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.
dHrxn = (n*dHproducts) - (dHreactants)
To calculate ΔH0 for the reaction 2 N2(g) + 5 O2(g) → 2 N2O5(g), we need to use the given data and apply Hess's Law, which states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual steps of the reaction.
Step 1: We need to find the enthalpy change for the formation of 2 N2O5(g) from its constituent elements.
- Start with the given reaction: N2(g) + O2(g) + O2(g) → N2O5(g)
- We can use the following reactions to calculate the enthalpy change:
Reaction 1: H2(g) + 1/2 O2(g) → H2O(ℓ) ΔH°f = -283 kJ/mol
Reaction 2: N2O5(g) + H2O(ℓ) → 2 HNO3(ℓ) ΔH° = -78.7 kJ/mol
Reaction 3: N2(g) + 3/2 O2(g) + 1/2 H2(g) → HNO3(ℓ) ΔH°f = -174.3 kJ/mol
- We multiply Reaction 1 by 2 to match the 2 H2O(ℓ) in Reaction 2:
2[H2(g) + 1/2 O2(g) → H2O(ℓ)] → 2H2O(ℓ)
ΔH1 = 2(-283 kJ/mol) = -566 kJ/mol
- We multiply Reaction 3 by 2 to match the 2 HNO3(ℓ) in Reaction 2:
2[N2(g) + 3/2 O2(g) + 1/2 H2(g) → HNO3(ℓ)] → 2HNO3(ℓ)
ΔH3 = 2(-174.3 kJ/mol) = -348.6 kJ/mol
- We reverse Reaction 2:
2HNO3(ℓ) → N2O5(g) + H2O(ℓ)
ΔH2 = 78.7 kJ/mol
Step 2: Calculate the overall enthalpy change (ΔH0) for the reaction.
ΔH0 = ΣΔH(products) - ΣΔH(reactants)
ΔH0 = [2ΔH2 + ΔH3] - [ΔH1]
ΔH0 = [2(78.7 kJ/mol) + (-348.6 kJ/mol)] - [-566 kJ/mol]
ΔH0 = 157.4 kJ/mol + 348.6 kJ/mol + 566 kJ/mol
ΔH0 = 1072 kJ/mol
Therefore, ΔH0 for the reaction 2 N2(g) + 5 O2(g) → 2 N2O5(g) is 1072 kJ/mol.