Verified answer

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< Data >

(1) N2(g) + O2(g) → 2 NO(g) ΔH = 180.6 kJ

(2) 2 CO(g) + O2(g) → 2 CO2(g) ΔH = -283.0 kJ

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You have to combine the two given equations to make appear the one you're looking for. As the enthalpy is a function of state, its change between initial state and final state does not depend on the path.

< 1 > The equation you want to have starts with CO(g) and NO(g). The first equation (1) in the data produces NO(g), you want to start from NO(g) as a reagent so you have to reverse it:

(-1) 2.NO(g) → N2(g) + O2(g) with ΔH = the opposite of the direct equation's enthalpy = - 180.6 kJ

< 2 > Now you can sum the two equations:

(-1)+(2) 2.NO(g) + 2 CO(g) + O2(g → N2(g) + O2(g) + 2 CO2(g) ΔHr

After simplification:

2 CO(g) + 2 NO(g) → 2 CO2(g) + N2(g) ΔHr

< 3 > Hess's law tells you that ΔHr is the sum of the enthalpies of (-1) and (2):

ΔHr = ΔH(-1) + ΔH(2) = - 180.6 - 283.0 = - 463.6 kJ