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Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: Burningkrome on August 08, 2012, 08:51:24 AM

Title: Equilibrium equation with solids
Post by: Burningkrome on August 08, 2012, 08:51:24 AM
So, discussing equilibrium reactions in my Chem book.  The equilibrium constant(k) is found by …

If aA  +  bB  ::equil:: cC + dD then keq = [C]c[D]d/[A]ab

Solids and solvents are NOT included in the keq calculation.
I.e. They give the example of…
MnO2(s) +  4H+(aq) +  2Cl-(aq)  ::equil::  Mn2+(aq)  +  Cl2(g) + 2H2O(l)
keq = [Mn2+][Cl2]/[H+]4[Cl-]2
…the solids and the liquid (solvent) left out of the equation.

But THEN they ask us to find keq for these…
AgCl(s)  ::equil:: Ag+(aq) +  Cl-(aq)
PCl5 (s)  ::equil:: PCl3 (g) + Cl2 (g)

What am I supposed to do with the PCl5 (s) and AgCl(s) in these equations?  Do I ignore them altogether and just calculate [PCl3][Cl2] and [Ag+][Cl-] respectively?

Thanks!
Title: Re: Equilibrium equation with solids
Post by: Ruminant on August 08, 2012, 07:20:16 PM
In general (and from my understanding) you can ignore solids and pure liquids as their concentrations remain constant. So in the equations given with AgCl(s) and PCl5 (s) proceed to write the Keq as if they were not there recalling that Keq = B]b[C]c

Since aA ::equil:: bB +cC

Perhaps someone can elaborate on it better than I can, hope this helps!

Title: Re: Equilibrium equation with solids
Post by: ramboacid on August 10, 2012, 12:57:41 AM
You can for all practical purposes "ignore" solids and pure liquids in calculations. This is because the activity or concentration of a solid or pure liquid is constant and thus taken to be "1". Therefore they don't affect calculation of the equilibrium constant.

Therefore in the second set of problems, the denominator of the equilibrium constant expression is 1, and we basically ignore it and just write out the numerator. That's why it looks as if we totally ignored the solids and pure liquids in the solubility equilibrium.

If you reversed the reaction so that the aqueous ions were on the left and the solid was on the right, then your numerator would be 1 and the denominator would be the product of the ion concentrations.
Title: Re: Equilibrium equation with solids
Post by: curiouscat on August 22, 2012, 11:18:41 AM
To elaborate a bit more on that; the entities in Keq are fundamentally activities of species. Those for a single phase multicomponent system reduce to conc.

For any single phase pure component that activity is one.

Lest you be surprised later this isn't so much to do with solids at the bottom of it. If you had a solid alloy you wouldn't set activities to one.