# hess' law - energetics

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#1
Hi the question i don't understand is the following:
i'm using this to represent delta ^
Calculate ^Hf of CCl4 (l) given the following data

CCl4 (l) -> CCl4 (g) = +31 kjmol-1

C(s) -> C(g) =+715 KJ mol-1

Bond enthalpy (Cl-Cl) =+242KJmol-1

Bond enthalpy (C-Cl) = +338 Kj mol-1

help
0
1 month ago
#2
Hi the question i don't understand is the following:
i'm using this to represent delta ^
Calculate ^Hf of CCl4 (l) given the following data

CCl4 (l) -> CCl4 (g) = +31 kjmol-1

C(s) -> C(g) =+715 KJ mol-1

Bond enthalpy (Cl-Cl) =+242KJmol-1

Bond enthalpy (C-Cl) = +338 Kj mol-1

help
Hi there

First I recommend drawing all your molecules out to see the number of each type of bond.
Then count the number of each type of bond and multiply this number by the bond enthalpy of that bond.
Do a total of reactant bond enthalpies and product bond enthalpies.
Enthalpy of formation = total of product bond enthalpies - total of reactant bond enthalpies.
Finally don't forget to add the units if they're not on the answer line.

Hope this helps. Any further help don't hesitate to ask.
Last edited by TriplexA; 1 month ago
1
#3
what do i do about the state conversions?
0
1 month ago
#4
what do i do about the state conversions?
I don't think they're needed for this Q.

Sometimes some Qs have unnecessary data that isn't used. I believe this is one of them.
If anyone could just confirm this - that would be great.
Last edited by TriplexA; 1 month ago
0
#5
0
#6
thats what ive done so far
0
1 month ago
#7
(Original post by TriplexA)
Hi there

First I recommend drawing all your molecules out to see the number of each type of bond.
Then count the number of each type of bond and multiply this number by the bond enthalpy of that bond.
Do a total of reactant bond enthalpies and product bond enthalpies.
Enthalpy of formation = total of product bond enthalpies - total of reactant bond enthalpies.
Finally don't forget to add the units if they're not on the answer line.

Hope this helps. Any further help don't hesitate to ask.
You can only use bond enthalpy terms for GASEOUS substances.

You must vaporise the carbon first (enthalpy of atomisation) AND turn the gaseous tetrachloromethane back to liquid at the end.

1. C(s) ==> C(g)
2. 2Cl2(g) ==> 4Cl(g)
3. C(g) + 4Cl(g) ==> (bond enthalpy terms) ==> CCl4(g)
4. CCl4(g) ==> CCl4(l)
Then add up all of the steps
Last edited by charco; 1 month ago
0
1 month ago
#8
(Original post by charco)
You can only use bond enthalpy terms for GASEOUS substances.

You must vaporise the carbon first (enthalpy of atomisation) AND turn the gaseous tetrachloromethane back to liquid at the end.

1. C(s) ==> C(g)
2. 2Cl2(g) ==> 4Cl(g)
3. C(g) + 4Cl(g) ==> (bond enthalpy terms) ==> CCl4(g)
4. CCl4(g) ==> CCl4(l)
Then add up all of the steps
Oh yes.
I've just realised this was a thermodynamics Q and not an enthalpy changes topic.

I do apologise. Thanks for helping bigbadger
1
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