Given the following data:

p4(s) + 6 cl2(g) → 4 pcl3(g) δh = −1225.6 kj

p4(s) + 5 o2(g) → p4o10(s) δh = −2967.3 kj

pcl3(g) + cl2(g) → pcl5(g) δh = −84.2 kj

pcl3(g) + 1/2 o2(g) → cl3po(g) δh = −285.7 kj

calculate δh for the reaction p4o10(s) + 6 pcl5(g) → 10 cl3po(g).

ΔH = -610.1 kJ

Explanation:

By the Hess Law, when a reaction is formed by various steps, the enthalpy change (ΔH) of the global reaction is the sum of the enthalpy change of the steps reactions. Besides, if it's necessary for a change in the reaction, ΔH will suffer the same change. If the reaction multiplied by a number, ΔH will be multiplied by the same number, and if the reaction is inverted, the signal of ΔH is inverted.

P₄(s) + 6Cl₂(g) → 4 PCl₃(g) ΔH = -1225.6 kJ

P₄(s) + 5O₂(g) → P₄O₁₀(s) ΔH = -2967.3 kJ (inverted)

PCl₃(g) + Cl₂(g) → PCl₅(g) ΔH = -84.2 kJ (inverted and multiplied by 6)

PCl₃(g) + 1/2O₂(g) → Cl₃PO(g) ΔH = -285.7 kJ (multiplied by 10)

P₄(s) + 6Cl₂(g) → 4 PCl₃(g) ΔH = -1225.6 kJ

P₄O₁₀(s) → P₄(s) + 5O₂(g) ΔH = +2967.3 kJ

6PCl₅(g) → 6PCl₃(g) + 6Cl₂(g) ΔH = +505.2 kJ

10PCl₃(g) + 5O₂(g) → 10Cl₃PO(g) ΔH = -2857.0 kJ

The bolded substances will be eliminated because have the same amount in product and reactant:

P₄O₁₀(s) + 6PCl₅(g) → 10Cl₃PO(g)

ΔH = -1225.6 + 2967.3 + 505.2 -2857.0

ΔH = -610.1 kJ

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.