Chemical Equilibrium: Free Energy, Reaction Quotient and Le Chatelier's Principle

Q: What is the difference between Q and K?

They share the same algebraic expression, but K is that expression evaluated only at equilibrium, while Q is evaluated at the current, possibly non-equilibrium, concentrations. Comparing Q to K tells you which direction the reaction still has to run.

Q: Does adding a catalyst shift the equilibrium position?

No. A catalyst speeds up both the forward and reverse reactions by exactly the same factor, so it changes how fast equilibrium is reached but not the equilibrium constant K or the final concentrations.

Q: Why does compressing a gas mixture shift equilibrium but adding an inert gas at constant volume does not?

Compression at constant temperature raises every reactive species' concentration, so the system responds by favouring the side with fewer gas moles. An inert gas added at constant volume does not change any reactive species' concentration or partial pressure, so Q is unchanged and there is nothing to shift.

Equilibrium is a balance of rates, not a stop

A reaction reaches chemical equilibrium not when it stops, but when the forward and reverse reactions run at the same rate, so the measurable concentrations stop changing. For a generic reaction aA + bB ⇌ cC + dD, that balance point is described by the equilibrium constant K, built from the concentrations raised to their stoichiometric coefficients:

K = [C]^c [D]^d / ([A]^a [B]^b)      (at equilibrium)
Q = [C]^c [D]^d / ([A]^a [B]^b)      (at any moment, same expression)

Q, the reaction quotient, has the identical algebraic form but is evaluated at whatever concentrations the system currently has. Comparing Q to K tells you which way the reaction must run: if Q < K there is not yet enough product, so the forward reaction dominates; if Q > K there is too much product and the reverse reaction dominates; if Q = K the system is at equilibrium and net change is zero.

live demo · Q relaxing toward K as concentrations shift● LIVE

Why equilibrium sits where it does: free energy

The thermodynamic reason Q chases K is that the Gibbs free energy of the reacting mixture, G, is a function of composition with a single minimum. At any composition, G(Q) = G° + RT ln Q, and the slope of that curve is the driving force. The reaction proceeds in whichever direction lowers G, and it stops exactly where the slope is zero — which by definition is the equilibrium composition. Setting ΔG = 0 at that point gives the bridge between thermodynamics and the equilibrium constant:

ΔG° = -RT ln K

A large negative ΔG° (a strongly favourable reaction) therefore corresponds to a large K, and a positive ΔG° corresponds to K less than 1 — the reaction barely proceeds before the free-energy curve turns back up.

Le Chatelier's principle, stated precisely

Le Chatelier's principle says that if a system at equilibrium is disturbed, it shifts in the direction that partially opposes the disturbance. That is a qualitative shortcut for a quantitative fact: any disturbance changes Q away from K, and the system then reacts in whichever direction restores Q = K.

Adding a reactant raises Q's denominator, so Q drops below K and the forward reaction runs to bring it back up. Removing product has the same effect. Compressing the volume of a gas-phase mixture raises every concentration, and Q shifts toward the side with fewer moles of gas because that side dilutes back to equilibrium fastest. Temperature is different in kind — it changes K itself, not just Q, because K depends on ΔH°, not on composition.

Building an ICE table

The standard way to solve for actual equilibrium concentrations is an ICE table: rows for Initial, Change and Equilibrium, one column per species, with the change row written in terms of a single unknown extent of reaction x, scaled by each species' stoichiometric coefficient.

        A       +   B   ⇌   C
I     [A]0        [B]0      0
C      -x          -x       +x
E    [A]0-x      [B]0-x      x

K = x / (([A]0-x)([B]0-x))   → solve for x, then read off [A],[B],[C]

Substituting the equilibrium row into the K expression turns the problem into a single algebraic equation in x — usually quadratic for a 1:1 reaction, cubic or worse for more complex stoichiometry, in which case numerical root-finding is faster than factoring by hand.

How K itself moves with temperature: the van't Hoff equation

K is not a universal constant of the reaction — it is a function of temperature. The van't Hoff equation follows directly from ΔG° = ΔH° - TΔS° combined with ΔG° = -RT ln K:

ln K = -ΔH°/R · (1/T) + ΔS°/R

A plot of ln K against 1/T is a straight line whose slope is -ΔH°/R, which is how ΔH° is measured experimentally for reactions that are awkward to run in a calorimeter. For an exothermic reaction (ΔH° < 0) raising T decreases K — the equilibrium shifts back toward reactants, exactly the outcome Le Chatelier's principle predicts by treating heat as a product being added. For an endothermic reaction the sign flips and raising T increases K.

Frequently asked questions

What is the difference between Q and K?

They share the same algebraic expression, but K is that expression evaluated only at equilibrium, while Q is evaluated at the current, possibly non-equilibrium, concentrations. Comparing Q to K tells you which direction the reaction still has to run.

Does adding a catalyst shift the equilibrium position?

No. A catalyst speeds up both the forward and reverse reactions by exactly the same factor, so it changes how fast equilibrium is reached but not the equilibrium constant K or the final concentrations.

Why does compressing a gas mixture shift equilibrium but adding an inert gas at constant volume does not?

Compression at constant temperature raises every reactive species' concentration, so the system responds by favouring the side with fewer gas moles. An inert gas added at constant volume does not change any reactive species' concentration or partial pressure, so Q is unchanged and there is nothing to shift.

Try it live

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