Question

The Org. Synth. prep reports the boiling point of the product diene to be 80°; at 0.15 mmHg.

1) What would you expect the boiling point to be at 760 mmHg? Please explain in detail how you arrived at your answer.

2) What would you expect its boiling point to be at 0.1 mmHg? Please explain in detail how you arrived at your answer.

Answer 1

According to van't Hoff's equation:

ln(P2/P1) = H_vaporization/R * (1/T1 - 1/T2) ........ Equation 1

Where P1 and P2 are the initial and final pressures, respectively.

T1 and T2 are the initial and final temperatures, respectively.

H_vaporization is the heat of vaporization of the product diene under atmospheric pressure.

'R' is the universal gas constant = 8.314 J mol^-1 K^-1

P1 = 0.15 mmHg, T1 = (80 + 273.15) K = 353.15 K

1) P2 = 760 mmHg, T2 = ?

2) P2 = 0.1 mmHg, T2 = ?

Note: H_vaporization of the product diene, under atmospheric pressure, should be given in order to calculate T2 in both the problems 1 and 2.

If you have that value, substitute in the Equation 1 to get the value of T2.