Clear my confusion about Average bond enthalpy question?

So if a reaction is endothermic then it would take energy in from its surroundings. Correct?

Say for example the total reactants equate to 400jkmol and the products are 800kjmol. Then this would clearly be an endothermic reaction as the products gain energy.

My question is, when you find out the enthalpy change it is a negative. ( reactants- products. Eg 400-800=-400) This indicates that it is an exothermic reaction. Hence my confusion. Please help, I know I must of misunderstood something. Thanks

(edited 8 years ago)

Reply 1

username986184

Original post by Questioness

So if a reaction is endothermic then it would take energy in from its surroundings. Correct?

Say for example the total reactants equate to 400jkmol and the products are 800kjmol. Then this would clearly be an endothermic reaction as the products gain energy.

My question is, when you fidnd out the enthalpy change it is a negative. ( reactants- products. Eg 400-800=-400) This indicates that it is an exothermic reaction. Hence my confusion. Please help, I know I must of misunderstood something. Thanks

Using your numbers.

Energy required to break bonds of reactants = 400
Energy released in forming bonds of products = 800

Therefore, nett energy change is a release of 400. Exothermic.

(edited 8 years ago)

Reply 2

abitpissy

Thank you.

I have one more question about Hess's law. Question 2b.

I keep getting -154 as the answer, however the answer is +154kj.

Attachment not found

^^This is my working out so you can see how I got it

I realise that some of the molecules aren't correctly put, please forgive me (it's hardly relevant to my answer)

Reply 4

username986184

Original post by Questioness

I realise that some of the molecules aren't correctly put, please forgive me (it's hardly relevant to my answer)

See below

image.jpeg158.4KB

Reply 5

Mayhem™

Original post by TeachChemistry

See below

At what level of education is this taught? I'm only in year 11 but it looks like something I could have a go at

Reply 6

abitpissy

Original post by Mayhem™

At what level of education is this taught? I'm only in year 11 but it looks like something I could have a go at

AS chemistry.

Reply 7

username986184

Original post by Mayhem™

At what level of education is this taught? I'm only in year 11 but it looks like something I could have a go at

Hess cycles are year 12

Reply 8

Mayhem™

Original post by Questioness

AS chemistry.

Original post by TeachChemistry

Hess cycles are year 12

Thanks!

Reply 9

abitpissy

Original post by TeachChemistry

See below

Does it matter which way the arrows are pointing? my book shows it pointing down for combustion and arrows point up for formation.

(edited 8 years ago)

Reply 10

rxns_00

Original post by Questioness

Hun its product-reactants not the other way round.

So 800-400=+400

Therefore deltaH + hence endothermic reaction

Reply 11

username986184

Original post by Questioness

Does it matter which way the arrows are pointing? my book shows it pointing down for combustion and arrows point up for formation.

The values in your book are formation.

Reply 12

abitpissy

Original post by RonnieRJ

Hun its product-reactants not the other way round.

So 800-400=+400

Therefore deltaH + hence endothermic reaction

I was talking about calculating enthalpy change in average bond enthalpies.

(edited 8 years ago)

Reply 13

abitpissy

Original post by TeachChemistry

The values in your book are formation.

Ahah right. That makes a whole lot more sense now. Thank you so much.

Reply 14

rxns_00

Original post by Questioness

I was talking about calculating enthalpy change in average bond enthalpies.

Yeah still. No offence that book is ****

Ok when you work out enthalpy change using bond enthalpies follow these:

- work out bond enthalpies of products
- work out bond enthalpies of reactants
- find the difference between the two (take whichever is larger away from the smaller)
- if dH products>reactants, stick a + to the enthalpy change
- if dH reactants>products, stick a - to the enthalpy change.

That is it

Reply 15

abitpissy

Original post by RonnieRJ