1.

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)

Answer»

`-304.1 kJ//mol`
` +304.1 kJ//mol`
`-403.1 kJ//mol`
`+403.1kJ//mol`

Solution :`NO+O_(3)toNO_(2)+O_(2) DeltaH=-198.9 "kJ//mole"`
`O_(3)to3/2O_(2)(G)DeltaH=-142.3 "Kj//mole"`
`O_(2)to2O_(2)(g)DeltaH=+495.0 "K J//mole"`
`DeltaH "of" NO+O(g)toNO_(2)(g)`
` "for this"i-ii-(III)/2`
`198.9-(-142.3)-(495)/2`
304.1 " KJ//mole"`


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