Question

In: Chemistry

Estimate the heat of vaporization of diethyl ether at its normal boiling point using Trouton’s rule...

Estimate the heat of vaporization of diethyl ether at its normal boiling point using Trouton’s rule and Chen’s rule and compare the results with a tabulated value of this quantity. Calculate the percentage error that results from using each estimation. Then estimate ΔH^v at 100°C using Watson’s correlation.

Solutions

Expert Solution

a)

Troutons rule:

dHvap / (R*Tb) = 10

so

dHvap = R*Tb*10

Tb = normal boiling point for diethyl ether = 307.4 K

R = ideal gas constant = 8.31 J/molK

dHvap = 8.314*(307.4)*10 = 25,557.236 J/mol

dHvap = 25.557 kJ/mol

b)

Apply Chen rule:

dHv = R*Tb*(3-974*(Tb/Tc) 3.958 + 1.555*ln(Pc)) / (1.07 - (Tb/Tc)

where, Tc and Pc are critical values

so

Tc = 466.7 K

Pc = 36.4 bar

Substitute data

dHv = (8.314*307.4) (3-974*(307.4/466.7) - 3.958 + 1.555*ln(36.4)) / (1.07 - (307.4/466.7)

dHva = 26400 J/mol

dHv = 26.4 kJ/mol

Actual value for Hvap is approx = 26.52 kJ/mol

for

%error = Real - Calcualted / real * 100

%error Trouton = (26.52- 25.57)/(26.52) * 100 = 3.6%

%error Chen = (26.52.-26.4)/(26.52) * 100 = 0.5 %

c)

Whatson correlation

dH 373 / dH 307 = ((1-Tr2) / (1-Tr1) ) ^0.38

Tr = reduced temperature i.e. Ti/Tc

Tr1 = 307.4/466.7 = 0.685

Tr2 = 373/466.7 = 0.80

So

dH at 373 = dH at 307 * ( 1-Tr2 ) /(1-Tr1)^0.38

dH @ 373 = 26.52* ( 1-0.08) / (1-0.658) ^0.38

dH @ 373 = 21.63 kJ


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