Use the following data to estimate the normal boiling point and the molar enthalpy of vaporization of lead: T (K) 1500 1600 1700 1800 1900 P (Torr) 19.72 48.48 107.2 217.7 408.2

ΔHvap  = 1.796 x 10⁵ J = 179.6 kJ

Tb = 1737 ºC

Explanation:

The expression we should use to solve this equation is given by

ln p = - ΔHvap/R x (1/T) +c

where ln p is the natural log of the pressues, ΔHvap is the enthalpy of vaporization, R is the gas constant (8.314 J/Kmol), T is the temperature and c is a constant. This is called the Clausius-Clayperon equation for the temperature dependence of pressure.

Note this equation has a form

y= mx + b

where m the slope is equal to - ΔHvap/R.

We have the values of T and P, therefore, we can determine the slope of these data to answer our question.

T(K)       1/T (K⁻¹)           P (torr)        lnp

1500    6.67 x 10⁻⁴        19.72        2.98

1600    6.25 x 10⁻⁴       48.48        3.88

1700     5.88 x 10⁻⁴      107.2         4.67

1800     5.56 x 10⁻⁴      217.7         5.38

1900     5.26 x 10⁻⁴      408.2       6.01

Using a comercial software or calculating by hand the best slope, one is able to determine the desired values. In excel we get the equation of line:

y = -21598x + 17.38

with R²= 1

Therefore

- ( 21598 ) = -ΔHvap/ 8.314

⇒ ΔHvap  = 1.796 x 10⁵ J = 179.6 kJ

and to calculate the boiling point we will use the equation for the best line just determined with p = 760 torr

ln p = -21598x + 17.38

where p is the pressure , and Tb is the normal boiling point:

ln ( 760 ) = -21598 ( 1/Tb ) + 17.38

Tb = 2009 K = 1737 ºC

thats how its supposed to be

can you attach the