Hess Cycle, determining enthalpy change of formation
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 $$\ce{N2~+1/2O2-&gt;N2O~~{\Delta}H_{f}=82}$$