According to Le Chatelier's principle, is it possible to fully consume a reactant if we increase the concentration of the other reactant(s)?

Suppose there is a hypothetical chemical equation: $$\ce{ A + B <=> C}$$


If we increase the concentration of B to a large extent with the intention of fully consuming A in the reaction, will A actually be fully consumed in the reaction (no traces of A left), and the reaction completely favors the forward reaction to produce C?


Because the formula for equilibrium constant is K= [C]/[A][B], where in this case, A is fully consumed, therefore K = [C]/[0][B], which is not possible because it is not divisible by 0.




At equilibrium, all species have to be present at some concentration other than zero. This is because at equilibrium, the forward and reverse rates have to be equal and non-zero. This is the property of a dynamic equilibrium.


Adding lots of B will not remove A completely. If the zero in the denominator bothers you, write the equation like this:


$$[A] \* [B] \* K = [C]$$


Unless [C] is zero, [A] and [B] have to be non-zero. [A] could get very small, but not zero.


On the other hand, if you continuously remove C until there are no atoms left, there will be no C and no A (assuming B is in excess). However, that situation is not called equilibrium.



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> Because the formula for equilibrium constant is K= [C]/[A][B], where in this case, A is fully consumed, therefore K = [C]/[0][B], which is not possible because it is not divisible by 0
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The division by zero is a technical problem of the description of the equilibrium constant, but not the underlying explanation. You can try to calculate the reaction quotient Q for a situation where one of the reactants is absent, and you run into that problem. This does not mean, however, that you can't have that situation. It just means that your model (concerning the definition of Q) is not able to capture it in a meaningful way.


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You can make [A] and [B] very small by removing [C]!


Conversely, you make C at a greater rate by increasing A and B.


Just increasing B will not help, except possibly in the case where you have a 2 phase reaction. Increasing [B] in one phase may make more of it available to A, which is dissolved in the other phase.


[C]/0 cannot exist precisely because the **equilibrium constant** =



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> K = [C]/[A][B]
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if you have alot of C and no A or B than [A] and [B] will increase until equilibrium is reached.


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