Question

The normal boiling point of methanol is 64.7∘c and the molar enthalpy of vaporization of 71.8kj/mol. the value of δs when 1.11 mol of ch3oh(l) vaporizes at 64.7∘c is ________ j/k.

Answer 1

The answer is 236.5 J/K

According to Δ G formula:

ΔG = ΔH - TΔS

when ΔG is the change in free energy (KJ)

and ΔH is the change in enthalpy (KJ)= ΔHvap * moles

                                                              = 71.8 KJ/mol * 1.11 mol
                                                             
                                                              =   79.7 KJ

and T is the absolute temperature (K)= 64 °C + 273°C = 337 K

Δ S is the change in entropy  KJ/K

by substitution:

when at equilibrium ΔG = 0 

∴ΔS = ΔH / T

       =79.7 KJ/ 337 K

     = 0.2365 KJ/K

     = 236.5 J/K