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? Express your answer using three significant figures.

Answer 1

Given :

Boiling point of acetic acid = 118 ⁰C

Delta H vap = 23.4 kJ /mol = 23400 J / mol

We have to find vapor pressure at 40 deg C

Solution:

We use Classius clayperon equation

Ln (P2/P1) = - Delta H / R ( 1 / T2 – 1 / T1)

Here delta H is heat of vaporization , R is gas constant = 8.314 J / ( K mol )

Here p1 and p2 are the vapor pressures at the T1 and T2 (temperature)

We know at boiling temperature atmospheric pressure = vapor pressure

So the at boiling point T1 = 118 deg C   ( 118 + 273.15 ) K , vapor pressure p1 = 1.0 atm

T2 = 40.0 deg C( 40+273.15) K , P2 = ?

Lets plug in the values.

Ln ( p2 / 1 atm ) = (- 23400 /8.314 ) x ( 1/ 391.15)

P2 = 0.167 atm

So the vapor pressure of acetic acid at 40 deg C = 0.167 atm

Conversion of vapor pressure to mmHg

= 0.167 atm x 760 mmHg / 1 atm

= 126.9 mmHg