A similar experiment to the one performed in lab was carried out by our instruct

Question

A similar experiment to the one performed in lab was carried out by our instructor. The data was collected and analyzed with the use of Excel. The equation for the best fit line was y=-5532.6xÇ=20.19. Using the equation of the best line, calculate the heat of vaporization (deltaHvap) and the normal boiling point(in degreeC) of this volatile liquid. Remember for this experiment pressure is measured in kPa not atm Using the answers from question 1 identify the volatile liquid based on the following data. Explain your reasoning. Ethanol has a normal boiling point of 78.3 degree C. Why is the normal boiling point higher than methanol but lower than isopropanol listed above? Would you expect the heat of vaporization of ethanol to follow the same trend? Explain.

Explanation / Answer

-5532.6 = -deltaHvap / R

-5532.6 = -deltaHvap / 8.314 J/molK

deltaHvap = 46000.25 J./mol = 46 kJ/mol

Pvap = 101325 Pa = 101.325 kPa

ln (101.325) = - 5532.6 * (1/T) + 20.19

4.6183 = -5532.6 / T + 20.19

-15.57167 = -5532.6/T

T = 355.299 = 82.15 ºC

Judging by the results, the substance is iso-propanol.

Ethanol has a higher boiling point than methanol because there is more attraction between hydrogens than in methanol. And it has a lower bp than isopropanol because isopropanol is a more compact molecule, and it is more diffficult to break interactions among them. And yes, it is expected to follow the trend of the heat of vaporization.

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