[Technicalhuman]

When HCl+NaOH NaCl+H₂O the enthalpy is more negative than CH₃COOH with NaOH. This is because more energy is used to dissociate the ethanoic acid to the acetate ion and H₃O⁺ ions. And since there is a certain equilibrium to the CH₃COOH CH₃COO^- + H₃O⁺ wouldn't it be impossible to fully react the ethanoic acid?

Also, if we were to add NaOH to the solution as a limiting reagent, so to find the number of moles of ethanoic acid, we would take the number of moles of ethanoic acid initially minus the number of moles of NaOH while for the acetate ion it would be the number of moles of NaOH produced only. we would draw out an ICE table to determine the pH. And lastly, for the H3O+ ions it would be 0.

However I don't quite understand this. How can be simply subtract the total number of moles of ethanoic acid with the number of moles of NaOH since some of the ethanoic acid would have dissociated into acetate ions? And similarly since I already have some acetate ions from the dissociation reaction of the ethanoic acid shouldn't the initial be more than 0? And also for the H3O+ how can it be 0 initially?

Thanks in advance for the help.

[Babcock_Hall]

A couple of initial thoughts.  One, ethanoic acid has a very unfavorable entropy of dissociation.  Other acids behave differently, and I would not make assumptions about the relative contributions of enthalpy and entropy (better to obtain or to measure the values).  Two, the reaction between a strong base and a weak acid goes to completion; I suppose one could look at the dissociation reaction of ethanoic acid (which barely proceeds at all), and then apply Le Chatelier's principle to see why. 

[Dan]

When HCl+NaOH NaCl+H₂O the enthalpy is more negative than CH₃COOH with NaOH. This is because more energy is used to dissociate the ethanoic acid to the acetate ion and H₃O⁺ ions. And since there is a certain equilibrium to the CH₃COOH CH₃COO^- + H₃O⁺ wouldn't it be impossible to fully react the ethanoic acid?

The ethanoic acid (AcOH) does not have to dissociate in order to react with hydroxide.

The difference in enthalpy is to do with the different bonds being broken and formed - in both cases you release energy from forming a HO-H bond, but in one case you are paying to break a H-Cl bond and in the other you are paying to break the AcO-H bond. Dissociation energy is a measure of how easy the X-H bond is to break heterolytically, but it does not mean that the X-H bond has to break before a base picks up H⁺.

How can be simply subtract the total number of moles of ethanoic acid with the number of moles of NaOH since some of the ethanoic acid would have dissociated into acetate ions? And similarly since I already have some acetate ions from the dissociation reaction of the ethanoic acid shouldn't the initial be more than 0? And also for the H3O+ how can it be 0 initially?

The degree of dissociation for AcOH in water is so low that it can be assumed to be 0 for most calculations.

[Technicalhuman]

When HCl+NaOH NaCl+H₂O the enthalpy is more negative than CH₃COOH with NaOH. This is because more energy is used to dissociate the ethanoic acid to the acetate ion and H₃O⁺ ions. And since there is a certain equilibrium to the CH₃COOH CH₃COO^- + H₃O⁺ wouldn't it be impossible to fully react the ethanoic acid?

The ethanoic acid (AcOH) does not have to dissociate in order to react with hydroxide.

The difference in enthalpy is to do with the different bonds being broken and formed - in both cases you release energy from forming a HO-H bond, but in one case you are paying to break a H-Cl bond and in the other you are paying to break the AcO-H bond. Dissociation energy is a measure of how easy the X-H bond is to break heterolytically, but it does not mean that the X-H bond has to break before a base picks up H⁺.

How can be simply subtract the total number of moles of ethanoic acid with the number of moles of NaOH since some of the ethanoic acid would have dissociated into acetate ions? And similarly since I already have some acetate ions from the dissociation reaction of the ethanoic acid shouldn't the initial be more than 0? And also for the H3O+ how can it be 0 initially?

The degree of dissociation for AcOH in water is so low that it can be assumed to be 0 for most calculations.

Hello Dan thank you for the informative post.

I was lead to believe that the ethanoic acid would have to dissociate into its ions before they react with the alkali.  Does this mean for the actual ethanoic acid molecules and acetate ion/H₃O⁺ ion pair can react with the alkali? But as they both get reacted away there would there be a change in equilibrium of the dissociation reaction of ethanoic acid? Because for the reaction CH3COOH CH3COO- + H+ the concentration of all the reactants and products decrease. So I would think the equilibrium shifts to the left? But that would mean more energy would be given out in the process?

If what i said was true, I would think that since the dissociation of the ethanoic acid is insignificant, majority of the ethanoic acid molecules would be the ones reacting. So the only comparison is the Ac-H and H-Cl as you mentioned? And since the Ac-H is stronger than H-Cl more less energy is released in the overall reaction compare to HCl?

Thanks so much for the help.

[Dan]

since the Ac-H is stronger than H-Cl more less energy is released in the overall reaction compare to HCl?

Energy is released when bonds form and is used to break them. You have to use more energy to break a stronger bond.

[Babcock_Hall]

For the dissociation of acetic acid ΔG° = 27.2 kJ/mol, ΔH° = -0.1 kJ/mol, and ΔS° = -91.6 J/K, where K stands for degrees Kelvin.

"For example, one source which gives the enthalpy change of neutralisation of sodium hydroxide solution with HCl as -57.9 kJ mol-1, gives a value of -56.1 kJ mol-1 for sodium hydroxide solution being neutralised by ethanoic acid."
http://www.chemguide.co.uk/physical/energetics/neutralisation.html

[Technicalhuman]

since the Ac-H is stronger than H-Cl more less energy is released in the overall reaction compare to HCl?

Energy is released when bonds form and is used to break them. You have to use more energy to break a stronger bond.

Hi thanks for the post.

But would any of the dissociated AcO-H react too? Because I was thinking if the equation is AcOH  AcO- +H+ if they do get reacted the products decreases more than the reactants. So the equilibrium shifts to the left giving out more energy.

Thanks so much for the help.