small chemistry doubt

[anushey]

guys can someone plz help me solve this question.i just cant do it.
it is question no 4 of oct/nov 2008 P4 part 3
4 (a) The viscosity of engine oil can be improved by the addition of certain medium chainlength


polymers.

A portion of the chain of one such polymer is shown below.



–CH2CH(CH2CH2CH3)CH2CH(CH2CH2CH3)CH2–


On average, the molecules of the medium-chain polymer contain 40 carbon atoms.

(b) Used car engine oil can be recycled for use as a fuel by the processes of distillation and



cracking.


(i) Assuming a typical molecule of engine oil has the formula C40H82, suggest an



equation for a cracking reaction that could produce diesel fuel with the formula


C16H34

and other hydrocarbons only.
(iii) Considering only the bonds broken and the bonds formed during the reaction, use

the Data Booklet to calculate the enthalpy change for the reaction you wrote in

(b)(i).
how is the answer 180.i cant understand the method used in m.s

 

[Eshika]

guys can someone plz help me solve this question.i just cant do it.
it is question no 4 of oct/nov 2008 P4 part 3
4 (a) The viscosity of engine oil can be improved by the addition of certain medium chainlength


polymers.

A portion of the chain of one such polymer is shown below.



–CH2CH(CH2CH2CH3)CH2CH(CH2CH2CH3)CH2–


On average, the molecules of the medium-chain polymer contain 40 carbon atoms.

(b) Used car engine oil can be recycled for use as a fuel by the processes of distillation and



cracking.
i) Assuming a typical molecule of engine oil has the formula C40H82, suggest an



equation for a cracking reaction that could produce diesel fuel with the formula


C16H34

and other hydrocarbons only.
(iii) Considering only the bonds broken and the bonds formed during the reaction, use

the Data Booklet to calculate the enthalpy change for the reaction you wrote in

(b)(i).
how is the answer 180.i cant understand the method used in m.s

(b) (i) C40H82 → C16H34 + 2 C12H24
(b) (iii) when cracking is used large hydrocarbon molecules breaks down to alkene and alkane as product here our compound is alkane thats why from the data booklet you use the value of (C-C) as bond breaking which is 350 and for bond formed the value will be 610 as (C=C) is formed in the product + it is an additional polymerisation as told in the question before thats why you don't use the whole compound for calculation but use only 2 monomers.

bonds broken: 4(C–C) = 4 × 350 = 1400 kJ mol–1
bond formed: 2 (C=C) = 2 × 610 = 1220 kJ mol–1
∴∆H = +180 kJ mol–1
i hope what i wrote is clear and you can understand it.

 

[anushey]

thankyou so much !!

 

[Eshika]

no prb