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Question: Enthalpy of polymerisation of ethylene, as represented by the reaction,\[nC{H_2} = C{H_2} \to {\left...

Enthalpy of polymerisation of ethylene, as represented by the reaction,nCH2=CH2→(−CH2−CH2)nnC{H_2} = C{H_2} \to {\left( { - C{H_2} - C{H_2}} \right)_n}​is -100 kJ per mole of ethylene. Given bond enthalpy ofC=CC = Cbond is 600kJmol−1600kJmo{l^{ - 1}} , enthalpy of C−CC - C bond (in KJ mol)\left( {in\,KJ\,mol} \right) will be:
(A) 116.7116.7
(B) 350350
(C) 700700
(D) Intermediates

Explanation

Solution

Hint - In this question we will come across the concept of thermodynamics. In this question we will be crossing path with many other important concepts like enthalpy, about reaction enthalpy ΔrH{\Delta _r}H and about standard enthalpy of formation ΔfHΘ{\Delta _f}{H^\Theta }. And then with the help of this information and the formulas required, we can easily approach our answer. Below here we have explained each topic very properly.

Complete step by step solution:
Enthalpy: We know that energy change occurring during the reaction at a constant pressure and constant volume is given by internal energy change that is, heat absorbed at constant volume is equal to change in internal energy that is ΔU=qv\Delta U = {q_v}. As atmospheric pressure is constant, therefore, such reactions may involve change in volume. It is a sum of internal energy and pressure-volume energy of the system at a particular temperature and pressure.
ΔH\Delta H Enthalpy change is the measure of heat change taking place during a process at constant temperature and pressure.
Reaction enthalpy ΔrH{\Delta _r}H: the enthalpy change accompanying a chemical reaction when the number of moles of reactants react to give the products as by the balanced chemical equation.
Standard enthalpy of formationΔfHΘ{\Delta _f}{H^\Theta }: The enthalpy change accompanying the formation of one mole of the compounds from its elements at standard conditions and all the substance being their standard states.
As a convection ΔfHΘ{\Delta _f}{H^\Theta } of every element is assumed to be zero.
Double bond of one molecule of​ CH2=CH2C{H_2} = C{H_2} cleaves then two −CH2−- C{H_2} - groups will come, that is, formation of two new single bonds will occur. So, C=CC = C for each, two new C−CC - C bonds are formed. So, in the complete polymer of ethylene there are 2 single bonds in every mole of ethylene.
ΔHpolymerisation=ΔH(CH2=CH2)−2ΔH(CH2−CH2) 2ΔH(CH2−CH2)=600−(−100)=700kJmol−1 2ΔH(CH2−CH2)=7002=350kJmol−1 \Delta {H_{polymerisation}} = \Delta {H_{\left( {C{H_2} = C{H_2}} \right)}} - 2\Delta {H_{\left( {C{H_2} - C{H_2}} \right)}} \\\ 2\Delta {H_{\left( {C{H_2} - C{H_2}} \right)}} = 600 - \left( { - 100} \right) = 700kJmo{l^{ - 1}} \\\ 2\Delta {H_{\left( {C{H_2} - C{H_2}} \right)}} = \dfrac{{700}}{2} = 350kJmo{l^{ - 1}} \\\
So, we can conclude that option (B) is correct. Hence, the correct answer is option B.

Note - In this question we have learned about enthalpy, about reaction enthalpy ΔrH{\Delta _r}H and about standard enthalpy of formation ΔfHΘ{\Delta _f}{H^\Theta }. Change in enthalpy is measured by calorimeter and the process is called calorimeter. This information we have learned in this question will help us in future.