Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: rosebudd88 on October 06, 2012, 11:16:45 PM
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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 ΔfH° for PCl3(l)
I know that that
ΔH° = ΣΔfH°(products) - ΣΔfH°(reactant)
Therefore
ΣΔfH°(products) = ΔH° + ΣΔfH°(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 ΔfH° (PCl3(l)) = -1278.8 kJ/mol is WRONG
Then could it be ΔfH° (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.
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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
ΔfH° = -319.7kJ/mol for 1 PCl3(l)
Maybe that is the answer
Can someone please give me feedback
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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 ;D