Heat of vaporization help?

Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 290.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this liquid.

Answer 1

Hello Alex!

 

This is a Clausius-Clapeyron equation situation.

 

ln (P₂/P₁) = (delta H vap /R) * (1/T₁-1/T₂)

P₁ = 92.0 Torr
T₁ = 23.0 degrees Celsius (we have to convert these to Kelvin) = 23.0 + 273.15 = 296.15 K

P₂ = 290.0 Torr
T₂ = 45.0 degree Celsius = 45.0 + 273.15 = 318.15 K

 

R is a constant at 8.314 J /K mol

 

Now we plug and solve! 

 

ln(P₂/P₁) = (delta H vap /R) * (1/T₁-1/T₂)

 

= ln (290.0 Torr / 92.0 Torr) = (delta H vap / (8.314 J /K mol) * (1/296.15K - 1/318.15K)

 

(This equation is always super intense! Just do one piece of it at a time, it'll keep you sane).

 

= ln (290.0 Torr / 92.0 Torr) = delta H vap / (8.314 J /K mol) * (1/296.15K - 1/318.15K)

 

= ln (290.0 Torr / 92.0 Torr) = 1.148                                                      =  (1/296.15K - 1/318.15K) = 2.334 x 10^-4 K

 

 

1.148 = delta H vap / (8.314 J /K mol) * 2.334 x 10^-4 K

 

I am going to rewrite this in a way that's easier to see how we solve algebraically: 

 

               delta H vap

1.148 =  ___________  * 2.334 x 10^-4 K

               

              8.314 J/K mol

 

 

delta H vap = 40876.42 J or 40.876 KJ