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

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