Question

The normal boiling point of liquid ethanol is 351 K. Assuming that its molar heat of...

The normal boiling point of liquid ethanol is 351 K. Assuming that its molar heat of vaporization is constant at 45.1 kJ/mol, the boiling point of C2H5OH when the external pressure is 0.602 atm is ? K.

Homework Answers

Answer #1

We know that the clausius-clapeyron equation is given by :

log (P2/P1) =(ΔHvap / 2.303*R) ( 1/T1 - 1/T2)

Where : P2 is pressure at temperature T2 and P1 is pressure at temperature T1

ΔHvap is enthalpy of vapourisation and R is universal gas constant (8.314 J/mol-K )

Substituting all the values we get :

log (0.602 atm / 1 atm ) = ( 45100 J mol-1 / 2.303 * 8.314 J mol-1 K-1 ) ( 1/351 K - 1/T2)

( 1/351 K - 1/T2) = -4.22 / 45100 K

On solving , we get T2 = 339.839 K

Therefore the boiling point of liquid ethanol at 0.602 atm pressure is 339.839 K.

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