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Enthalpy

Polymerisatio...

Question

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.

A

72 kJ

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B

-72 kJ

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C

144 kJ

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D

-144 kJ

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Solution

The correct option is D
-144 kJ

In the polymerisation of $C H_{2} = C H_{2}$ into $(- C H_{2} - C H_{2} -)_{n}$ , one C = C double bond is changed into 2C - C bonds.
Hence, $Δ H = Δ H_{C = C} - 2 Δ H_{C - C} = 590 - 2 \times 331$
$= - 72 k J m o l^{- 1}$
Number of moles of $C_{2} H_{4}$ polymerised $= \frac{56}{28} = 2$
Enthalpy change $= 2 \times (- 72) = - 144 k J$

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