2H5OH = mass/molar mass when the molar

mass of C2H5OH = 46 g/mol and mass = 42.2 g 

∴ moles of C2H5OH = 42.2 g / 46 g/mol = 0.92 moles

when 1 mol of C2H5OH turns in liquid entropy change →-109.76 J/K/mol

∴ 0.92 mol of C2H5OH → X

∴ X entropy change when 0.92 mol = -109.76 *0.92 mol / 1 mol

                                                             = 84.64 J/K