`Delta_(f)H^(Theta)` per mole of `NH_(3)(g). NO(g)`, and `H_(2)O(l)` are `-11.04 + 21.60` and `-68.32 kcal`, respectively. Calculate the standard heat of reaction at constant pressure and at a constant volume for the reaction:
`4NH_(3)(g) +5O_(2)(g) rarr 4NO(g) +6H_(2)O(l)`

Explanation

`Deltan_(g) = 4 - 9 = 5`
`Delta_(f)H^(Theta)O_(2) = 0`
`DeltaH = [4Delta_(f)H^(Theta)(NO)+6Delta_(f)H^(Theta) (H_(2)O)] -4 Delta_(f)H^(Theta)(NH_(3))`
`=4 xx 21.6 +6 xx -(68.32-(-11.04) =- 279.36 kcal`
`DeltaU = DeltaH - Deltan_(g)RT`
`=- 279.36 -(-5) xx2xx 10^(-3) xx 298`
`=- 276.38 kcal`