Question

Use equations II and III to calculate the ΔHrxn of equation I.

P4(s)+6Cl2(g)→4PCl3 ΔH=-1280 kJ

3.P4(s)+10Cl2(g)→4PCl5 ΔH= -1775 kJ

Answer 1

ΔHrxn of equation I = -123.5 kJ

Further explanation

Complete question :

Reaction 1 :

PCl₃ + Cl₂ → PCl₅

Reaction 2 :

P₄ + 6Cl₂ → 4PCl₃ ΔH = −1280 kj

Reaction 3 :

P₄ + 10Cl₂ → 4PCl₅ ΔH = −1774 kj

Based on the principle of Hess's Law, the change in enthalpy of a reaction will be the same even though it is through several stages or ways

Reverse reaction 2 :

4PCl₃ ⇒ P₄ + 6Cl₂ ΔH = +1280 kj (sign change)

Add to reaction 3 :

4PCl₃ ⇒ P₄ + 6Cl₂ ΔH = +1280 kj

P₄ + 10Cl₂ ⇒ 4PCl₅ ΔH = −1774 kj

-------------------------------------------------------- +

Canceled the same compound in the different side :

4PCl₃ + 4Cl₂ ⇒ 4PCl₅ ΔH = -494 kJ ⇒divide by 4

PCl₃ + Cl₂ ⇒ PCl₅ ΔH = -123.5 kJ