The enthalpy of the reaction P₄O₆(s) + 2O₂(g) → P₄O₁₀(s) calculated from the enthalpies of the reactions P₄(s) + 3O₂(g) → P₄O₆(s) and P₄O₁₀(s) → P₄(s) + 5O₂(g) is -1300 kJ.
We need to find the enthalpy of the following reaction:
P₄O₆(s) + 2O₂(g) → P₄O₁₀(s) (1)
And we know the enthalpies of the reactions:
P₄(s) + 3O₂(g) → P₄O₆(s) ΔH₁ = -1640.1 kJ (2)
P₄O₁₀(s) → P₄(s) + 5O₂(g) ΔH₂ = 2940.1 kJ (3)
To calculate the enthalpy of reaction (1) using the values of enthalpies of reactions (2) and (3), we need to make the following changes for these two reactions:
1. Invert reaction (3):
P₄(s) + 5O₂(g) → P₄O₁₀(s) ΔH₂ = -2940.1 kJ (4)
Now we have the P₄O₁₀ on the side of the product as in reaction (1). The inversion changed the sing of enthalpy ΔH₂.
2. Invert reaction (2):
P₄O₆(s) → P₄(s) + 3O₂(g) ΔH₁ = 1640.1 kJ (5)
The compound P₄O₆ is now on the side of the reactant as in reaction (1). The inversion changed the sing of enthalpy ΔH₁.
3. Addition of reactions (4) and (5)
To get the reaction (1) we need to add reactions (4) and (5)
P₄(s) + 5O₂(g) + P₄O₆(s) → P₄O₁₀(s) + P₄(s) + 3O₂(g)
P₄O₆(s) + 2O₂(g) → P₄O₁₀(s) (1)
The enthalpy value of the reaction (1) can be calculated by the sum of the enthalpies of the reactions (4) and (5):
[tex]\Delta H_{t} = -2940.1 kJ + 1640.1 kJ = -1300 kJ[/tex]
Therefore, the enthalpy is -1300 kJ.
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I hope it helps you!
