given `NO(g)+O_(3)(g)toNO_(2)(g)+O_(2)(g) " "DeltaH=198.9kJ//mol`
`O_(3)(g)to3//2O_(2)(g) " " DeltaH=-142.3 kj//mol`
`O_(2)(g)to2O(g) " " DeltaH=+495.0 kj//mol`
The entalpy change `(DeltaH)` for the following reaction is
`NO(g) + O (g) to NO_(2) (g)`
A. `-304.1 kJ//mol`
B. ` +304.1 kJ//mol`
C. `-403.1 kJ//mol`
D. `+403.1kJ//mol`