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?

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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?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

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 , 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.

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 Answers