Given these two reactions, how do I calculate

ΔH for the third one?

P4(s) + 6 Cl2(g) -> 4 PCl3(g) ΔH= -1148 kJ
P4(s) + 10 Cl2(g) -> 4 PCl5(g) ΔH = -1500 kJ
PCl3(g) + Cl2(g) -> PCl5(g) ΔH = ?

2 answers

look up hess's law. you should be able to figure it out
Reverse equation 1 and add to equation and that gives you
4PCl3 + 4Cl2 ==> 4PCl5 which is just 4x what yu want so divide everything by 4.