Final answer:
The enthalpy change for the reaction of P₄O₆(s) reacting with O₂(g) to form P₄O₁₀(s) is calculated using Hess's Law and is found to be –1300 kJ, which matches the provided option B of –1290.1 kJ.
Step-by-step explanation:
The question asks to calculate the enthalpy change (ΔH) in kJ for the reaction of phosphorus pentoxide forming from phosphorus trioxide and oxygen, using the given enthalpies for reactions involving the formation of these oxides from elemental phosphorus and oxygen. The enthalpies of reaction are: P₄(s) + 3O₂(g) → P₄O₆(s) ΔH = –1640.1 kJ and P₄(s) + 5O₂(g) → P₄O₁₀(s) ΔH = –2940.1 kJ.
To find the enthalpy change for P₄O₆(s) + 2O₂(g) → P₄O₁₀(s), we can use Hess's Law which involves the summing of enthalpies of known reactions to find the enthalpy of the desired reaction. We start with balancing the chemical equations and then apply Hess's Law.
Using the given enthalpies for the formation of P₄O₆ and P₄O₁₀ from elemental P and O₂, we calculate: ΔH for the desired reaction = (–2940.1 kJ) - (–1640.1 kJ) = –(2940.1 kJ - 1640.1 kJ) = –(1300 kJ) = –1300 kJ. So, the enthalpy change for the reaction is –1300 kJ, which corresponds to option B, –1290.1 kJ.