Respuesta :
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.
What is enthalpy?
Enthalpy is a thermodynamic property of a system. It is the sum of the internal energy added to the product of the pressure and volume of the system. Enthalpy is denoted as H.
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.
Now, the 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)
Now, reaction (1) will be
P₄O₆(s) + 2O₂(g) → P₄O₁₀(s)
The enthalpy value of the reaction (1) can be calculated by the sum of the enthalpies of the reactions (4) and (5):
ΔH = ΔH₁ + ΔH₂
ΔH = 1640.1 + (-2940.1)
ΔH = -1300 kJ
Therefore, 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.
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Answer:
option B
p4O6(s) + 2O2(g) Right arrow. P4O10(s)
Explanation:
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