Question

Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °C, the vapor pressure of CCl4 is 0.530 atm, and its enthalpy of vaporization is 33.05 kJ/mol. Use this information to estimate the normal boiling point for CCl4.

please show work

Answer 1

Answer : Carbon tetrachloride, CCl₄, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 ⁰C, the vapor pressure of CCl₄ is 0.530 atm, and its enthalpy of vaporization is 33.05 kJ/mol. By using this information we have estimated the normal boiling point for CCl₄ and found it to be 76.3 ⁰C. This has been calculated as follows :

As we know to determine the normal boiling point of Carbon tetrachloride, CCl₄, we can use the Clausius Clayperon equation which is shown as follows ;

In (P2 / P1) = (- Hvap / R) x (1/T2 - 1/T1)

where

In = the natural logarithm

P2 = 0.530 atm

P1 = 1 atm since the boiling point is the point when the vapour pressure of compound is equal to 1 atm

Hvap = 33.05 kJ/mol

R = 0.00831447 kJ / mol Kelvin (the universal gas constant)

T2 = 273.15 + 57.8 ⁰C= 330.95 Kelvin

T1 = ? Kelvin (that we have to find out)

Now keeping these value in the above equation we get

In (P2 / P1) = (- Hvap / R) x (1/T2 - 1/T1)

= In (0.530 / 1 atm) = (- 33.05 kJ/mol / 0.00831447 kJ / mol Kelvin ) x (1/330.95 - 1/T1)

= - 0.634878 = - 3975 Kelvin x (1/330.95 - 1/T1)

Now cancelling negative sign from both sides we have

0.634878 = 3975 x 1/330.95 - 3975/T1

= 3975/T1 = 3975 /330.95 - 0.634878

= 3975/T1 = 12.01 - 0.634878 = 11.375122

Thus T1 = 3975 / 11.375122

= 349.4468 = 349.447 Kelvin

Or we can say 349.447 Kelvin - 273.15 Kelvin = 76.297 ⁰C

Or after rounding off we can say Ti would be equal to 76.3 ⁰C.

Thus, at 57.8 ⁰C, the vapor pressure of CCl₄ is 0.530 atm, and its enthalpy of vaporization is 33.05 kJ/mol. By using this information we have estimated the normal boiling point for CCl₄ and found it to be 76.3 ⁰C.