Bond Enthalpies Problem

Basically, I'm stuck on a question and would very much appreciate your help. It's page 134 question 2 of the Nelson Thornes AS AQA Chemistry Book.

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The table below contains some mean bond enthalpy data.

H-O 463 kJ mol^-1
O-O 146 kJ mol^-1
O=O 496 kJ mol^-1

The bonding in hydrogen peroxide can be represented by H-O-O-H. Use these data to calculate the enthalpy change for the following reaction.

H₂O₂(g) -> H₂O₂(g) + ½O₂(g)

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My understanding is that is that in this reaction the only bonds that are broken and formed are that half a mol of O=O formed.

However, the mark scheme suggests that on top of this, a mol of O-O bonds is broken. Can someone tell me how they worked that out?

Reply 1

Creative

Something wrong with the equation you've posted... it doesn't balance?

That H2O2 on the RHS is sposed to be a water I guess...

H2O2(g) -> H2O(g) + ½O2(g)

So the enthalpy change of the reaction = sum of enthalpies of bonds broken - sum of enthalpies of bonds formed.

So, the bonds broken are H-O, O-O, O-H

The bonds formed in H2O are O-H, O-H

The bonds formed in the 1/2O2 are 1/2(O=O)..

I did Chemistry last year so if I got something wrong don't burn me at the stake :biggrin:

.

I guess the O-O bond is broken because it's a bond contained within the hydrogen peroxide molecule ^^.

Reply 2

electriic_ink

Creative

Something wrong with the equation you've posted... it doesn't balance?

That H2O2 on the RHS is sposed to be a water I guess...

H2O2(g) -> H2O(g) + ½O2(g)

You're probably right. It definitely says H2O2 in the book but that's just a misprint.

Thanks!

Reply 3

emsmith182

In case anyone is looking at this now, exam pro has the same mistake on this question!Think it should be H20 as the product not H2O2 like the question says.