Question

The vapor pressure, P, of a certain liquid was measured at two temperatures, T. The data is shown in this table.   

T(K) P(kPa)

325 3.54

675 8.12

If you were going to graphically determine the enthalpy of vaporizaton, ΔHvap, for this liquid, what points would you plot?

Answer 1

We need to plot lnK on y axis and 1/T on x axis

P = 3.54

ln P = ln (3.54) = 1.26413

T = 325.0

1/T = 0.00308

point 1 is:

x = 0.00308

y = 1.26413

P = 8.12

ln P = ln (8.12) = 2.09433

T = 675.0

1/T = 0.00148

point 2 is:

x = 0.00148

y = 2.09433

Answer: we are going to plot these point 1 and 2

In other words, we should plot ln K on y axis and 1/T on x axis

----------------------------------------------------------------------------

rise = (y1-y2) = 1.26413 - 2.09433 = -0.8302

rise = (X1-X2) = 0.00308 - 0.00148 = 0.0016

slope = rise/run

= -0.8302/0.0016

= -520.35965

----------------------------------------------------------------------------

use

slope = - Hvap/R

= -520.35965 = -Ea/8.314

Hvap = 4326.3 J/mol