Find the heat of formation for a mole of hydrazine (N2H4) given:
2N2H4(l) + 2O2(g) → 2N2(g) + 4H2O(g) ∆ Ho = -1068.4 kJ
H2(g) + 1/2O2(g) → H2O(g) ∆ Ho = -241.8 kJ
Use kJ and 3 significant figures for your answer.
Heat of formation?
2017-04-25 7:23 pm
回答 (2)
2017-04-25 7:35 pm
We can rewrite the given two thermochemical equations as :
N₂(g) + 2H₂O(g) → N₂H₄(l) + O₂(g) …. ΔH° = -(1/2)(-1068.4) kJ = +534.2 kJ
2H₂(g) + O₂(g) → 2H₂O(g) …. ΔH° = 2(-241.8) = -483.6 kJ
Add the above two thermochemical equations, and cancel 2H₂O(g) and O₂(g) on the both sides :
N₂(g) + 2H₂(g) → N₂H₄(l) …. ΔH°f[N₂H₄(l)]
ΔH°f[N₂H₄(l)]
= [(+534.2) + (-483.6)] kJ
= +50.6 kJ
N₂(g) + 2H₂O(g) → N₂H₄(l) + O₂(g) …. ΔH° = -(1/2)(-1068.4) kJ = +534.2 kJ
2H₂(g) + O₂(g) → 2H₂O(g) …. ΔH° = 2(-241.8) = -483.6 kJ
Add the above two thermochemical equations, and cancel 2H₂O(g) and O₂(g) on the both sides :
N₂(g) + 2H₂(g) → N₂H₄(l) …. ΔH°f[N₂H₄(l)]
ΔH°f[N₂H₄(l)]
= [(+534.2) + (-483.6)] kJ
= +50.6 kJ
2017-04-25 7:25 pm
ert
收錄日期: 2021-04-18 16:10:41
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