4
$\begingroup$

The question states the following equation: $$ \ce{C(s) + 2H2(g) -> CO2(g) + 2H2O}$$ And then asks you to calculate the enthalpy change for it, given enthalpies of combustion.

The stated values are $C(s)$ -393, $\ce{H2}(g)$ -286, and $\ce{CH4}(g)$ -890

I'm curious as to why oxygen atoms have suddenly appeared on the products side. The only thing I have been able to think is that it refers to the enthalpy of formation of methane, and that the answer would be -393 + (2*-286) + 890. But it still doesn't make sense to me considering the equation provided. I understand the products are the products expected from combustion, but no oxygen has been added on the reactants side.

$\endgroup$
  • 1
    $\begingroup$ There's a serious problem with the equation. You've already noted that. If the problem came from a book, check the book. If it came from a homework assignment, ask about that. $\endgroup$ – Paul J. Gans Feb 11 '13 at 2:47
1
$\begingroup$

They probably want you to calculate the enthalpy change required to convert the reactants into the products. Another way of looking at it is the enthaply difference between the products and reactants.

Remember, $\ce{O2(g)}$ has 0 enthalpy of formation in standard state. (so do $\ce{C(s)}$ and $\ce{H2(g)}$, but we don't need that)

So, adding $\ce{2O2}$ to the reactants side won't change $\Delta H^o=H^o_{products}-H^o_{reactants}$. So calculating $\Delta H^o$ for the given "reaction" is the same as calculating it for $$\ce{C(s) + 2H2(g) +2O2 -> CO2(g) + 2H2O}$$. And $\Delta S^o$ for this reaction can be calculated very easily using enthalpies of combustion. You don't even need to use the given value for methane.

So, while the question is wrong in calling it a reaction, it is still asking us to do something reasonable -- that is, calculate $$\Delta H^o=H^o_{\ce{CO2(g) + 2H2O}} - H^o_{\ce{C(s) + 2H2(g)}}$$. Since we can subtract $H^o_{\ce{2O2(g)}}=0$ easily, the problem can be solved by taking a combustion path.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.