# AS Energetics & Enthalpy's HELP!

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#1
Hey guys.
So i was doing some a level chemistry questions (http://www.a-levelchemistry.co.uk/AQ...2.1%20home.htm)
exercise 2.
We are given bond disassociation energies and are asked to calculate an enthalpy of formation of HF? I am unsure of how to go about this.. The question before asks to calculate an enthalpy of reaction where HF is a product
All help is greatly appreciated
0
5 years ago
#2
Hey guys.
So i was doing some a level chemistry questions (http://www.a-levelchemistry.co.uk/AQ...2.1%20home.htm)
exercise 2.
We are given bond disassociation energies and are asked to calculate an enthalpy of formation of HF? I am unsure of how to go about this.. The question before asks to calculate an enthalpy of reaction where HF is a product
All help is greatly appreciated
ooooookay so im not the best at chemistry (especially at 2am) but here goes first i wrote up the equation ''H2 + F2 ---> 2HF'' using the values given in the table you can calculate the sum of bonds broken (593) and the sum of bonds formed (1130). applying these values to the equation ''deltaH = sum of bonds broken - sum of bonds formed'' would give you an answer of -537 kjmol-1.. this is for 2HF ... so then you would just divide this answer by 2 to get the value for HF which is -268.5 kJmol-1 0
#3
(Original post by tayyabk64)
ooooookay so im not the best at chemistry (especially at 2am) but here goes first i wrote up the equation ''h2 + f2 ---> 2hf'' using the values given in the table you can calculate the sum of bonds broken (593) and the sum of bonds formed (1130). Applying these values to the equation ''deltah = sum of bonds broken - sum of bonds formed'' would give you an answer of -537 kjmol-1.. This is for 2hf ... So then you would just divide this answer by 2 to get the value for hf which is -268.5 kjmol-1:d
thank you so much you absolutely amazing person!!!!!
0
5 years ago
#4
thank you so much you absolutely amazing person!!!!!
im really not haha i kinda totally flopped AS chem 0
#5
so im stuck on the next one - enthalpy formation of propene
i have : 3C + 3H2 ----- C3H6
bond breaking - making = -554
What should i do next?
thanks for all the help
0
5 years ago
#6
so im stuck on the next one - enthalpy formation of propene
i have : 3C + 3H2 ----- C3H6
bond breaking - making = -554
What should i do next?
thanks for all the help
your equation is correct however bond breaking - making is wrong. bonds formed should = 3444 bonds broken should = 3436
3444 - 3436 = 8 kjmol-1 .. which is the correct answer...
0
#7
(Original post by tayyabk64)
your equation is correct however bond breaking - making is wrong. bonds formed should = 3444 bonds broken should = 3436
3444 - 3436 = 8 kjmol-1 .. which is the correct answer...
OMG THANKS SO MUCH. Lol everyone must think im a wierdo for doing chemistry revision at this time hahah. I SWEAR I HAVE A LIFE GUYS 0
5 years ago
#8
OMG THANKS SO MUCH. Lol everyone must think im a wierdo for doing chemistry revision at this time hahah. I SWEAR I HAVE A LIFE GUYS if thats the case then im equally as weird -__- chem5kills haha.
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