In organic chemistry lab you are doing vacuum distillations. The normal boiling

Question

In organic chemistry lab you are doing vacuum distillations. The normal boiling point of a liquid occurs when the vapor pressure matches the external pressure under normal conditions (1 atm -760 mmHg 101325 Pa). To what pressure value must the system pressure be reduced to boil iodine at l 15.00 °C (Tboil-457.4 K. molar AHvap41.57 kJ/mol) [mmHg]? If instead, you want to purify your iodine (Tm 386.8 K) by sublimation at a pressure of 100 Pa (equilibrium vapor pressure over the solid = 10 Pa at-885 °C, molar AHfus 15.52 kJ/mol), what temperature is required PC

Explanation / Answer

Use clausis clapeyron equation

ln (P1/P2) = [detlaH / R] * (1/T2 - 1/T1)

P ---> pressure and T in K

P1 = 1 atm

T1 = 457.4K

T2 = 115C in Kelvin we needed

R = 8.314

delta Hvap = 41.57 KJ

ln (1/P2) = [41.57*1000 / 8.314] *[ 1/(273+115) - 1/457.4 ]

P2 = 0.1415 atm = 107.56245 mm Hg

2nd part

Same formula apply

Equilibrium vapor pressure over the solid at 386.8 K ?

ln (10/P2) = [15.52*1000/8.314] [1/386.8 - 1/(273-8.85) ]

P2 = 94 Pa

Equilibrium vapor pressure over the solid at 386.8 K is 94 Pa

ln (100 / 94) = [15.52*1000/8.314] [1/386.8 - 1/T ]

T = 391.8235 K = 118.823 C

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