Question

The enthalpy of vaporization, DH_vap^o, 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.

Answer 1

Given,

H_vap[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] = H_vap/ 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