Question

The normal melting point of H2O is 273.15 K, and Hfusion= 6010 J mol–1. Calculate the decrease in the normal freezing point at 100 and 500 bar assuming that the density of the liquid and solid phases remains constant at 997 and 917 kg m–3, respectively.

Answer 1

using the clapeyron equation for the solid-liquid boundary,

p=p* + (d_fusH/d_fusV) &ln(T/T*)

solving it for T,

ln(T/T*)=(p-p*)X dfusV/dfusH

or T=T* e^((p-p*)*dV/dH)

dV= Vm(l) - Vm(s)

since Vm = M/density,

Vm(l) = 18.02*10^-3/997 = 1.8074*10^-5 m^3/mol

Vm(s)=18.02/917 = 1.9651*10^-5 m^3/mol

dfusV = -1.577*10^-6 m^3/mol

so,

for p=100, T=273.15

so,

putting the values in the equation, we get

T=269.6 K

so decrease = 273.15-269.6

= 3.55 K

so the melting point decreases by 3.55 K