Question

Please help me
N2O5(g) + H2O(l) --> 2HNO3(l) delta H° = -76.2 kJ
H2O(l) --> H2(g) + 1/2O2(g) delta H° = 286.0 kJ
1/2N2(g) + 3/2O2(g) + 1/2H2(g) --> HNO3(l) delta H° = -174.0 kJ

Calculate delta H° for the reaction
2N2O5(g) --> 2N2(g) + 5O2(g)

Answer 1

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Answer 2

PLEASE

Answer 3

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Answer 4

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 = −285.6 kJ/mol
N2O5(g) + H2O(ℓ) −→ 2 HNO3(ℓ)
∆H0 = −75.9 kJ/mol
1
2
N2(g) + 3
2
O2(g) + 1
2
H2(g) −→ HNO3(ℓ)
∆H0
f = −173.4 kJ/mol
Answer in units of kJ.