How can oxygen atoms suddenly appear on the product side of an equation? - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-10-17T20:17:40Z https://chemistry.stackexchange.com/feeds/question/4069 https://creativecommons.org/licenses/by-sa/4.0/rdf https://chemistry.stackexchange.com/q/4069 4 How can oxygen atoms suddenly appear on the product side of an equation? WestSidePotato https://chemistry.stackexchange.com/users/1182 2013-02-03T02:22:39Z 2014-11-04T08:02:24Z <p>The question states the following equation: \$\$ \ce{C(s) + 2H2(g) -&gt; CO2(g) + 2H2O}\$\$ And then asks you to calculate the enthalpy change for it, given enthalpies of combustion.</p> <p>The stated values are \$C(s)\$ -393, \$\ce{H2}(g)\$ -286, and \$\ce{CH4}(g)\$ -890</p> <p>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.</p> https://chemistry.stackexchange.com/questions/4069/-/4070#4070 1 Answer by ManishEarth for How can oxygen atoms suddenly appear on the product side of an equation? ManishEarth https://chemistry.stackexchange.com/users/22 2013-02-03T08:19:19Z 2013-02-03T08:19:19Z <p>They probably want you to calculate the enthalpy change required to <em>convert</em> the reactants into the products. Another way of looking at it is the enthaply <em>difference</em> between the products and reactants.</p> <p>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)</p> <p>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 -&gt; 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.</p> <p>So, while the question is wrong in calling it a <em>reaction</em>, 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.</p>