Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 400.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this liquid. And also calculate the normal boiling point in celcius.

Answer 1

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)

ln(400/92) = dHvap/8.314*(1/(23+273)-1/(45+273))

dHvap = ln(400/92)*8.314 / (1/(23+273)-1/(45+273))

dHvap = 52279.059 J/mol = 52.28 kJ/mol

b)

normal BP = P = 760 torr; T = ?

substitute in

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

ln(760/92) = 52279.059 /8.314 * (1/(23+273) - 1/T2))

ln(760/92) *8.314 / 52279.059 - 1/(23+273) = -1/T2

-0.00304 = -1/T2

T2 = 1/0.00304

T2 = 328.947 K

T2 = 328.947-273.15 = 55.797°C