Question

Enthalpy of polymerization of ${{SN}}2$ ethylene , as represented by the reaction , ${{nC}}{{{H}}_{{2}}}{{ = C}}{{{H}}_{{2}}}{{ - (C}}{{{H}}_{{2}}}{{ - C}}{{{H}}_{{2}}}{{{)}}_{{n}}}{{ - }}$ is ${{ - 100 kJ mo}}{{{l}}^{{{ - 1}}}}$ of ethylene. Given bond enthalpy of ${{C = C}}$ bond is ${{600 kJ mo}}{{{l}}^{{{ - 1}}}}$. Determine enthalpy of ${{C - C}}$ bond (in ${{KJ mo}}{{{l}}^{{{ - 1}}}}$).
A. ${{350}}$
B. ${{323}}$
C. ${{565}}$
D. ${{453}}$

Answer 1

Hint: Before moving towards the calculation part, first we will know about polymerization and which thermodynamic factors affect the polymerization. There are two types of thermodynamic contributions. They are enthalpy contributions and entropy contributions.

Complete step by step answer:
Polymerization is the mechanism by which the monomers chain themselves into macro-molecule. The polymerization mostly should be done in cooler conditions as it maximizes the yield. Certain polymerization for example ring opening of functional groups are done at a little higher temperature to cross the energy barrier of the reaction. The entropy for polymerization should be positive and the reaction should be exothermic as these two factors will make the reaction spontaneous. The enthalpy for an alkene polymerization is mostly \[ - 20{{kcalmo}}{{{l}}^{ - 1}}\]. The strength of ${{C = C}}$ is mostly ${{147kcalmo}}{{{l}}^{ - 1}}$ and strength of \[{{C - C}}\] is \[{{kcalmo}}{{{l}}^{ - 1}}\] so the difference between two values is ${{64kcalmo}}{{{l}}^{ - 1}}$ which is the value of ${{\pi }}$ bond.
So, when we break a double bond in the polymerization it is replaced by two sigma bonds means two new ${{C - C}}$ bonds are formed. So, in whole polymer chain of ethylene there is addition of two single bonds per mole
${{\Delta }}{{{H}}_{{{pol}}}}{{ = \Delta }}{{{H}}_{{{(C}}{{{H}}_{{2}}}{{ = C}}{{{H}}_{{2}}}{{)}}}}{{ - 2\Delta }}{{{H}}_{{{( - C}}{{{H}}_{{2}}}{{ - C}}{{{H}}_{{2}}}{{ - )}}}}$
${{2\Delta }}{{{H}}_{{{( - C}}{{{H}}_{{2}}}{{ - C}}{{{H}}_{{2}}}{{ - )}}}}{{ = 600 - ( - 100 ) = 700 KJ mo}}{{{l}}^{{{ - 1}}}}$
${{\Delta }}{{{H}}_{{{( - C}}{{{H}}_{{2}}}{{ - C}}{{{H}}_{{2}}}{{ - )}}}}{{ = }}\dfrac{{{{700}}}}{{{2}}}{{ = 350 KJ mo}}{{{l}}^{{{ - 1}}}}$

So, the correct answer is Option A.

Additional information:
There are many factors that affect the change in enthalpy. As the hybridization changes the bond strength changes and intra – molecular steric interactions are also different compared to monomers.

Note: Especially in olefin polymerization the phase of reactant is either liquid or gas but the product we get is in solid phase and mostly it is an exothermic reaction. The rate of reaction of this polymerization will be affected by the activators which are used in the chemical reaction.