I need help in finding the heat change (in KJ) for the following reaction: 2P(g) 5Cl2(g)---->2PbCl5(g)?

Question

I need help in finding the heat change (in KJ) for the following reaction: 2P(g) 5Cl2(g)---->2PbCl5(g)?

Answer 1

I'm sure the Pb thing was a typo.  What is being produced is phosphorous pentachloride.

2P(g) +  5Cl2(g) ---> 2PCl5(g)

Unfortunately, we can't just say it is the heat of formation of PCl5 because P(g) is not in its standard state.  So we need the molar heat of vaporization of phosphorous, as well as the heat of formation of PCl5.  Of course, chlorine is in its standard state, so its heat of formation is zero.

The heat of vaporization is found at the Wikipedia article on phosphorous.

http://en.wikipedia.org/wiki/Phosphorus

DHrx = SUM(DHf products) - SUM(DHf reactants)

DHrx = 2DHf PCl5 - 2DHf P(g)

DHrx = 2 mol(-398.9 kJ/mol) - 2 mol(12.4 kJ/mol) = -822.6 kJ

Answer 2

How do you get lead pentachloride from phosphorus and chlorine?