I am working a practice problem out Ebbing's General Chemistry and for part of the problem I needed to calculate $ΔH.$ The problem considers the burning of ethane in oxygen to yield carbon dioxide and water vapor.

The coefficients used in the solution manual were 1 for ethane 7/2 for oxygen 2 for carbon dioxide and 3 for the water vapor. The coefficients I used were 2, 7, 4, and 6, respectively.

Naturally, it follows the $ΔH$ value I got is twice as large. Can anyone provide clarification here?


1 Answer 1


There is not much to clarify: you determined dimolar enthalpy of combustion, whereas standard enthalpy of combustion is determined for a complete oxidation of one mole of the substance:

$$ \begin{align} \ce{C2H6(g) + 7/2 O2(g) &-> 2 CO2(g) + 3 H2O(l)} &\quad &Δ_\mathrm{c}H^\circ\\ \ce{2 C2H6(g) + 7 O2(g) &-> 4 CO2(g) + 6 H2O(l)} &\quad &Δ_\mathrm{c}H^\circ ×2 \end{align} $$

When composing the corresponding reaction equation, make sure to balance it in such a way that the stoichiometric number $ν$ of the substance the enthalpy is determined for is always $-1.$

  • 1
    $\begingroup$ Excellent answer, andselisk, very enlightening. Thank you! $\endgroup$ Commented Nov 24, 2019 at 10:18

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