1. A similar experiment was performed in lab and was carried out by your instruc
ID: 820629 • Letter: 1
Question
1. A similar experiment was performed in lab and was carried out by your instructor. THe data was collected and analyzed with the use of Excel. The equation of best fit line was y= -4058.7x + 16.10 . Using the equation of the best fit line, calculate the heat of vaporization ( Delta H vap) and the normal boiling point (in degrees C) of this volitile liquid. (The experiment pressure was measured in kPa not atm).
(Hint: What is the value of P vap when a liquid boils at its normal boiling point?)
I've calculated H vap, it is 33.7440318 kJ/mol
However, I need help in calculating the normal boiling point.
Also,
2. Using the answers from question 1, identify the volitile liquid based on the following data. Explain your reasoning.
Compound Normal Boiling Pt. ( Degrees C) Delta H vap (kJ/mol)
Ethanol 78.3 42.4
iso- Propanol 82.3 45.4
Cyclohexane 80.7 33.1
Please help me, or guide me to figuring out these questions... any help would be greatly appreciated. Thank you
Explanation / Answer
(1) The straight line (y = mx + c) is a plot of the Clausius-Claperyon equation:
ln(P) = (-DHvap/R) x (1/T) + c
where y = ln(P), x = 1/T, slope = m = (-DHvap/R) and y-intercept = c
Comparing with equation for best fit: y = -4058.7x + 16.10:
Slope = -DHvap/R = -4058.7
Heat of vaporization DHvap = -slope x R
= 4058.7 x 8.314
= 3.37 x 10^4 J/mol = 33.7 kJ/mol
At normal boiling point, vapor pressure P = atmospheric pressure = 1 atm = 101.325 kPa
Substituting value into equation:
ln(101.325) = -4058.7 x (1/T) + 16.10
1/T = 0.0028289
Boiling point T = 1/0.0028289 = 353.49 K = 80.3 deg C
(2) Experimental DHvap = 33.7 kJ/mol
Experimental boiling point = 80.3 deg C
Comparing experimental values with theoretical values, the closest match for the volatile liquid is
Cyclohexane:
Theoretical DHvap = 33.1 kJ/mol
Theoretical boiling point = 80.7 deg C
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