Question

Given the following thermochemical data: ∆H°f (kJ)

I. P4(s) + 6 Cl2(g) ⟶ 4 PCl3(g) –1225.6

II. P4(s) + 5 O2(g) ⟶ P4O10(s) –2967.3

III. PCl3(g) + Cl2(g) ⟶ PCl5(g) –84.2

IV. PCl3(g) + ½ O2(g) ⟶ POCl3(g) –285.7

Calculate the value of ∆H° for the following reaction: P4O10(s) + 6 PCl5(g) ⟶ 10 POCl3(g)

Why/how is the answer 610.1 kJ?

Answer 1

Given data

I. P₄(s) + 6 Cl₂(g) ⟶ 4 PCl₃(g) –1225.6

II. P₄(s) + 5 O₂(g) ⟶ P₄O₁₀(s) –2967.3

III. PCl₃(g) + Cl₂(g) ⟶ PCl₅(g) –84.2

IV. PCl₃(g) + ½ O₂(g) ⟶ POCl₃(g) –285.7

Required equation

P₄O₁₀(s) + 6 PCl₅(g) ⟶ 10 POCl₃(g)   ∆H°=?

Arrange the equations I, II, III and IV as per the above equation by miltiplying with suitable coeffecient and reverse it if necessary (depending on the reactants and products in the required equation)

Reverse II       ---->.P₄O₁₀(s) ⟶ P₄(s) + 5 O₂(g) + 2967.3

Reverse III x 6 ----> 6 PCl₅(g) ⟶ 6PCl₃(g) + 6Cl₂(g) +505.2

        IV x 10   ----> 10PCl₃(g) + 5 O₂(g) ⟶ 10 POCl₃(g) –2857

                I    ----> P₄(s) + 6 Cl₂(g) ⟶ 4 PCl₃(g) –1225.6

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Adding above 4 eqs --> P₄O₁₀(s) + 6 PCl₅(g) ⟶ 10 POCl₃(g) -610.1

                                 ∆H°= 610.1 KJ