Given the equations below, what is the total enthalpy change for the formation of phosphorus pentachloride, PCl5, when phosphorus is burned in excess chlorine?

P4 (s) + 6Cl2 (g) → 4PCl3 (g), ∆H = - 1224 kJ

4PCl3 (g) + 4Cl2 (g) → 4PCl5 (g), ∆H = - 372 kJ

P4 (s) + 10Cl2 (g) → 4PCl5 (g)

A. 852 kJ

B.-852 kJ

C.-1596 kJ

D. 1596 kJ

We are given the following balanced chemical equations:

P4 (s) + 6Cl2 (g) → 4PCl3 (g), ∆H = - 1224 kJ

4PCl3 (g) + 4Cl2 (g) → 4PCl5 (g), ∆H = - 372 kJ

adding the two equations, we arrive at a final equation:

P4 (s) + 10Cl2 (g) → 4PCl5 (g)

Now, we need to find the total enthalpy of this reaction. Add both enthalpies of the two initial equations:

∆H = - 372 kJ + - 1224 kJ

∆H = - 1596 kJ

Therefore, the total enthalpy change of burning phosphorus with chlorine is - 1596 kJ, meaning the reaction is exothermic and produces heat.