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In the calculation for equilibrium constant, we take the active mass of solids and liquids as unity. Intuition tells that more amount of a solid reactant should result in more product being formed, but according to the formula for the equilibrium constant it has no role in determining that. Is it unity in all possible cases? How can we assume it to be one and have it not affect the equilibrium calculation?

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    $\begingroup$ Chemists don't call it the active mass, but rather the activity. The activity of a solid is taken as unity if the reaction is happening in a liquid or gas. The activity of a liquid is taken as unity if the reaction is happening in a gas. // Consider putting some calcium carbonate in pure water. A tiny amount will dissolve. But once the solution is saturated it doesn't matter if there is one microgram or 1 kilogram of calcium carbonate excess, no more calcium carbonate will dissolve. $\endgroup$ – MaxW Mar 28 at 6:50
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You might view the relevant principle as follows: components whose activity changes during a reaction tend to put a stop to a reaction before they are completely consumed. Components with constant activity (independent of the amount of other reactants), on the other hand, can be completely consumed (exhausted): they can be limiting reagents, provided they are less abundant (stoichiometrically) than the other reagents. If a reagent with constant activity is exhausted before equilibrium is reached the reaction stops - it is then a limiting reagent. Reagents whose activities change during a reaction (because their concentration changes, for instance) are in principle not exhausted this way because their activity increases too much for the reaction to proceed any further. Reactions involving only reagents with variable activity tend to go to an equilibrium in which none of the species is completely exhausted. If reagents with constant activity are present in excess (such that their initial amount exceeds that at equilibrium), then they will not limit the extent of reaction. Note also that if the increase in activity with dilution of a reagent with variable activity is negligible, its concentration at equilibrium may be so low that it can be considered exhausted (and limiting).

A good example of the relevant principle is a mixture of two reactive solids or two immiscible liquids of constant activity that react at the interface between them. In either case, if the reaction is favorable ($Q\gt K_{eq}$) then it will proceed until one or both of the substances has been completely consumed (both being completely consumed if present in stoichiometric proportion). On the other hand, when a solid dissolves into a solvent, the activity of the solid remains constant and it will continue to dissolve provided the activity of the dissolved species does not increase above that of the solid. If it doesn't, then the solid is completely dissolved (exhausted).

If a reaction is already at equilibrium, adding more of a solid won't make a difference (provided there are no surface effects). But if you are not at equilibrium, it will if the amount present is limiting. A good example again is solubilization of a solid in a solvent. Added solid will continue dissolving until you reach the saturation point. Past this point the concentration of the compound in solution does not change any further. In addition, with saturation the activity of the compound in the solid and solution phases becomes equal.

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  • $\begingroup$ In case of solid being the limiting reagent, If we add more solid reactant, the reaction should proceed further. But according to le chatelier's principle, which depends on the equilibrium equation, it has no effect. Am I understanding something wrong here? $\endgroup$ – Micelle Mar 28 at 15:10
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    $\begingroup$ @AnanyaPateriya If the reaction is already at equilibrium, adding more of a solid won't make a difference. But if you are not at equilibrium, it will if the amount present is limiting. $\endgroup$ – Buck Thorn Mar 28 at 16:14
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    $\begingroup$ A good example again is solubilization of a solid in a solvent. Added solid will continue dissolving until you reach the saturation point. Past this point the concentration of the compound dissolved in solution does not change any further. In addition, with saturation the activity of the solid in the solid and solution phases becomes equal. $\endgroup$ – Buck Thorn Mar 28 at 16:19

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