Question

Benzene is heated to boiling under a pressure of 1.0 atm with a 12 V source operating an electric current of 0.50 A. For how long would a current need to be supplied in order to vaporize 10 g of benzene? The molar enthalpy of benzene at its boiling point (353.25 K) is 30.8 kJ/mol

Answer 1

Molar mass of benzene = 78 g/mol

Mass of benzene needs to vaporize = 10 g

Moles of benzene needs to vaporize = 10/78 = 0.13 mol

Molar enthalpy of benzene = 30.8 kJ/mol.

Energy require to vaporize 0.13 mol benzene = 30.8 x 0.13 = 4 kJ/mol

Electrical energy E = VIt

Where V = voltage = 12 V

I = Current = 0.50 A

t = time

Now, in order to vaporize 10 g benzene energy requirement = 4 kJ = 4000 J

Therefore,

E = VIt = 4000

or, 12 x 0.50 x t = 4000

or, t = 4000/ (12 x 0.5)

or, t = 4000/6 = 667 s

Therefore the current need to be supplied for = 667 s