Question

In: Chemistry

Estimate the boiling point of benzene given that its vapour pressure is 20 kPa at 35C...

Estimate the boiling point of benzene given that its vapour pressure is 20 kPa at 35C and 50.0 kPa at 58.8C. Please show work.

Expert Solution

Solution :-

Lets first calculate the enthalpy of vaporization of benzene

P1 = 20 kpa

T1 = 35 C +273 = 308 K

P2 = 50.0 kpa

T2 = 58.8 C +273 = 331.8 K

Delta H vap = ?

ln [P2/P1] = Delta H vap / R [ (1/T1)-(1/T2)]

ln[50.0 kpa / 20.0 kpa] = Delta H vap / 8.314 J per mol K [(1/308)-(1/331.8)]

0.9163 = [ Delta H vap / 8.314 J per mol K ]* 0.000233

0.9163 * 8.314 J per mol K / 0.000233 K = Delta H vap

32711 J per mol = Delta H vap

Now lets calculate the boiling point of the benzene

At the boiling point the vapor pressure is 101.3 kpa

So

P2 = 101.3 kpa

T2 = ?

ln [P2/P1] = Delta H vap / R [ (1/T1)-(1/T2)]

ln[101.3 kpa / 20.0 kpa] = 32771.3 J per mol / 8.314 J per mol K [(1/308)-(1/T2)]

1.62235 = 3934.48 K * [0.003247 – 1/T2]

1.62235 / 3934.48 = 0.003247 – 1/T2

0.000412 = 0.003247 – 1/T2

0.000412 – 0.003247 = -1/T2

-0.00283 = - 1/T2

-1/-0.00283 = T2

352.8 K = T2

352.8 K – 273 = 79.8 C

So the boiling point of benzene = 79.8 ^oC or we can round it to 80 ^oC

queen_honey_blossom answered 2 years ago

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