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Equation for bond enthalpy of H-Br

How is the equation for bond enthalpy of H-Br written like this: HBr(g) ---> H(g) + Br(g) ? I get that they are all gaseous because they are under standard condition but I just don't understand why they are shown this way and why is it not Br2. Thanks.

Reply 1

RMNDK

Original post by coconut64

It's just a way of helping us to calculate the enthalpy change of a reaction by considering a reaction as breaking all the bonds to get free atoms, and then rearranging them to get our products by forming new bonds.

For instance, say I want to find the enthalpy change of the reaction of HBr and CH₂=CH₂

I break each bond.
If you say, "Why not show it as Br₂ and H₂?"
Well look what I do.

2HBr(g) ---> H₂(g) + Br₂(g)

So I break the bonds in HBr, and then make new bonds to form Hydrogen and Bromine.

But I don't want to make Hydrogen and Bromine.

Spoiler

I want to form bromoethane. So I'm going to have to break the bonds between hydrogen and bromine again.
See how I've pretty much reversed my steps. I formed hydrogen and bromine, and broke the bonds again. There's no point to writing those additional steps, so I just forget it altogether and write them as separate atoms. Now I only need to worry about the new bonds between all these atoms in the product I want to make.

Reply 2

Jingbah

the definition for bond enthalpy is the energy required to break one mole of a bond in a gaseous molecule, what you suggested means that we would have to break 2 moles of HBr which is incorrect