2
$\begingroup$

I read in a page that ionisation is an irreversible process. But, as equilibrium is always reversible would that mean equilibrium does not involve ionisation process.

And than dissociation is a reversible process therefore only dissociation process is seen in ionic equilibrium.

Now if the above statement is correct would that mean only covalent compounds take part in ionic equilibrium?

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ A ionic compound like NaCl(s) performs dissociatiating/associating dissolution equilibrium. A covalent compound like acetic acid performs ionizating/recombining acido-basic equilibrium. Both reversible in common and thermodynamic sense. $\endgroup$
    – Poutnik
    Apr 16, 2021 at 5:57

2 Answers 2

Reset to default
2
$\begingroup$

An analogy might help. Solubilization of a salt in a solvent (say NaCl in water) is similar to condensation of a gas.

Imagine you have an amount of gas in a cylinder with a piston to regulate the volume and pressure. Beginning at a pressure below the vapor pressure of the substance (which means that all is gas, none is liquid), the pressure is increased (while temperature is constant) until eventually the boiling point is encountered. At this point the pressure is equal to the vapor pressure (at that temperature) and the gas and liquid can coexist (all of this by the way is simpler to illustrate with a T-p phase diagram). Further compression condenses more gas until all of the substance is liquid. Below the boiling point pressure none of the substance is liquid, and above that pressure none is gas.

Next consider solubilization of a substance in a fixed volume of liquid (with T,p constant). One could traverse the solubility diagram by changing the temperature, but it is for the present discussion simpler to consider changing the concentration. At very low concentrations (small amounts of substance) all of the substance solubilizes. Exactly none (yes, none) is present as the solid. But as you increase the amount of added substance a concentration is reached at which solid remains. That exact point is the solubility limit of the substance at that temperature. It can be thought of as equivalent to the boiling point pressure (coexistence point) of a gas and liquid at a given temperature. If you increase the amount of added substance at that point, no more substance will solubilize (exactly none).

So when a salt is added below its solubility concentration limit no solid remains, all of the substance is in the liquid, and there is no equilibrium only in the sense that there cannot be solid below the solubility limit.

This is also why the equilibrium equation for solubilization of a substance is written as

$$K_{\mathrm{sp}} = \frac{a(\textrm{soln})}{a(\textrm{s})}= a(\textrm{soln})$$

where a's are activities (for small concentrations, the activity of the solubilized substance is equal to concentration ). We ignore the activity of the solid (it is constant and assumed equal to 1) and above the solubility concentration limit the activity of the solubilized substance is also constant (equal to the solubility product).

Improve this answer
$\endgroup$
1
$\begingroup$

For most salts the equilibrium constant for ionization is very large and thus the reaction practically proceeds only in the forward direction and so we generally say it is irreversible.

But in reality no chemical reaction is irreversible and some backward reaction always takes place.

So, in a strict sense ionization is a reversible equilibrium process but for practical we assume it to be irreversible.

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ Note that in context of thermodynamics, reversibility versus irreversibility is not both way versus one way processes. The process occuring in system is reversible, if returning the system surrounding to its initial state returns the system to the initial state as well. This happens, if during all the procedure we can consider everything being all the time in equilibrium. In reality, no process is fully reversible in this sense, as reaching equilibrium takes infinity time. $\endgroup$
    – Poutnik
    Apr 16, 2021 at 5:40
  • $\begingroup$ @Poutnik I understand that but I didn't mean reversible in thermodynamic terms but in the common terms. As per the OP's question it seems like he is not referring to reversible and irreversible from thermodynamic terms but with regard to one way or two way process. Or in simple words I believe OP is asking about the extent of reaction when referring to reversible and irreversible. $\endgroup$
    – Nisarg Bhavsar
    Apr 16, 2021 at 5:43
  • $\begingroup$ well, equilibrium involves thermodynamics. $\endgroup$
    – Poutnik
    Apr 16, 2021 at 5:44
  • $\begingroup$ @Poutnik I do accept that the terms I have used are a little ambiguous. But I wanted to convey the answer to the OP in a layman's manner. This is a very standard question from high school or secondary school where they still don't understand the thermodynamic meaning of reversible and irreversible but use the term for the extent of reaction. $\endgroup$
    – Nisarg Bhavsar
    Apr 16, 2021 at 5:50
  • 1
    $\begingroup$ Thank you guys...so let me just sum it up..all reactions are reversible(In a strict sense as stated) and therefore can be considered in equilibrium.(and yes I am a high school student) $\endgroup$
    – Roy Joseph
    Apr 16, 2021 at 6:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.