Question

# The enthalpy of vaporization, DHvapo, of ethylene glycol (HOCH2-CHOH-CH2OH) is 58.9 kJ/mol, and the vapor pressure...

The enthalpy of vaporization, DHvapo, of ethylene glycol (HOCH2-CHOH-CH2OH) is 58.9 kJ/mol, and the vapor pressure of ethylene glycol at 100°C is 14.9 mm Hg. Calculate the normal boiling point of ethylene glycol, reported in °C. Show all work and units and circle your final answer.

Given,

Hvap[ethylene glycol] = 58.9 kJ/mol x (1000 J / 1 kJ) = 58900 J/mol

Vapour pressure(P1) = 14.9 mmHg x ( 1 atm /760 mmHg) = 0.0196 atm

Temperature(T1) = 100 oC + 273.15 = 373.15 K

We know,

Vapoure pressure(P2) at normal boiling point = 1 atm

We know, the Clausius-clayperon equation,

ln [P1/P2] = Hvap/ R [1/T2 - 1/T1]

Here, R = 8.314 J/K.mol

Substituting the known values,

ln [0.0196 atm/1 atm] = 58900 J/mol/8.314 J/K.mol [1/T2 - 1 /373.15]

-3.932 = 7084.436 [ 1/T2 - 0.00268 ]

[ 1/T2 - 0.00268 ] = -0.00056

1/T2 = 0.002125

T2 = 470.6 K

Converting to oC,

= 470.6 -273.15

= 197.5 oC

Thus, normal boiling point of ethylene glycol = 197.5 oC

#### Earn Coins

Coins can be redeemed for fabulous gifts.