Question

1)

How many kJ of energy are required to transform 30.00 moles of solid benzene at -20.00 °C to

gaseous benzene at 140.0 °C?

2)

Consider He, CaS, CH4, LiF, CaO, diamond, H2O, CH3CH2OH, LiCl, and CH3OCH3. Arranged

in increasing melting point, these are:

Answer 1

Melting point of the benzene is 5.5 deg C

Boiling point of benzene = 80.1 deg C

Total q in the process would be divided into following transitions.

1^st : From – 20.0 deg C to 5.5 deg C   

2^nd : From 5.5 to 5.5 deg C ( fusion)

3^rd: From 5.5 deg C to 80.1 deg C

4^th : From 80.1 deg C to 80.1 deg C ( vaporization)

5^th : From 80.1 deg C to 140 deg C

We know q = m c delta T

And for fusion , q= heat of fusion * mol

For vaporization q = heat vaporization * mol

q is heat absorbed/given , m= mass in g , c = specific heat ,

Delta T is change in T

Specific heat solid benzene is 1.516 J / g C

Liquid phase, 1.726 J/g degC

Gas phase 1.055 J / g deg C 1.055 J/g deg C

Heat of fusion:

9.9 kJ/ mol = 9.9 E3 J/ mol

Heat of vaporization = 30.77 E3 J /mol

Conversion of mole of benzene to mass

= 30.00 mol benzene * molar mass of benzene

= 30.00 mol benzene * 78.11 g per mol

= 2343.3 g benzene

1^st :

q = 2343.3 g * 1.516 J / g C * (5.5 – ( - 20.0) )

= 90587.29 J

2^nd

q = Delta Hf * mol

= 9.9 J / mol * 30.00 mol

=297000 J

3 rd :

q = 2343.3 g * 1.726 J/g degC * (80.1-5.5)

= 301722.37 J

4^th:

q = molar heat of vaporization * mol

= 30.77 E 3 J / mol * 30.00 mol

= 923100 J

5^th

q = 2343.3 g * 1.055 J/g deg C * ( 140.0 – 80.1 )

= 148083.67 J

Lets add 1 to 5

q = 90587.29 J +297000 J + 301722.37 J+923100 J +148083.67 J

q = 1760493 J

heat in kJ

= 1760.5 kJ

Heat required to transform = 1760.5 kJ

2)

Melting point order is increases from covalent molecule to polar molecule to ionic compound

Order is :

He< CH4 < CH3OCH3<CH3CH2OH<LiCl<LiF<CaS<CaO<diamond