Question

32. Given the following data: A. P4(s)+ 6 Cl2(g)4 PCls(g) B. P4(s)+ 5 02(g) P4O10(s)AH=- 2967.3 C. PCls(g)+Cl2(g) PCIS(g) D. PCI3(g)+O2(g) ClaPO(g) AH-285.7 AH 1225.6 kJ AH=-84.2 Calculate AH for the reaction: P4O10(s)+6 PCI5(g) 10 CI3PO(g). Note when I solved this problem, I used 3 of the 4 data reactions, added them together, then used the fourth data reaction to get to the desired reaction. -B-6C+10D+A

Answer 1

P₄ + 5 O₂ ====> P₄O₁₀   ΔH = -2967.3

Let us reverse the reaction

P₄O₁₀ ====> P₄ + 5 O₂   ΔH = +2967.3 KJ -------------- Eq 1

PCl₃ + Cl₂ ====> PCl₅  ΔH = -84.2 KJ

Let us reverse the reaction

PCl₅ ====> PCl₃ + Cl₂ ΔH = +84.2 KJ

Multiply by 6

6PCl₅ ====> 6PCl₃ + 6Cl₂ ΔH = +505.2 KJ ----------------Eq 2

PCl₃ + 1/2 O₂  ====>  POCl₃  ΔH = -285.7 Kj

Multiply by 10

10PCl₃ + 5 O₂  ====> 10POCl₃  ΔH = -2857 KJ ----------------Eq 3

P₄ + 6 Cl₂  ====> 4 PCl₃   ΔH = -1125.6 KJ -------Eq 4

Sum up Eq 1 -4

P₄O₁₀ ====> P₄ + 5 O₂   ΔH = +2967.3 KJ

6PCl₅ ====> 6PCl₃ + 6Cl₂ ΔH = +505.2 KJ

10PCl₃ + 5 O₂  ====> 10POCl₃  ΔH = -2857 KJ

P₄ + 6 Cl₂  ====> 4 PCl₃   ΔH = -1125.6 KJ

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P₄O₁₀ + 6PCl₅ ====> 10POCl₃   ΔH = 510.6 KJ

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Enthalpy of the reaction is  510.6 KJ