A system at equilibrium is placed under stress by adding more reactant. If this reaction has a small equilibrium constant (Keq), how will the addition of this stress affect the equilibrium of this system?

1 Answer
Aug 19, 2017

I can't tell you the multiple choice answer, but that should not matter...

Since #Q < K_(eq)# after the stress, #Q uarr# to resolve the stress by making more products.


Recall that an equilibrium constant for the reaction

#aA + bB -> cC + dD#

is

#K_(eq) = ([C]^c[D]^d)/([A]^a[B]^b)#,

where #a,b,c,d# are the stoichiometric coefficients of #A,B,C,D#, respectively, and #[" "]# indicates molar concentration.

If an equilibrium constant is small, i.e.

#K_(eq) < 1#,

then that means there are more reactants than products before the equilibrium is disturbed.

(Note that in principle, the actual size of #K_(eq)# does not affect which direction the equilibrium shifts given a certain induced stress.)

Adding more reactants initially decreases the reaction quotient #Q# so that #Q < K_(eq)#. This is the stress that was induced.

Since #Q < K_(eq)#, in accordance to Le Chatelier's principle, the equilibrium shifts so that #Q# increases to equal #K_(eq)# again, going against the disturbance. The equilibrium always tries to undo a given disturbance.

That means it will shift to consume more reactants to generate more products.