Question

Acetic acid has a normal boiling point of 118 ?Cand a ?Hvap of 23.4 kJ/mol. Part A What is the vapor pressure (in mmHg) of acetic acid at 35 ?C? Express your answer using three significant figures.

Answer 1

Answer:

The Clausius Clapeyron equation is

ln(P2/P1)=(?Hvap/R)/(1/T2 - 1/T1)

Given P1=?(let us say x), T1=35°C= 308K, P2=760 mmHg at T2=118°C=391 K.

(Since boiling point of acetic acid is 118°C, where vapor pressure is equal to atmospheric pressure)

?Hvap=23.4 kJ/mol=23.4 x10^-3 J/mol, R=8.314 J/mol.K

Then

ln(P1/P2) = (?Hvap / R)(1/T2 - 1/T1)
ln (x mmHg/ 760 mmHg) = (23.4x10^3 J/mol / 8.314 J/molK) (1/391K - 1/308K)

ln(x/760 mmHg)=-1.94

x/760 mmHg=e^-1.94

x=760 mmHg x 0.1437

x=109.24 mmHg.

Therefore the vapor pressure of acetic acid at 35°C ~ 109 mmHg.

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