Question

The polymerisation of ethylene to linear polyethylene is represented by the reaction

nCH₂=CH₂→(−CH₂−CH₂−)n
where n has a large integral value. Given that average enthalpies of bond dissociation for C=C &
C−C at 298 K are + 590 & + 331 kJ mol−1 respectively. Then the enthalpy of polymerisation per mole of ethylene at 298 K is:
A. − 62 kJ mol⁻¹
B. − 72 kJ mol⁻¹
C. − 82 kJ mol⁻¹
D. − 52 kJ mol⁻¹

Answer 1

The enthalpy of polymerization per mole of ethylene at 298 K is calculated by considering the enthalpy of bond dissociation for a C=C bond and the formation of a C-C bond. The result is −72 kJ/mol, matching option B.

Step-by-step explanation:

The question asks us to calculate the enthalpy of polymerization per mole of ethylene at 298 K. We start by understanding that in the polymerization reaction of ethylene (nCH2=CH2 → (−CH2−CH2−)n), each C=C double bond in the monomer ethylene is converted into a C-C single bond in the polymer polyethylene.

Using the average enthalpies of bond dissociation for C=C (590 kJ/mol) and C−C (331 kJ/mol) bonds provided, we can calculate the enthalpy change per mole of ethylene.

The enthalpy change for the breaking of one mole of C=C double bonds is +590 kJ, and the enthalpy change for the formation of one mole of C−C single bonds is −331 kJ. The overall enthalpy change per mole of ethylene (∆H) would be the sum of these values:

∆H = Enthalpy of bond formation - Enthalpy of bond breaking

= (2 × −331 kJ) - (590 kJ)

= −662 kJ - 590 kJ = −72 kJ/mol

Thus, the enthalpy of polymerization per mole of ethylene is −72 kJ/mol, which corresponds to option B.