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Standard XII

Chemistry

Enthalpy

The polymeriz...

Question

The polymerization of ethylene to linear polyethylene is represented by the reaction,
$n (C H_{2} == C H_{2}) ⟶ - - (- - C H_{2} - - C H_{2} - -) - -_{n}$
where n has large integral value. Given that the average enthalpies of bond dissociation for $C == C$ and $C - - C$ at $298 K$ are $+ 590$ and $+ 331 k J m o l^{- 1}$ respectively.

Calculate the enthalpy of polymerization per mole of ethylene at $298 K$ :

Δ H = - 72 k J m o l^{- 1}

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Δ H = + 72 k J m o l^{- 1}

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Δ H = 1252 k J m o l^{- 1}

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None of these

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Solution

The correct option is A $Δ H = - 72 k J m o l^{- 1}$
Polymerisation of ethylene to linear polyethylene is
$n (C H_{2} == C H_{2}) ⟶ - - (- - C H_{2} - - C H_{2} - -) - -_{n}$
In this polymerization reaction, every molecule of ethylene involves breaking of one C = C (double bond) and formed two C-C (single bonds).

$E n e r g y r e l e a s e d = E n e r g y d u e t o f o r m a t i o n o f t w o s i n g l e b o n d s$
$= 2 \times 331$
$= 662 k J m o l e^{- 1}$ of ethylene.
$Δ H_{p o l y m e r i s a t i o n} = 590 - 662$
$= - 72 k J m o l^{- 1}$

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