Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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feedback to help with question - Most of you managed this well once you realised that for Henry’s Law, the temperature should be approximated as the boiling temperature of the bulk component. This allowed the pure component vapour pressure of the dilute component to be evaluated. For the comparison with the x-y diagram, the determination of the initial gradient was approximate, but the value you found should certainly have corresponded more closely to the non-ideal Henry’s Law constant. This comment applies both to part (a) and part (b).
ans = a) (i) 3.48 atm; (ii) 7.21 atm, b) (i) 0.242 atm; (ii) 0.449 atm
Transcribed Image Text:Data:
Antoine coefficients (P in mmHg, Tin K, log to base e):
Margules parameters:
1. a) Calculate the Henry's Law constant for dilute methanol in water at 1 atmosphere:
i) assuming ideal solution behaviour;
ii) using the Margules Coefficients provided to account for the non-ideality of the solution.
Compare your answers with the value roughly obtained from the gradient in the x-y diagram below.
Ym (molefraction of methanol in vapour)
b) Now repeat part a) but this time to determine the Henry's Law constant for dilute water in
methanol at p=1 atm. Again, compare your results to the constant roughly obtained from the x-y
diagram.
0.8
0.6
0.4
Methanol: A = 18.588, B = 3626.6, C = -34.29
Water: A = 18.304, B = 3816.4, C = -46.13
Ethanol: A = 18.9119, B = 3803.98, C = -41.68
0.2
0
Methanol-Water: Awm = 0.6174, Amw = 0.7279
Ethanol-Water: Awe = 0.7917, Aew = 1.6366
0
x-y diagram for methanol/water at 1 atm
0.2
0.4
Xm (molefraction of methanol in liquid)
0.6
0.8
1
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