Asked by jane
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 = −283 kJ/mol
N2O5(g) + H2O(ℓ) −→ 2 HNO3(ℓ)
∆H0 = −78.7 kJ/mol
1
2
N2(g) + 3
2
O2(g) + 1
2
H2(g) −→ HNO3(ℓ)
∆H0
f = −174.3 kJ/mol
Answer in units of kJ.
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 = −283 kJ/mol
N2O5(g) + H2O(ℓ) −→ 2 HNO3(ℓ)
∆H0 = −78.7 kJ/mol
1
2
N2(g) + 3
2
O2(g) + 1
2
H2(g) −→ HNO3(ℓ)
∆H0
f = −174.3 kJ/mol
Answer in units of kJ.
Answers
Answered by
DrBob222
dHrxn = (n*dHproducts) - (dHreactants)
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