Chemistry: Ethalpy change.

Question

Could someone please explain this to me:

I do a neutralisation reaction between 0.025L 1M HCl & NaOH.

change in T = 7 degrees C.

i.e. H OH --> H2O

Therefore:

deltaH = 50 x 4.186 x 7 = 1465.1 J = 58.6 kJ per mol.

But the bond enthalpy for O-H is 463 kJ per mol.

How can it be so high, if I only get 58.6 from my experiment?

Is it something I'm doing or is my collected data wrong?

Answer 1

Nothing is wrong, and you have asked a very good question.

You have found the enthalpy for the reaction

Hplus(aq) plus OH-(aq) = H2O (liq)

Bond enthalpy relates to the process by which you simply break the bond, to make a hydrogen atom and an OH fragment (or radical, as it is sometimes called). This requires a lot more energy, and only happens at very high temperatures indeed, or using energy from light.

Answer 2

u have: H3O^+ and HO^- result2* H2O

there are 3 types of bonds: O^+_H(crack), O^-_H(crack), O_H(form)

bond:H3O^+ =3 breaking

bond:HO^- =1 braking

bond:H2O =4 forming

this is like:an accountant at IBM deals with money,but in the end the difference from net output(billions) and inputs(billions)

is a meagre  sum(thousands)=his salary