chem 2 enthalpy change?

Question

chem enthalpy change?
Sadly, I have no idea how to do this problem. I have one more like it and an explanation in "idiots" terms would be great, but just an answer would be fine too...

PCl3(g) + 1/2O2(g) -----> Cl3PO(g) (delta)Hrxn= -286kJ

P4(s) + 6 Cl2(g) -----> 4PCl3(g) (delta)Hrxn=-1226kJ

(delta)Hf [P4O10(s)] = -2967kJ/mol

(delta)Hf[PCl5(g)]= -84kJ/mol

_________________________... Calculate the enthalpy change for the reaction: P4O10(s) + 6 PCl5(g) -----> 10 Cl3PO(g)


The numbers after the letters (eg, H2O2) are the subscripts...the numbers before (2 H...) are the regular numbers. Any help would be appreciated...

Answer 1

P4O10(s) + 6 PCl5(g) -----> 10 Cl3PO(g)

-2967kJ/mol + 6[-84kJ/mol] + enthalpy change = 10[-286kJ]

[P4O10(g)]

deltaHf1 + deltaHf2 + deltaHf3 = deltaHf0



By hess's law:

enthalpy change = 2967 + 504 - 2860 = 611kJ/mol

Answer 2

The key to this kind of problem is to multiply the various equations you're given by an appropriate number, write them backwards or forwards as you need, then add them all up to get the desired equation with nothing left over. Finally add up all the kJ's. Whenever you multiply an equation by 2, 3,..., you multiply the kJ's too. Whenever you write an equation backwards, -kJ's become +kJ's. By the way, the heat of formation of P4O10 is P4 + 5O2 ===> P4O10 -2967kJ. The heat of formation of PCl5 is P + 5/2Cl2 ===> PCl5 --84kJ. First write:

P4O10 ===> P4 + 5O2 +2967

That gets you P4O10 on the left. Next add to that:

6PCl5 ===> 6P + 15Cl2 +504 (multiplied by 6 and written bacwards)

That gets you 6PCl5 on the left. Next add:

10PCl3 + 5O2 ===> 10Cl3PO -2860kJ (x 10)

Now you have 10Cl3PO on the right. Next comes:

10P4 + 15Cl2 ===> 10PCl3 -3065kJ (x 2.5)

That cancels out the P and Cl2 on the right and the PCl3 on the left. +2967 + 504 - 2860 - 3065 = -2454kJ, which is the answer.