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AS Chemistry - Average Bond Enthalpies?!

How the hell do I calculate the average bond enthalpy of this bond?!

The standard molar enthalpy change of formation of H20 is -242kJmol^-1. Calculate the average bond enthalpy of the O-H bond.

O=O = 496
H-H = 436
Reply 1
Avatar for VokeA
VokeA
3
Enthalpy of formation = total bonds broken - total bonds made



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Reply 2
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VokeA
3
So first you need to write out the equation for the formation of water and then draw out the bonds included on each side of the equation so you know what is being broken/formed. Then you can work backwards from there. And if you get a negative value you know to try again as bond enthalpies are endothermic


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(edited 8 years ago)
1/2 O2 + H2 -> H2O so the bond enthalpy is 1/2 * 496 + 436 - 242 = 442 I think. However, I have a feeling that the signs should be reversed and bond enthalpies should be negative (as bonds are more energetically stable than radicals and the like) but maybe someone who's done AS more recently can confirm.
Reply 4
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VokeA
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Original post by AmateurRocketman
1/2 O2 + H2 -> H2O so the bond enthalpy is 1/2 * 496 + 436 - 242 = 442 I think. However, I have a feeling that the signs should be reversed and bond enthalpies should be negative (as bonds are more energetically stable than radicals and the like) but maybe someone who's done AS more recently can confirm.


I got +463 as the answer


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Reply 5
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AndrewKn0x
OP
5
Original post by AmateurRocketman
1/2 O2 + H2 -> H2O so the bond enthalpy is 1/2 * 496 + 436 - 242 = 442 I think. However, I have a feeling that the signs should be reversed and bond enthalpies should be negative (as bonds are more energetically stable than radicals and the like) but maybe someone who's done AS more recently can confirm.


That's what I did and it was wrong according to the book:frown:
Reply 6
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AndrewKn0x
OP
5
Original post by VokeA
I got +463 as the answer


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That's right! How did you get that answer?
Reply 7
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VokeA
3
Original post by AndrewKn0x
That's right! How did you get that answer?


ImageUploadedByStudent Room1460285975.563277.jpg


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Reply 8
Avatar for sue99
sue99
10
Original post by AndrewKn0x
How the hell do I calculate the average bond enthalpy of this bond?!

The standard molar enthalpy change of formation of H20 is -242kJmol^-1. Calculate the average bond enthalpy of the O-H bond.

O=O = 496
H-H = 436


so the equation: 2H2O -----> 2H2 + O2

H-O-H H-H O=O
H-O-H H-H

standard enthalpy change of formation = enthalpy of products - enthalpy of reactants
-242 = +1368 - x (total enthalpy of reactants)
x = +1368 + 242 = +1610
O-H = +1610 / 2 (because 2 moles in equation)
= +805 kJmol^-1
Honestly, I don't know if this is correct or not - it may actually be +805 / 2 = +402.5 kJmol^-1 = +403 kJmol^-1. Hope this helps though!:biggrin:
Reply 9
Avatar for AndrewKn0x
AndrewKn0x
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5
Original post by VokeA

Ok thanks! I see it now. Looking at the spec I think we should've been taught this, but we haven't but I get the idea now.
Reply 10
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VokeA
3
Original post by AndrewKn0x
Ok thanks! I see it now. Looking at the spec I think we should've been taught this, but we haven't but I get the idea now.


You're welcome. And sorry for my handwriting. Its not usually like that


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Reply 11
Avatar for AndrewKn0x
AndrewKn0x
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5
Original post by VokeA
You're welcome. And sorry for my handwriting. Its not usually like that


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Your writing is v neat don't worry :wink:

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