2P(s) 3CI2(g) 2PCI3 ∆H= 151.8 kJ PCl3+ CI2(g) , PCI5(g); ∆H = -32.8 kJFrom the following data, the heat of formation of PCI5 comes out to be :

Explanation

To find the heat of formation of PCl5, we need to add the enthalpy changes of the given reactions. Using Hess's Law, we have:2P(s) + 3Cl2(g) → 2PCl3(g); ΔH = 151.8 kJPCl3(g) + Cl2(g) → PCl5(g); ΔH = -32.8 kJBy combining these reactions, we obtain the enthalpy for the formation of 2 moles of PCl5. The total enthalpy change is 151.8 kJ + (-32.8 kJ) = 119 kJ. Since we are considering the formation of 1 mole of PCl5, the heat of formation is +119 kJ.The other options result from incorrect calculations or sign errors.