Equilibrium shifts: separate the reaction quotient from the constant

MCAT content library / Chemistry & physics / Equilibrium shifts and reaction quotients

Write the mass-action expression correctly, compare Q with K immediately after a perturbation, and reserve changes in K for changes that alter the equilibrium constant itself.

About 17 minutes, including concentration and compression perturbations.

K describes equilibrium at a given temperature

For the reaction

aA + bB ⇌ cC + dD,

a concentration-based equilibrium expression has the form

K_c = [C]^c[D]^d/([A]^a[B]^b)

for species whose activities are represented by concentration in the model. Pure solids and pure liquids are omitted from the ordinary expression because their activities are treated as constant.

At one fixed temperature, K is a property of the reaction under the chosen standard-state convention. Changing concentrations or adding a catalyst does not change K. Temperature can change K because it changes the thermodynamic balance between reactants and products.

Concept: Q uses the same form before equilibrium is restored

The reaction quotient Q has the same algebraic form as K but uses the current, not necessarily equilibrium, composition.

• Q < K: the forward reaction is favored as the system moves toward equilibrium.
• Q > K: the reverse reaction is favored.
• Q = K: the system is at equilibrium under the modeled conditions.

This makes perturbation problems mechanical: calculate how the perturbation changes Q immediately, then compare that new Q with the unchanged K if temperature has not changed.

Worked application: perturb a gas equilibrium

An original gas reaction is:

A(g) ⇌ 2B(g).

At the stated temperature, an equilibrium mixture has [A] = 1.0 M and [B] = 2.0 M, so

K_c = [B]²/[A] = 4.0.

Experiment 1 suddenly adds B so that [B] becomes 3.0 M while [A] is momentarily unchanged:

Q = 3.0²/1.0 = 9.0.

Because Q > K, the system shifts in the reverse direction, consuming B and forming A until Q returns to 4.0.

Pattern: volume changes act through concentration or pressure

Halving the container volume at constant temperature immediately doubles both concentrations in the original equilibrium mixture. The new quotient is:

Q = 4.0²/2.0 = 8.0 > K.

The reaction therefore shifts left, toward the side with fewer gas molecules in the balanced equation. This is the quantitative basis for the familiar Le Châtelier shortcut about compression favoring fewer moles of gas.

If both sides had the same total gas stoichiometric coefficient, a uniform volume change could leave Q/K unchanged. And if a species is a pure solid, changing its amount while some solid remains does not enter the ordinary equilibrium expression.

Rule: catalysts change the route to equilibrium, not the equilibrium constant

A catalyst lowers the activation barrier for forward and reverse reaction pathways. It can make equilibrium establish more quickly but does not change the equilibrium free-energy difference or K at fixed temperature.

Temperature is different. For an endothermic forward reaction, increasing temperature typically raises K; for an exothermic forward reaction, increasing temperature typically lowers K. Use the reaction enthalpy or thermodynamic information supplied rather than treating heat as though it were simply another concentration term in Q.

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Continue learning

• Reaction free energy and equilibrium
• Activation barriers and catalysis
• Browse Chemistry & Physics MCAT lessons

Sources

1. AAMC. Content Category 5E.
2. OpenStax Chemistry 2e. Equilibrium Constants.