Hess Cycle, determining enthalpy change of formation I am asked to find the enthalpy change of formation of the following:
$$\ce{N2 + 1/2O2 -&gt; N2O}$$

I am given the following enthalpies of reaction:
\begin{align}
\ce{C + N2O &amp;-&gt; CO + N2} &amp;\Delta H_f =-193 \tag{1} \\
\ce{C + 1/2O2 &amp;-&gt; CO} &amp;\Delta H_f=-111 \tag{2}
\end{align}

How do I calculate enthalpy change of formation of the nitrous oxide?

I thought it would just be adding the two enthalpies together because I substitute the second equation into the first one, but this is wrong.
If you start by reversing equation 1, you get $$\ce{CO + N2 -&gt;C + N2O~~ {\Delta}H_{f}=+193}$$

Now, add this result with equation 2 and you get $$\ce{CO~+N2~+C~+1/2O2-&gt;C~+N2O~+CO~~{\Delta}H_{f}=+193-111=82 }$$

Cancelling like terms that exist on both sides of this equation yields $$\ce{N2~+1/2O2-&gt;N2O~~{\Delta}H_{f}=82}$$