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.
Your post makes no sense with all of the spaces and starts on new lines.
To calculate ∆H0 for the reaction 2 N2(g) + 5 O2(g) ⟶ 2 N2O5(g), you can use the given data and apply the Hess's Law of heat summation.
Hess's Law states that the enthalpy change of a reaction is independent of the pathway between the initial and final states, as long as the initial and final conditions are the same.
Let's break down the reaction into two steps that we have data for:
1. N2(g) + 3/2 O2(g) + 1/2 H2(g) ⟶ HNO3(ℓ)
∆H0 = -174.7 kJ/mol
2. N2O5(g) + H2O(ℓ) ⟶ 2 HNO3(ℓ)
∆H0 = -77.8 kJ/mol
Now, we need to manipulate these equations so that when added together, they give us the target equation:
2 N2(g) + 5 O2(g) ⟶ 2 N2O5(g)
To do that, we can multiply equation 1 by 2 and equation 2 by 2:
2(N2(g) + 3/2 O2(g) + 1/2 H2(g) ⟶ HNO3(ℓ)) ⟶ 2(-174.7 kJ/mol)
2 N2O5(g) + 2 H2O(ℓ) ⟶ 4 HNO3(ℓ) ⟶ 2(-77.8 kJ/mol)
Now, we can add both equations together to cancel out the intermediates (HNO3):
2 N2(g) + 5 O2(g) + 2 N2O5(g) + 2 H2O(ℓ) ⟶ 2 N2O5(g) + 4 HNO3(ℓ)
Simplifying the equation gives us:
2 N2(g) + 5 O2(g) ⟶ 2 N2O5(g)
Now, let's add up the ∆H0 values:
2*(-174.7 kJ/mol) + 2*(-77.8 kJ/mol) = -349.4 kJ/mol + (-155.6 kJ/mol) = -505 kJ/mol
Therefore, ∆H0 for the reaction 2 N2(g) + 5 O2(g) ⟶ 2 N2O5(g) is -505 kJ/mol.
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.)