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le chatelier principle with solids

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johndoe3344

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

If I add more A, does it shift the equilibrium to the right?

EDIT: What about now? A(s) <-> B(aq) + C(aq)
 

Dr Gerrard

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I don't think changing the solid would.

However, what if you increase the concentration of B or C.

Or what if this is an exothermic reaction, and you add heat.

Will the reaction shift towards the solid?
 

chemtopper

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If you add heat ,exothermic reaction would give more solid.

Solid has no effect on the equilibrium.
 

soby10

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Always exclude pure liquids or pure solids in equib. only (aq) and (g) are in equilbrium expressions. So answer is no for both. The expression is written Keq= B(aq) + C(aq)/1.
 

sv3

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i get that solids don't go into the Keq expression but if Keq is a constant at a given temperature, then how is it constant if you add more B (product) let's say if the reaction doesn't progress in the reverse direction? Without going in reverse, the Keq would change if a product that is part of the expression increases. I would have thought that the reaction does in fact go in reverse due to Le chat and this would also keep Keq constant

of course if Keq is not constant at a given temperature my question makes no sense....and in that case please let me know!

thanks
 

BenUstudent

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Always exclude pure liquids or pure solids in equib. only (aq) and (g) are in equilbrium expressions. So answer is no for both. The expression is written Keq= B(aq) + C(aq)/1.

I agree with SOBY10, You do Exclude pure liquids and Solids.
 

Biochemistry

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

If I add more A, does it shift the equilibrium to the right?

EDIT: What about now? A(s) <-> B(aq) + C(aq)

-Le chataleir's principle that chemical equilibrium should be obtiained in a reversible chemical reaction, if more products are put then the reaction will shift to make more reactants until an equilibrium is reached under a close controlled system, if more reactants are present then the reaction is shifted whereore products are produced until an equilibrium is reached....
 

peyman2002

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I don't think changing the solid would.

However, what if you increase the concentration of B or C.

Or what if this is an exothermic reaction, and you add heat.

Will the reaction shift towards the solid?

Since concentrations of solids are constants and do not appear in expressions for Q or K, removing or adding some solid does not cause shifts. However, shifts in the equilibrium do change the amount of solid present!
 

sv3

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Since concentrations of solids are constants and do not appear in expressions for Q or K, removing or adding some solid does not cause shifts. However, shifts in the equilibrium do change the amount of solid present!

That was a nice post.
So shifts in equilibrium can change the amount of solid but not its concentration. For solutions, shifts change their concentration.
Now I wonder, does a change in concentration automatically mean a change in amount? For example, if you had an all solution reaction and increased the amount of product at equilibrium, we know the reaction reverses and amount of product decreases. Does this mean the amount and concentation of the reactant increases? After all this solid talk im confused about solutions!!

thanks
 

peyman2002

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That was a nice post.
So shifts in equilibrium can change the amount of solid but not its concentration. For solutions, shifts change their concentration.
Now I wonder, does a change in concentration automatically mean a change in amount? For example, if you had an all solution reaction and increased the amount of product at equilibrium, we know the reaction reverses and amount of product decreases. Does this mean the amount and concentation of the reactant increases? After all this solid talk im confused about solutions!!

thanks

if we have an all solution rxn, and if we change the amount of product at equilibrium we are increasing the amount of reactants and by M=moles solute/Liters solution the concentration goes up since the volume is cte.
 
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