This is how my chemistry professor compared reaction of a weak and a strong acid with water:

$$\ce{HCl + H2O -> H3O+ + Cl- + Heat}$$ In this case $\ce{H2O}$ breaks the bond between $\ce{H}$ and $\ce{Cl}$ in $\ce{HCl}$ which results in he formation of $\ce{H+}$ and $\ce{Cl-}$ ions. The $\ce{H+}$ ion combines with $\ce{H2O}$ to form $\ce{H3O+}$ (also known as hydronium). Since adding water to an acid causes its hydration, which is an exothermic process, the reaction also results in the emission of heat. This reaction favors going to the right hand side of the equation. $$\ce{CH3COOH + H2O <=> H3O+ + CH3COO- + Heat}$$ In this case, the size of the acid is greater than the size of $\ce{HCl}$. It means that the process of hydration will be slower. This will lead to a comparatively less number of $\ce{H+}$ ions being combined with water. So, considering Arrhenius' definition of acids, $\ce{CH3COOH}$ is a weaker acid than $\ce{HCl}$. Also, this reaction is going to be reversible.

Now, my questions are as follows:

  1. Why the first reaction favors going to the right hand side of the equation? I think that it has something to do with heat as the heat is lost to the surroundings and the energy level on right hand side decreases. I am still not sure.
  2. Why is the second reaction reversible? I have no idea of why it is reversible so I can't provide my opinions.
  3. Why does water causes ionization of acids? This question might look dumb to some but I am wondering what special water has got that it ionizes acids (or maybe other species too).
  4. Why is there a release of heat? I have read that breaking of bonds is an endothermic process.

1 Answer 1


$$\ce{HCl + H2O -> H3O+ + Cl- + Heat}$$

This reaction owes its ability to go to completion to the great stability the ions it forms when dissolved in water ($\ce{H+}$ and $\ce{Cl-}$). This ionization is caused by stronger attractive forces between the water and the atoms in a molecule (or ions in a salt) than between the atoms within the molecule have themselves, an is an endothermic process. In the dissociation of salts like $\ce{KNO3}$, the solution becomes notably cooler. When an acid dissociates, however, the hydration of $\ce{H+}$ to $\ce{H3O+}$ releases a great deal of heat.

$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO- + Heat}$$

This reaction also results in the formation of $\ce{H+}$ ions, though I wouldn't say that $\ce{CH3COO-}$ is less stable because it is larger than the $\ce{Cl-}$ ion. Consider that $\mathrm{p}K_\mathrm{a}(\ce{HCl})=-7$, $\mathrm{p}K_\mathrm{a}(\ce{CH3COOH})=4.75$, and $\mathrm{p}K_\mathrm{a}(\ce{CH3COOH})=0.75$. In case you are unfamiliar, the $\mathrm{p}K_\mathrm{a}$ is a way of expressing how much an acid dissociates in water, and a drop in $\mathrm{p}K_\mathrm{a}$ by $1$ means an acid releases $10$ times as many $\ce{H+}$ ions. $\ce{HCl}$ releases $10^{11.75}$ times as many $\ce{H+}$ ions as $\ce{CH3COOH}$, but when acetic acid's $\ce{C-H}$ bonds are swapped out for $\ce{C-F}$ bonds to make $\ce{CF3COOH}$ (an even larger molecule), its acidity increases by $10^4$. This shows that ion stability has more to do with bond strength and the inductive effect than ion size.

Both of these reactions are reversible. This just means that the forward reaction is occurring as well as the backward reaction (when occurring at the same rate this is known as equilibrium), but the first reaction lies more towards the formation of the products due to its greater stability. All reactions proceed in the backward direction, unless they are irreversible. Though the dissociation of $\ce{HCl}$ is often considered irreversible due to how shifted its equilibrium is, it is not truly irrversible because it has the potential to reform a molecule of $\ce{HCl}$. A truly irreversible chemical reaction is usually achieved when one of the products exits the reacting system, which does not happen in this case.

  • $\begingroup$ I got your point about why the rate of reversibility is more in case of acetic acid. But I still don't understand why does the reaction go backward? $\endgroup$
    – Parth
    Commented Apr 19, 2016 at 14:42
  • $\begingroup$ @unknownCoder Addressed. $\endgroup$
    – ringo
    Commented Apr 19, 2016 at 15:25

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