Question

Calculate ∆H0
for the reaction
2 N2(g) + 5 O2(g) −→ 2 N2O5(g)
given the data
H2(g) + 1
2
O2(g) −→ H2O(ℓ)
∆H0
f = −290 kJ/mol
N2O5(g) + H2O(ℓ) −→ 2 HNO3(ℓ)
∆H0 = −77.8 kJ/mol
1
2
N2(g) + 3
2
O2(g) + 1
2
H2(g) −→ HNO3(ℓ)
∆H0
f = −174.7 kJ/mol
Answer in units of kJ.

Answer 1

Your post makes no sense with all of the spaces and starts on new lines.

Answer 2

Here’s a repost …
Determine the enthalpy of reaction for 2N₂(g) + 5O₂(g) => 2N₂O₅(g)
from the following reactions using Hess’s Law.

Rxn 1: H₂(g) + 2O₂(g) => H₂O(l); ΔH₁ = 2(-290)-Kj = -580-Kj
Rxn 2: N₂O₅(g) + H₂O(l) => 2NHO₃(l); ΔH₂ = -77.8-Kj
Rxn 3: N₂(g) + 3O₂(g) + H₂(g) => ΔH₃ = 2(-174.7)-Kj = -349.4-Kj

Work these type reactions in 'pairs', that is, add the 1st two reactions above to get a net reaction ( say, Rxn 1,2) then add to Rxn 3 to obtain the 'target' reaction. Remember you can use multiples of the given reactions and/or reverse the reactions of the given but don't forget to change signage changing from exothermic (-) to endothermic (+). Give a try, if ya get stuck post your work.

(Oh, the formatting is using MS Word with symbols, then copy and paste into post field.)

Answer 3

i need answer help