Considering the Hess's Law, the enthalpy change for the reaction is 68 kJ/mol.
Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.
In this case you want to calculate the enthalpy change of:
N₂ + 2 O₂ → 2 NO₂
which occurs in two stages.
You know the following reactions, with their corresponding enthalpies:
Equation 1: N₂ + O₂ → 2 NO ΔH = 180 kJ/mol
Equation 2: 2 NO + O₂ → 2 NO₂ ΔH = –112 kJ/mol
Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.
In this case, first, to obtain the enthalpy of the desired chemical reaction you need one mole of N₂ on reactant side and it is present in first equation so let's write this as such.
In the same way, you need two moles of NO₂ on product side and it is present in second equation so let's write this as such.
Finally, in this way the two moles of NO cancel.
Then, it is not necessary to modify the given equations. So, adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:
N₂ + 2 O₂ → 2 NO₂ ΔH= 68 kJ/mol
Finally, the enthalpy change for the reaction is 68 kJ/mol.
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