Calculating ΔH with non-integer coefficients in the balanced equation

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

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 $$-1.$$

• Excellent answer, andselisk, very enlightening. Thank you! Commented Nov 24, 2019 at 10:18