The enthalpy of the change of the reaction P₄O₆ + 2O₂ ------ P₄O₁₀ is -1300.1kJ.
The reactions are as follows:
P₄O₆ + 2O₂ ------> P₄O₁₀
P₄ + 3O₂ ------->P₄O₆
Here, ΔHₐ = -1640.1kJ
P₄ + 5O₂ --------> P₄O₁₀
Here, ΔHₓ = -2940.1kJ
The enthalpy change of any type of reaction depends on the initial and final states of the reactions and is independent of the path taken by the system to reach the product.
To find the overall change in enthalpy, we need to sum up the reactions at each stage. On reversing the reactions we get,
P₄O₆ ------> P₄ + 3O₂.................(1)
Here, ΔHₐ = +1640.1kJ
P₄O₁₀------> P₄ + 5O₂.................(2)
Here, ΔHₓ = -2940.1kJ
By adding (1) and (2) we get,
P₄O₆ + 2O₂ ------> P₄O₁₀
By reversing (1) the sign of the enthalpy is changed. The enthalpy is given by
ΔH = ΔHₐ + ΔHₓ
ΔH = 1640.1 + (-2940.1)
ΔH = -1300.1kJ
Therefore the enthalpy of the reaction, P₄O₆ + 2O₂ ------> P₄O₁₀ is -1300.1kJ
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