The the reaction enthalpy for the following reaction is +1,220 kJ (Option D)
2N₂H₄g)+ 2NO₂(g) → 3N₂(g) + 4H₂O(g)
ΔHrxn = ∈ΔfH(Product ) - ∈ΔfH(reactant)
= (3 x ∈ΔfH(N₂) +4 x ΔfH(H₂O) - (2 x (N₂H₄ΔHF) + 2 ΔfH(NO²)
= (3 x 0 + 4 x (-241.8)) - (2 x (95.4) + (2 x 33.1))
= -1224.3Kj
Thus D is the most approximate answer.
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Full Question:
Given the standard enthalpies of formation for the following substances, determine the reaction enthalpy for the following reaction. (Δ Hrₓₙ = ? kJ)
2 N2H4( g) + 2 NO2( g) → 3 N2( g) + 4 H2O(g)
A) -967 kJ/mol
B) −1,220 kJ
C) +257 kJ/mol
D) +1,220 kJ
E) +967 kJ/mol
