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https://i.stack.imgur.com/uUcQr.jpg

Suppose we have this polymer chain. Now, we cross link some chains ( not all ) , so some chains get linked, and others remain as they are. What we now have is a 'mixture' of polymer chains which are linked, and those which aren't. My question is how does this increase the melting point? I've read that intermolecular forces between chains is replaced by the covalent bonds of cross-linking, which are much stronger, and thus melting requires much more energy. But melting doesn't involve the breaking of covalent bonds unless the structure we have is 'one giant molecule', like Diamond, in which there are no 'intermolecular forces' because it's just one molecule.

In our case, we most certainly will have more than one molecule, and so the melting point should depend on the intermolecular forces between the chains. As these forces are largely unchanged, the melting point should be unaffected. When we heat this molecule, the weakest of the forces ( intermolecular ) should be broken first, and thus the melting point should remain unchanged.

I've been confused about this for about a week now, and I'm surprised google cannot find a similar question.

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  • $\begingroup$ You're right, melting point does depend on the intermolecular forces. But the term intermolecular isn't quite applicable in the case of branched polymers like Bakelite, since the whole plastic just a very large molecule. In that case, you have to consider the branching bond energies to melt the plastic. $\endgroup$ – Pritt Balagopal Apr 25 '17 at 3:14
  • $\begingroup$ I see what you mean, but consider the following case : There are 5 chains of a polymer in a sample, and 3 are crosslinked to each other. Now, if we start heating it up, then the two which were not cross-linked will soon move freely, and the 3 which were cross-linked, though they cannot move independently of each other, can also move freely as a whole. The only bonds we've broken are the intermolecular forces between these chains, so the melting point should be unchanged. Or is it only 'melting' if the 3 cross-linked chains also move freely of each other? $\endgroup$ – Saad Apr 26 '17 at 9:53

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