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Le Chatelier's principle in a reaction with solid reactant


Raghav Dua

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According to Le Chatelier's Principle Increasing the conc. of a reactant shifts the reaction in forward direction to re establish the equilibrium.Now if one of the reactant is solid and let the reaction be:

 

 

A(s)+B(g)------->C(g)+D(g)

 

In such a case if we apply the equilibrium constant equation in which conc. of a solid reactant at equilibrium is taken as 1,then according to this increasing to decreasing the conc. of solid will have no effect on equilibrium so there should not be any shift in the reaction

 

Why rule should be applied in such a case ?

 

 

 

 

 

 

Thanks In advance

 

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The simple rate equation relies on the reactants being intimately mixed and free to move.

 

This is not the case with solids.

Reactions involving solids are multistep.

An important step brings the reactants into contact and is often the rate determining step.

Here is a guide to solid kinetics.

 

http://www.fhi-berlin.mpg.de/acnew/groups/nanostructures/pages/teaching/pages/teaching__malte_behrens__solid_state_kinetics.pdf

Edited by studiot
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Hold on... wouldn't there be a noticeable increase in product concentration due to a higher entropy on the right side, making the reaction thermodynamically favored toward the products... regarldess of Le Chateliers principle.

 

 

The crux of this question is "What is the concentration of a solid?"

Edited by studiot
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delta H, delta G and so forth tell you nothing about the rate of reaction, only whether the process will produce or consume energy (heat energy, free energy etc)

They do not even describe the activation energy that may be necessary.

 

I just watched the latest Tom Cruise and he made great foofah about Horatius at the Bridge.

The 'rate determining step' is a bit like that - only soe many molecules at a time can participate.

 

Yes a reaction will continue to proceed if new material (reactant) is supplied or if products are removed eg as a precipitate in a solution reaction.

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