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bryanmcmillianjr
23.05.2020 •
Chemistry
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
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Ответ:
- 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
Ответ:
Hydrogen (H)
Helium (He)
Lithium (Li)
Beryllium (Be)
Boron (B)
Carbon (C)
Nitrogen (N)
Oxygen (O)
Fluorine (F)
Neon (Ne)
Explanation: