# 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?

Aug 19, 2017

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

Since $Q < {K}_{e q}$ after the stress, $Q \uparrow$ to resolve the stress by making more products.

Recall that an equilibrium constant for the reaction

$a A + b B \to c C + \mathrm{dD}$

is

${K}_{e q} = \frac{{\left[C\right]}^{c} {\left[D\right]}^{d}}{{\left[A\right]}^{a} {\left[B\right]}^{b}}$,

where $a , b , c , d$ are the stoichiometric coefficients of $A , B , C , D$, respectively, and $\left[\text{ }\right]$ indicates molar concentration.

If an equilibrium constant is small, i.e.

${K}_{e q} < 1$,

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

(Note that in principle, the actual size of ${K}_{e q}$ does not affect which direction the equilibrium shifts given a certain induced stress.)

Adding more reactants initially decreases the reaction quotient $Q$ so that $Q < {K}_{e q}$. This is the stress that was induced.

Since $Q < {K}_{e q}$, in accordance to Le Chatelier's principle, the equilibrium shifts so that $Q$ increases to equal ${K}_{e q}$ 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.