Diffference in enthalpy of weak acid and strong acid.

Enthalpy of neutralisation for enthanoic acid/sodium hydroxide is 97% of the enthalpy of neutralisation for hydrochloric acid/sodium hydroxide.

Question goes on to say:

Ethanoic acid is only about 1% dissociated, whereas hydrochloric acid is completely dissociated.
By considering what happens to ions as they are formed in aqueous solution, suggest why the enthalpy of neutralisation for enthanoic acid/sodium hydroxide is only slightly less exothermic than the value for hydrochloric acid/sodium hydroxide, in spite of the very different degrees of dissociation of the acids.

Mark Scheme says:

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Ions are hydrated

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Bonds formed so exothermic

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Which compensate for O-H bond breaking/dissociation and hence enthalpy of neutralisation is only slightly less than strong acid

Well i understand the ions are hydrated and stuff, but the last bullet point i have no idea about. How is the enthalpy of neutralisation slightly less? How is the hydration enthalpy related to it?

Help would be much appreciated and rep would be up for grabs if someone could tell me how to give it :smile:

Reply 1

charco

Study Forum Helper

energy changes are the sum of all the processes involved.

Although it requires energy to dissociate the weak acid (endothermic process) this is compensated for by the energy released (exothermic process) when the ions are hydrated.

The overall energy change on neutralisation must take these two factors (that nearly cancel out) into account

Reply 2

GJS

as if its to neutralisation all acid molecules are reacted, it doesnt matter whther there partially dissociated of not, however, in a strong acid the bonds break with very little resistance, thats why they are strong acid due to there "full dissociation" however it required energy to break weak acid bonds from C ---- O ----- H --> C-----O +H

so by doing sum of bonds broken - sum of bonds made, some energy is lost in breaking the C ----O bond

hope it helped

Reply 3

spacko101

Thanks for the replies guys.

When they says "ions are hydrated", does that mean they react with the OH- or by the water in the solution?

Am i right in thinking:
Imagine if there was 1 mole of ethanoic acid, which dissociated to give 0.01 moles of H+, and 1 mole of hydrochloric acid, which gave 1 mole of H+. Therefore with the weak acid, only 0.01 moles of H+ were "hydrated". Therefore the energy released would be be 100 times smaller than the energy released from hydrochloric acid.

But as only 0.01 moles had reacted, i would divide the energy change by the number of moles to give the enthalpy of neutralisation. This would produce values which are similar to the energy change for hydrochloric acid, but less exothermic due to the bond breaking.

Reply 4

GJS

well you know water is polar, so the -ve O is attracted to the H+ and the positive H in water is attracted to the -ve O in the alcohol group

just realised i drew the diagram wrong, changed it now

they kinda surround it,

no the weak acid is, completely neutralised, therefore all weak scid molecules have also been dissociated, the difference in energy comes from the energy taken to break then bond to release the H+

Reply 5

spacko101

Right, i'm cool with this surrounding thing now :smile:

But if only 1% dissociates, does that mean that once the 1% has reacted, more H+ ions are produced until eventually there is no more H+ to be dissociated? Due to the loss in H+, the equilibrium shifts so that more H+ ions are formed?

Sorry im i sound a bit like a spacko. :redface:

Reply 6

GJS

you mean the equilibrium CH3CH2OH <--> CH3CH2O- + H+

yes there is a store of H+ ready to be released and resotre eq, le chateliers principle "i think", yes until no more H+ availiable then it is neutralised :smile:

Reply 7

spacko101

so does the enthalpy of neutralisation have anything to do with the OH- from sodium hydroxide?

or is it only to do with the hydration enthalpy

or is it a combination of both, eg. neutralisation enthalpy = enthalpy of (H+ + OH-) + hydration enthalpy - dissociation enthalpy

Thanks for all of this so far GJS

Reply 8

GJS

hmmmmm sorry thats a bit further than i could predict, if i had to guess it would be due the formation of water and also from the hydration of Na+ and hydration of CH3COO-, and then minus the bonds broken, yeah id say that seems reasonable

lol your making me think and i dont like it