Answer to Question #131730 in General Chemistry for jade

Question

A liquid has its vapor pressure measured at several temperatures. A graph of the ln P vs 1/T is prepared and appears below. If the equation for the line is..

y = -4699.3 k X + 2.38

What is ΔHvap for the liquid in kJ/mol?

Answer 1

We know that lnp =-ΔH/RT+I

When plot the ln P vs 1/T

Then slope = =-ΔH/R

y = -4699.3 K X + 2.38, From this equation slope = -4699.3k

-ΔH/R =-4699.3K      R=8.314 J.K^-1.mol^-1

ΔH =4699.3K×8.314 J.K^-1.mol^-1

     = 39069.98 J mol^-1

     = 39.07 k J mol^-1  (Answer)