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Question: The polymerisation of ethylene to linear polyethene is represented by the reaction \[nC{H_2} = C{H_2...

The polymerisation of ethylene to linear polyethene is represented by the reaction nCH2=CH2(CH2CH2)nnC{H_2} = C{H_2} \to {( - C{H_2} - C{H_2} - )_n} where n has a large integral value. Given that average enthalpies of bond dissociation for C=C  C = C\; & CC  C - C\; at 298 K{\text{298 K}} are +590 + 590 and  + 331 kJ  mol - 1  {\text{ + 331 kJ\;mo}}{{\text{l}}^{{\text{ - 1\;}}}}respectively. Then the enthalpy of polymerisation per mole of ethylene at 298 K{\text{298 K}} is:
A.62kJ mole - 1 - 62\,{\text{kJ mol}}{{\text{e}}^{{\text{ - 1}}}}
B.72kJ mole - 1 - 72\,{\text{kJ mol}}{{\text{e}}^{{\text{ - 1}}}}
C.82kJ mole - 1 - 82\,{\text{kJ mol}}{{\text{e}}^{{\text{ - 1}}}}
D.52kJ mole - 1 - 52\,{\text{kJ mol}}{{\text{e}}^{{\text{ - 1}}}}

Explanation

Solution

To answer this question, you should recall the concept of polymerization. Polymerization refers to the process of molecules called monomers reacting together and produces three-dimensional networks or polymer chains is called polymerization. We shall calculate the energy required for the dissociation of the double bond and the formation of the new single bonds in polyethylene.

Complete step by step solution:
During the polymerization reaction: nCH2=CH2(CH2CH2)nnC{H_2} = C{H_2} \to {( - C{H_2} - C{H_2} - )_n}, one mole of ethylene breaks, i.e. one C=C  C = C\; double bond breaks and the two CH2C{H_2} - groups are linked with CCC - Csingle bonds thus forming three single bonds. But in the whole unit of the polymer, the number of single CCC - C bonds formed per mole of ethylene is 2.
Energy due to the formation of 2 CC2{\text{ }}C - Csingle bonds =2 × 331 = 662 kJ/mol2{\text{ }} \times {\text{ }}331{\text{ }} = {\text{ }}662{\text{ kJ/mol}}.
Energy due to dissociation of 1 C=C  C = C\;double bond = 590 kJ/mol.   = {\text{ }}590{\text{ kJ/mol}}{\text{.}}\;
Thus, the enthalpy of polymerisation per mole of ethylene at 298 K = (590662)  = 72 kJ/mol  298{\text{ }}K{\text{ }} = {\text{ }}\left( {590 - 662} \right)\; = {\text{ }} - 72{\text{ kJ/mol}}\;.

Hence, the correct answer to this question is option B.

Note: You should know about the industrial methods of free radical polymerization:
1.Bulk polymerization: reaction mixture contains only initiator and monomer, no solvent.
2.Solution polymerization: reaction mixture contains a solvent, initiator, and monomer.
3.Suspension polymerization: reaction mixture contains an aqueous phase, water-insoluble monomer, and initiator soluble in the monomer droplets (both the monomer and the initiator are hydrophobic).
4.Emulsion polymerization: similar to suspension polymerization except that the initiator is soluble in the aqueous phase rather than in the monomer droplets (the monomer is hydrophobic, and the initiator is hydrophilic). An emulsifying agent is also needed.