Consider the following intermediate chemical equations. no(g)+o(g) --> no2(g)+o2(g) δh1= - 198.9kj 3/2 o2(> )3(g) δh2= 142.3kj o(> 1/2 o2(g) δh3= -247.5kj what is the enthalpy of the overall chemical equation no(g) + o(> no2(g)? a) -305 kj b) -304.1 kj c) -93.7 kj d) 588.7 kj

Considering the first reaction there is no NO on the left and NO2 on the right we start with adding equation 1. To remove O3 we subtract equation 2, then here is 1/2O2 left which we remove by subtracting  1/2 equation 3.
That is; Eqn 1 - Eqn 2 - 1/2(Eqn 3)
NO(g) + O3(g) + 3/2 O2(g) + O(g) = NO2(g) + O2(g) +  O3(g) + 1/2 O2(g)
This gives; 
 NO(g) + O(g) = NO2(g) as required, since O3(g) + 3/2O2(g) is on both sides and thus subtracts out.
Hence; ΔH = ΔH1-ΔH2-1/2 ΔH3
= (-198.9 + 142.3 - 0.5×495.0) kJ = -304.1 kJ

Yes i think so not 100 percent sure though

Answer is to avoid experimental error