Watching the assignment without knowing how to solve the questions is the worst thing to experience hope someone can help with this question and the privous ones too

- 622.5kJ

Explanation:

1)      2NH3 + 3N2O > 4N2 + 3H2O    ΔH° 1= - 1010kJ

-  3 (N2O  + 3H2     --->  N2H4 + H2O    ΔH° 2 = - 317 kJ)

2)      2NH3 + 3N2O > 4N2 + 3H2O        ΔH° 1= - 1010kJ

     - 3N2O   - 9 H2     > - 3N2H4 - 3H2O      - 3ΔH° 2 = 3*317 kJ

2NH3  + 3N2H4 --->4N2+9H2,     ΔH° 1 -  3*ΔH° 2

2NH3  + 3N2H4  --->4N2+9H2,        ΔH° 1 -  3*ΔH° 2

3) -(2NH3 +1/2O2 ---> N2H4 + H2O,     ΔH°3 = -143 kJ)

   -2NH3 - 1/2O2 ---> - N2H4 - H2O,    - ΔH°3 = 143 kJ

 2NH3  + 3N2H4 --->4N2+9H2,        ΔH° 1 -  3*ΔH° 2

 -2NH3 - 1/2O2              ---> - N2H4 - H2O,             - ΔH°3 = 143 kJ

 4N2H4 + H2O  ---> 4N2  + 9H2 + 1/2O2,    ΔH° 1 -  3*ΔH° 2 - ΔH°3

4)

H2 + 1/2O2 ---> H2O,    ΔH° 4 = - 286 kJ

9*(H2 + 1/2O2 ---> H2O,    ΔH° 4 = - 286 kJ)

9H2+ 9/2 O2 ---> 9H2O,  9*ΔH° 4 = 9*(- 286) kJ

4N2H4 + H2O  ---> 4N2  + 9H2 + 1/2O2,    ΔH° 1 -  3*ΔH° 2 - ΔH°3

9H2+ 9/2 O2    ---> 9H2O,                             9*ΔH° 4 = 9*(- 286) kJ

4N2H4 +4O2 --->4N2+8H2O,         ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

5)

1/4*(4N2H4 +4O2 --->4N2+8H2O,    ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4

N2H4 +O2 --->N2+2H2O,        1/4( ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)

6)

N2H4 +O2 --->N2+2,        

1/4( ΔH° 1 -  3*ΔH° 2 - ΔH°3 + 9*ΔH° 4)=

=1/4(-1010kJ - 3*(-317kJ) - (-143kJ) + 9*(-286kJ))= - 622.5 kJ

it is stable as it is in group 0 which have full outer shells of electrons

the answer is c i took the test