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Question: If bond energy of \[C = C\]and \[C - C\] at 298 K are 600 and 330 \[KJmo{l^{ - 1}}\]respectively, th...

If bond energy of C=CC = Cand C−CC - C at 298 K are 600 and 330 KJmol−1KJmo{l^{ - 1}}respectively, then the magnitude of enthalpy change when 112 gm of ethylene changes into polythene is:
A.240 KJ
B.270 KJ
C.1080 KJ
D.540 KJ

Explanation

Solution

When ethylene molecule gets dissociated and forms polythene, there is dissociation of 1 double bond and formation of 2 single bonds and this energy of formation and dissociation could be calculated and their difference gives enthalpy for 1 mole and further could be calculated for 112 gm (converted to moles).

Complete step by step answer:
In the given question, ethylene is converted into polythene and in this process the double bond is converted into a single bond and the energy will be released which we will have to calculate.
The reaction of the polymerization is represented as: -
nCH2=CH2→ (−CH2−CH2−)nnC{H_2} = C{H_2} \to {\text{ }}{\left( { - C{H_2} - C{H_2} - } \right)_n}
So, in this reaction, we see that a double bond is breaking i.e. C=CC = C and forming 2 single bond units of CH2C{H_2}and we will find out the energy released in this bond.
So, in first part the energy because of dissociation of double bond C=C = 600 KJmol−1C = C{\text{ }} = {\text{ }}600{\text{ }}KJmo{l^{ - 1}} (because 1 double bond present)
Energy due to formation of single bond C−C = 2 × 330 KJmol−1= 660 KJmol−1C - C{\text{ }} = {\text{ }}2{\text{ }} \times {\text{ }}330{\text{ }}KJmo{l^{ - 1}} = {\text{ }}660{\text{ }}KJmo{l^{ - 1}} (because 2 single bonds formed by dissociation of 1 double bond)
Therefore, magnitude of enthalpy for 1 mol of ethylene converting to polythene = 600 − 660 KJmol−1 = −60 KJmol−1600{\text{ }}-{\text{ }}660{\text{ }}KJmo{l^{ - 1}}{\text{ }} = {\text{ }} - 60{\text{ }}KJmo{l^{ - 1}}
In the question, it is given that 112gm of ethylene is converting to polythene and here we have calculated for 1 mole. So, converting 112 gm into moles
Mole of ethylene =Given weightmolecular weight=11228.05=3.99 moles = \dfrac{{Given{\text{ }}weight}}{{molecular{\text{ }}weight}} = \dfrac{{112}}{{28.05}} = 3.99{\text{ }}moles
So, for 1 mole, enthalpy released = 60 KJmol−160{\text{ }}KJmo{l^{ - 1}} (negative sign indicates that energy is released)
For 3.99 mole, enthalpy released = 60 ×3.99 = 239.57≈240 KJ60{\text{ }} \times 3.99{\text{ }} = {\text{ }}239.57 \approx 240{\text{ }}KJ

So, the correct answer is A i.e. 240 KJ240{\text{ }}KJ .

Note:
Enthalpy of formation is defined as when 1 mole of compound formed from its most stable element and similarly there are other enthalpies also like enthalpy of neutralization, enthalpy of dissociation, enthalpy of atomization.