Question

Given the following reactions the enthalpy of reaction for 2N2O(g) --> 2NO(g) +N2(g) is __________ kJ? N2 (g) + O2(g) --> 2NO(g) ΔH = +180.7 kJ 2N2O(g) --> O2 (g) + 2N2(g) ΔH = -163.2 kJ

Answer 1

17.5 KJ

Explanation:

In all this kind of problems the first step is to balance the equations, in this case this is a very simple one, because if you add the two equations without any change you would obtain the asked one.

(1) `N_(2(g)) +O_(2(g)) rarr 2NO_((g)) Delta H_1 = +"180.7 kJ"`

(2) `2N_2O_((g)) rarr O_(2(g)) + 2N_(2(g)) Delta H_2 = -"163.7 kJ"`

(3) `2N_2O_((g)) rarr 2NO_((g)) + N_(2(g)) Delta H_2 = "? kJ"`

then `(1) + (2) = (3)`

as `DeltaH_1 + DeltaH_2 = DeltaH_3`

`:.DeltaH_3 = +17.5KJ`