`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) 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)/n= (DeltaH)/2= 180.6/2 + 90.3KJmol^(-1)`
At`25^(@)C` and 1 atm one mole of gas occupies 24.4 liters.
Heat required to from 24.4 litre of `NO=90.3KJ`
Heat reqyuired to from one litre of `NO=90.3/24.4=3.7KJlit^(-1)` .