ΔHrxn of equation I = -123.5 kJ
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
