Question

6.4. The vapor pressure of ethylene glycol at several temperatures is given below:



T (C) 79.7 105.8 120.0 141.8 178.5 197.3

p* (mmHg) 5.0 20.0 40.0 100.0 400.0 760.0


Use a semilog plot based on the Clausius-Clapeyron equation to derive an equation for p* (mm Hg) as a function of T (C). From the plot, estimate the heat of vaporization of ethylene glycol in kJ/mol.

Clausius-Clapeyron equation: ln p* = -Hv/RT + B

Answer 1

Answer:

BY_A.P.E.X

Explanation:

Plot P*(log scale) Vs        T + 273.2  (Vect)on semilog paper

Straight line = -7076k

Intercept = 21.67

InP*(mmHg) = $\frac{-7076}{T(C)+273.2} +21.67$

P(mmHg) = $Exp[\frac{-7076}{T(C)+273.2}+21.67 ]$

$\frac{\Delta Hv}{R} =7076K \Rightarrow \Delta Hv$

$= 7076 \times (\frac{8.314}{10^3} )\\\\=58.8kJ/mol$

b) InP*(atm) = -7074(1/T) + 15.03

c) T°C               P(atm)

   50°C            1.048 * 10⁻

   80°C             6.7302 * 10⁻    by A.p.e.x

   110°C            0.03229

The vapour pressure of Ethylene glycerol at 50⁰C, 80⁰C and 110⁰ C is shown in the above values