`N_(2(g)) + O_(2(g)) + 180.6 kJ to 2NO_((g))`, calculate (a) heat of reaction, (b) heat of formation of nitric oxide and (c) heat required to form one litre of nitric oxide at `25^@C`. 

Explanation

Thermochemical equation, `N_(2(g)) + O_(2(g)) to 2NO_((g)) , DeltaH = +180.6 kJ.` The heat of the given reaction `DeltaH = +180.6 kJ`.Since 2 mole of nitric acid is formed  in the reaction, the heat of formation of nitric oxide
`DeltaH_f = (DeltaH)/(2) = (180.6)/2 = 90.3 kJ^(-1)`
At `25^@C` and 1 atm one mole of a gas occupies 24.4 litres.
Heat required to form 24.4 litre of `NO = 90.3 kJ.`
Heat required to form one litre of `NO= (90.3)/(24.4) = 3.7 kJ lit^(-1)`.