Calculating standard combustion enthalpy of dodecane from that of heptane & octane

These are the problem from UKChO-2015.



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> Given the standard enthalpy changes of combustion ($\Delta\_\mathrm{c}H$) for heptane ($n = 7$) and octane ($n = 8$) are $-4816$ and $-5470~\mathrm{kJ~mol^{-1}}$ respectively, calculate $\Delta\_\mathrm{c}H$ of dodecane ($n = 12$).
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Any hint on regarding this problem? My efforts are not worth mentionable here




Heptane is $\ce{CH3-CH2(5)-CH3}$


Octane is $\ce{CH3-CH2(6)-CH3}$


Given combustion enthalpies for each of $-4816$ and $-5470~\mathrm{kJ~mol^{-1}}$ respectively, we now have a simple set of algeraic equations:


5A + 2B = $-4816~\mathrm{kJ~mol^{-1}}$


6A + 2B = $-5470~\mathrm{kJ~mol^{-1}}$


First we can solve for A as the combustion enthaply of a CH2 in a linear hydrocarbon and we get the value $-654~\mathrm{kJ~mol^{-1}})$.


We can now solve for B,


In the first equation we get 2B = ($-4816 - (654 \* 5)~\mathrm{kJ~mol^{-1}}$, A = $-773~\mathrm{kJ~mol^{-1}}$


In the second equation we get 2B = ($-5470 - (654 \* 6)~\mathrm{kJ~mol^{-1}}$ , A = $-773~\mathrm{kJ~mol^{-1}}$


dodecane has CH3-CH2(10)-CH3 , for 10A + 2B.


10 \* A + 2 \* B = ($-6540$ + $-1546$)$~\mathrm{kJ~mol^{-1}}$ = $-8086~\mathrm{kJ~mol^{-1}}$


 3

About $-8086~\mathrm{kJ~mol^{-1}}$.


Take the difference of the two givens and multiply by the difference of carbons between dodecane and octane.


 2

I wasn't first sure about this concept. But after seeing a gsurfer04's comment and working out a little bit, it became clear.


Alkanes combustion enthalpy is roughly proportional to the number of methylene(-CH2) groups it has. (You can simply work out the example given in the question) 


So, number of methylene groups in dodecane 10
number of methylene groups in octane 6


So, dodecane's enthalpy= octane's enthalpy\*(10/6)


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