# Given the following thermochemical equations: 2KCl(s)+3O2(g)=2KClO3(s) H=78.0kJ, P4(s)+6Cl2(g)=4PCl3(g) H=?

-1148.0kJ, P4(s)+2O2(g)+6Cl2(g)=4PClO3(g) H=-2168.8kJ, Calculate H for the reaction KClO3(s)+3PCl3(g)=3POCl3(g)+KCl(s)

• Since the Hf given are for the given moles of each specific compound in a formation reaction, to work with a different number of moles, the Hf is divided by the coefficient in the formation reaction and then multiplied the moles in the problem reaction. Also, if the product is a reactant in the problem reaction, the SIGN needs to be reversed. If that is confusing, just fill in the chart as we go along:

Chemical .....Moles...... Role........H, Kj/mole .........H in Reaction

PCl3...............3.......... Reactant... -287kJ/mole........ 861 kJ

POCl3............3........... Product..... -542kJ/mole........-1626 kJ

For the other reactants, they appear in the first given reaction. If you write in reverse, 2KClO3 -> 2KCl + 3O2(g). The formation of oxygen gas has no Hf value associated with it. So we can use this info in the problem. We switch the sign and divide by 2 to get -39kJ/mole.

Now we add up the three enthalpies to get -804 kJ for the rxn. (I hope)