[rosebudd88]

Calculate the standard head of formation for PCl3 (l)

given these two equations:

P4(s) + 6Cl2(g) = 4PCl5(s)     ΔH° = -1774.0kJ

PCl3(l) + Cl2(g) = PCl5(s)      ΔH° = -123.8 kJ

What I have already done:

1st : I wrote a balanced equation to get PCl3(l)

P4(s) + 6Cl2(g) = 4PCl3(l)

2nd: I applied Hess's law to find the ΔH° for "PCl3" equation (just above this sentence)
- I reversed the second (given) equation, and then multiplied it by 4
ΔH° = -1278.8 kJ

3rd: I need to find the Δ_fH° for PCl3(l)

I know that that
ΔH° = ΣΔ_fH°(products) - ΣΔ_fH°(reactant)

Therefore

ΣΔ_fH°(products) = ΔH° + ΣΔ_fH°(reactant)

And since P4 and Cl2 are in their elemental states, they are both zero, and the answer is simply what i first calculate -1278.8 kJ/mol

But since there are 4 products of PCl3 (l) then the answer should be -5115.2kJ

I have been told that Δ_fH° (PCl3(l)) = -1278.8 kJ/mol is WRONG

Then could it be Δ_fH° (PCl3(l)) =-5115.2kJ     


Some one Please Tell me where I am going wrong with this equation

And thank you very much for your time.

[rosebudd88]

I have been thinking about this

and if i get
P4(s) + 6Cl2(g) = 4PCl3(l)   ΔH°=-1279 k/J

Then it must mean that i get -1279 kJ for 4 moles of PCl3(l)
Therefore if I divide that by 4 then I will get the actual answer which is
Δ_fH° = -319.7kJ/mol for 1 PCl3(l)

Maybe that is the answer

Can someone please give me feedback

[rosebudd88]

Yay I solved my own problem. No need for replies any more
Doesn't really matter cause no one helped me figure this one out any ways !!!

Haha Yay Brain POWER