Question

Consider the two reactions. 2NH3(g)+3N2O(g)4NH3(g)+3O2(g)⟶4N2(g)+3H2O(l)⟶2N2(g)+6H2O(l)  Δ?∘=−1010 kJΔ?∘=1531 kJ2NH3(g)+3N2O(g)⟶4N2(g)+3H2O(l) ΔH∘=−1010 kJ4NH3(g)+3O2(g)⟶2N2(g)+6H2O(l) ΔH∘=1531 kJ Using these two reactions, calculate...

Consider the two reactions.

2NH3(g)+3N2O(g)4NH3(g)+3O2(g)⟶4N2(g)+3H2O(l)⟶2N2(g)+6H2O(l)  Δ?∘=−1010 kJΔ?∘=1531 kJ2NH3(g)+3N2O(g)⟶4N2(g)+3H2O(l) ΔH∘=−1010 kJ4NH3(g)+3O2(g)⟶2N2(g)+6H2O(l) ΔH∘=1531 kJ

Using these two reactions, calculate and enter the enthalpy change for the reaction below.

N2(g)+12O2(g)⟶N2O(g)N2(g)+12O2(g)⟶N2O(g)

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Answer #1

If the reaction is reversed then the value of ∆H° is doesn't change the sign of the value is reversed.

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Consider the two reactions. 2NH3(g)+3N2O(g)4NH3(g)+3O2(g)⟶4N2(g)+3H2O(l)⟶2N2(g)+6H2O(l)  Δ?∘=−1010 kJΔ?∘=1531 kJ2NH3(g)+3N2O(g)⟶4N2(g)+3H2O(l) ΔH∘=−1010 kJ4NH3(g)+3O2(g)⟶2N2(g)+6H2O(l) ΔH∘=1531 kJ Using these two reactions, calculate...
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