Question

15 g of S are added to 100 g of chloroprene. What is the maximum fraction of crosslinks that can be obtained assuming only 12% of the crosslinking atoms are effective. B) What conditions have to be satisfied for polybutadiene to exhibit elastomeric behavior? Will it be an elastomer at -10C?

Answer 1

A.)

All the sulphur added is not utilised for cross linking. Only 12% is utilised for cross linking .

Atomic weight of chloroprene C₄H₅Cl = 88.5365 g/mol

Atomic weight of sulphur S = 32.065 g/mol

Only 12% is utilised so make 1 Complete cross linking 100/12 = 8.33 times extra sulphur should be added.

So 88.5365 gram of chloroprene requires 32.065 8.33 = 267.198 gram

In the problem it is given that 15 gram of sulphur is added means

15 12% = 1.8 gram is utilised in cross linking.

Amount of sulphur present per mer weight of chloroprene is

x = 1.8 gram

Fraction of cross linking = 1.8 / 32.065 = 0.0561 = 5.614 %

For your practice purpose I am giving you one more example below

B.) Polybutadiene is not an elastomer . It has a melting point of 80°C.

But Polybutadiene can exhibit the elastomeric behaviour!.

The temperature range for the elastic behaviour of elastomers is limited by the
glass transition temperature (Tg). At temperatures lower than glass transition
temperature the movement of molecule chains is very restricted and the large
elastic deformations are not possible. Elastomers are rigid and fragile materials
below the glass transition temperature.

So glass transition temperature of Polybutadiene ranges from -20°C to 110°C for different addition of monomers. At -10°C Polybutadiene does not exhibits elastomeric behaviour.