Question

Acetic acid has a normal boiling point of 118 ∘C and a ΔHvap of 23.4 kJ/mol.What is the vapor pressure (in mmHg) of acetic acid at 40 ∘C?

Answer 1

Acetic acid has a normal boiling point of 118 ∘C and a ΔHvap of 23.4 kJ/mol.

What is the vapor pressure (in mmHg) of acetic acid at 40 ∘C?

Recall that in equilibrium; especially in vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.

The equation is given as:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Where

P2,P1 = vapor pressure at point 1 and 2

dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol

R = 8.314 J/mol K

T1,T2 = Saturation temperature at point 1 and 2

Therefore, we need at least 4 variables in order to solve this.

Substitute all known data:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Change negative signs

ln(P2/P1) = dHvap/R*(1/T1-1/T2)

substitute known data

T1 = 118°C = 118+273 = 391 K; P1 = 760 m Hg

T2 = 40°C = 40+273 = 313 K; P2 = X

ln(P2/P1) = dHvap/R*(1/T1-1/T2)

ln(X/760) = (23.4*10^3)/8.314*(1/391 -1/313 )

ln(X/760) = -1.7938

X = 760*exp(-1.7938)

X = 126.408 mm Hg

clealy, vapor pressure must decreaes a T decreass