Calculate the delta H for the reaction using hess' law:
N2H4 + O2 --> N2 + 2 H2O
Given the following data:
2NH3 + 3N2O--> 4N2 + 3H2O
delta H: -1010kJ
N2O + 3H2 ---> N2H4 + H2O
delta H: -317 kJ
2NH3 + 1/2O2 --> N2H4 + H2O
delta H: -143 kJ
H2 + 1/2O2 --> H2O
delta H: -286 kJ
Help with a chemistry problem please?
2018-02-22 1:07 pm
回答 (2)
2018-02-22 4:19 pm
Rewrite the four given thermochemical equations:
3N₂H₄ + 3H₂O → 3N₂O + 9H₂ …… ΔH = -3(-317 kJ) = +951 kJ
N₂H₄ + H₂O → 2NH₃ + (1/2)O₂ …… ΔH = -(-143 kJ) = +143 kJ
2NH₃ + 3N₂O → 4N₂ + 3H₂O …… ΔH = -1010 kJ
9H₂ + (9/2)O₂ → 9H₂O …… ΔH = 9(-286 kJ) = -1623 kJ
Add the above four thermochemical equations, and cancel 4H₂O, 3N₂O, 2NH₃, (1/2)O₂ and 9H₂ on the both sides :
4N₂H₄ + 4O₂ → 4N₂ + 8H₂O …… ΔH = 951 + 143 - 1010 - 1623 kJ = -1539 kJ
Divided the above thermochemical equation by 4 :
N₂H₄ + O₂ → N₂ + 2H₂O …… ΔH = -1539/4 kJ = -385 kJ (to 3 sig. fig.)
3N₂H₄ + 3H₂O → 3N₂O + 9H₂ …… ΔH = -3(-317 kJ) = +951 kJ
N₂H₄ + H₂O → 2NH₃ + (1/2)O₂ …… ΔH = -(-143 kJ) = +143 kJ
2NH₃ + 3N₂O → 4N₂ + 3H₂O …… ΔH = -1010 kJ
9H₂ + (9/2)O₂ → 9H₂O …… ΔH = 9(-286 kJ) = -1623 kJ
Add the above four thermochemical equations, and cancel 4H₂O, 3N₂O, 2NH₃, (1/2)O₂ and 9H₂ on the both sides :
4N₂H₄ + 4O₂ → 4N₂ + 8H₂O …… ΔH = 951 + 143 - 1010 - 1623 kJ = -1539 kJ
Divided the above thermochemical equation by 4 :
N₂H₄ + O₂ → N₂ + 2H₂O …… ΔH = -1539/4 kJ = -385 kJ (to 3 sig. fig.)
2018-02-22 1:08 pm
this brings up bad memories...
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