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Standard XII

Chemistry

Enthalpy

The polymeris...

Question

The polymerisation of propene to linear polypropene is represented by the reaction
where n ahs large integral value, 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. The enthalpy of polymerisation is $- 360 k J m o l^{- 1}$ . Find the value of n.

n = 5

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n = 2

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n = 7

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None of these

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Solution

The correct option is A $n = 5$
$n (C H_{2} = C H_{2}) ⟶ - - (- - C H_{2} - C H_{2} - -) - -_{n}$
in polymerisation we can see that one double bond breaks and two single bond forms so enthalpy of polymerisation per mole $= 590 - 2 \times 331$
$= - 72 k J / m o l$
here, as enthalpy of polymerisation is $- 360 k J$
so value of n $= 360 / 72$
$= 5$

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

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Polymerisation of ethene to poly - ethene is represented by the equation
$n (C H_{2} = C H_{2}) \to (- C H_{2} - C H_{2} -)_{n}$
Given that average enthalpies of C = C and C - C bonds at 298 K are 590 and 331 kJ $m o l^{- 1}$ respectively, predict the enthalpy change when 56g of ethene changes to polyethylene.

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The correct option is A n=5n(CH2=CH2)⟶−−(−−CH2−CH2−−)−−nin polymerisation we can see that one double bond breaks and two single bond forms so enthalpy of polymerisation per mole =590−2×331 =−72kJ/molhere, as enthalpy of polymerisation is −360kJso value of n =360/72 =5