question_answer

The polymerization of ethylene to linear polyethylene is represented by the reaction: \[nC{{H}_{2}}=C{{H}_{2}}\to {{[-\,C{{H}_{2}}-C{{H}_{2}}-]}_{n}}\] where n has large integral value. Given that the average enthalpies of bond dissociation for C = C and \[C-C\] at 298K are 590 and \[331\,\,kJ\,\,mo{{l}^{-1}}\]respectively, calculate the enthalpy of polymerization per mol of ethylene at 298 K.

A) \[-\,160\text{ }kJ/mol\]               

B) \[-\,72\text{ }kJ/mol\]

C) \[{{H}_{2}}{{O}_{2}}\]

D) \[-\,36\text{ }kJ/mol\]

Correct Answer: B

Solution :

In polymerization of ethane\[n\,({{H}_{2}}C=C{{H}_{2}})\xrightarrow{Polymerisation}{{(-\,C{{H}_{2}}-C{{H}_{2}}-)}_{n}}\]

Since, the no. of \[C-H\] bonds on both sides are equal. But n(C=C) changes to \[2n\,\left( C-C \right)\] bonds + 1.

or a C = C double bond changes to 3 \[\left( C-C \right)\]single bond

one bond each by \[C{{H}_{2}}\] group \[C-C\] and one bond extra.

\[n\,(C=C)=2n\,(C-C)+1\]

Since n is very large as compared to

\[n\,(C=C)\to 2n\,(C-C)\]

Enthalpy of polymerization

\[=n\,\,(\Delta \,{{H}_{c=c}})-2n\Delta {{H}_{c\,-\,c}}\]

\[=n\,\,\times \,\,590-2n\,\,\times \,\,331=-\,72n\]

Enthalpy of polymerization per molecule

\[=\frac{\Delta \,{{H}_{polymerisation/mol}}}{n}=-\frac{72n}{n}=-\,72\,kJ/mol\]