These are the problem from UKChO-2015.

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 ($n = 12$).

Any hint on regarding this problem? My efforts are not worth mentionable here

  • 3
    $\begingroup$ It seems it expects you to assume that the enthalpy is proportional to the number of CH2 units in the alkane. $\endgroup$
    – gsurfer04
    Feb 14, 2017 at 12:51
  • $\begingroup$ It's just computed from changes in bonding. Write out the balanced equations. Figure out the number of bonds broken, and solve some algebra. $\endgroup$
    – Zhe
    Feb 14, 2017 at 19:35
  • $\begingroup$ @ Zhe It's an Olympiad problem man. They don't expect you to do it so easily. The problem's needs to be done without bond enthalpy just using the information given in the question. $\endgroup$ Feb 15, 2017 at 5:12

3 Answers 3


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 = ($-6540$ + $-1546$)$~\mathrm{kJ~mol^{-1}}$ = $-8086~\mathrm{kJ~mol^{-1}}$

  • $\begingroup$ By way of comparison, the enthalpy of combustion that can be calculated from the enthalpy of formation given in the CRC Handbook is $-8087\ \mathrm{kJ\ mol^{-1}}$. $\endgroup$
    – user7951
    Feb 16, 2017 at 13:16

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

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

  • $\begingroup$ So you mean the difference in enthalpy is proportional to difference of number of carbon atoms between two alkanes? $\endgroup$ Feb 16, 2017 at 11:24
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    $\begingroup$ @Mockingbird I think this answer makes much more sense. Your answer neglects the fact that there are $\ce{CH3}$ groups at both ends which also get oxidised. To a first approximation those contribute a constant amount to $\Delta_\mathrm{c}H$. How much is this constant amount? You can work it out from the data given, but in fact you don't really need to know. All you need to know is each $\ce{CH2}$ group contributes 5470 - 4816 = 654 kJ/mol and hence dodecane is 5470 + 4(654) = 8086 kJ/mol. (With the negative signs, of course.) $\endgroup$ Feb 16, 2017 at 12:39
  • 1
    $\begingroup$ @user41388 I'd recommend you edit your answer to include at least a minimal amount of working. $\endgroup$ Feb 16, 2017 at 12:40
  • 1
    $\begingroup$ I'd put it this way. Take five octanes. Cut CH2 fragments from four of them and paste those into the fifth one. So, 5*octane = 4*heptane+dodecane. The same goes for enthalpies. $\endgroup$ Feb 16, 2017 at 13:00

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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