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.