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Find the heat of combustion for 1 mol of $\ce {C21H44}$ in the reaction where some wax is oxidized to elemental C (soot) and water vapor. The Heat of Formation of $\ce {C21H44}$ is -476 kJ/mol.

When it says some, does that necessarily change the molar ratio? I thought the balanced equation would be:

$\ce {C21H44 + 11O2 -> 21C (graphite) + 22H2O}$

$\ce {O2}$ and elemental carbon are zero for heat of formation

22(-241.818 kJ/mol) - (-476 kJ/mol) = -4843.996 kJ/mol

Is this right?

Thanks

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The question is ambiguous, unfortunately. There are two possibilities that I can see. Either the question:

  1. is asking about a reaction in which some quantity of wax undergoes the reaction you give, in which case you are 100% correct.

  2. is asking about a reaction in which part of the paraffin is being oxidized to C and part is being oxidized to CO2, the usual reaction that occurs. Though that is more likely to occur in real life, in my opinion, it also is impossible to solve without further information. The best you could do would be to set a lower limit (the value you've already found) and an upper limit, the heat of reaction for $\ce{C21H44 + 32 O2 -> 21 CO2 + 22H2O}$.

    (I guess that would be an upper limit and a lower, numerically, since the values are both negative, but this reaction would release far more heat.)

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  • $\begingroup$ OR you could write a pair of equations, one describing the reaction giving carbon, the other carbon dioxide. You could then derive an answer where the heat of combustion depended on the proportion of the reaction going down either path. $\endgroup$ – matt_black Jan 2 '15 at 13:05

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