How to cancel once-mentioned element in thermochemical reactions

If I have three thermochemical reactions in a series(in order to find the enthalpy change) how can I cancel an element having not included in the reaction that's being asked about?


and only mentioned in one of the reactions ( either products or reactants)


e.g: $$\ce{NO + O3 \rightarrow NO2 + O2}$$
 $$\ce{O3 \rightarrow 3/2 O2}$$
 $$\ce{O2 \rightarrow 2O}$$


**The reaction that is being asked**: $$\ce{2NO2 \rightarrow 2NO + O2}$$




We can make use of this formula: $$\Delta H = \Sigma ~\Delta H\_f~(\text{products}) - \Sigma ~\Delta H\_f~(\text{reactants}) $$


Take $\ce{O3 \rightarrow \frac{3}{2} O2}$ into consideration. Ozone molecule is the less stable form of O. If you know that how much the $\Delta H$ for that reaction is, you can in fact predict the $\Delta H\_f$ for $\ce{O3}$. If you're interested, [wikipedia has documented](http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_%28data_table%29) some of the enthalpies of formation. Notice that the most thermodynamically stable form of the element is considered to have $\Delta H\_f = 0$.


Now that the $\Delta H\_f$ for ozone is determined, you can apply the formula above for the first reaction: ($x$ is $\Delta H$) $$x = (\Delta H\_f \ce{(NO2)} + 0) - (\Delta H\_f (\ce{NO}) + \Delta H\_f (\ce{O3}))$$


So, from the equation above we can conclude that $$x + (\Delta H\_f (\ce{NO}) + \Delta H\_f (\ce{O3})) = (\Delta H\_f \ce{(NO2)}) $$


Now write the master formula for the last reaction: $$(2 \* \Delta H (\ce{NO}) + 0) - 2 \* \Delta H\_f (\ce{NO2})$$


What did we conclude from the last reaction? $x + (\Delta H\_f (\ce{NO}) + \Delta H\_f (\ce{O3})) = (\Delta H\_f \ce{(NO2)}) $ Assuming that $x$ is at hand, do you really need to know the $\Delta H\_f$ for $\ce{NO}$ & $\ce{NO2}$?


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If you're just interested in the reaction enthalpy of the reaction


$$\ce{2NO2 <=> 2NO + O2}$$


and one set of equations for the calculation of $\Delta H\_f(\mathrm{products})$ and/or $\Delta H\_f(\mathrm{reactants})$ doesn't work out, just use another one.


**It does not matter! The Hess Law is invariant of the path.**


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