Le Chatelier's principle

Le Chatelier's principle: when a system at equilibrium is subjected to a change in concentration, pressure, volume, or temperature, the equilibrium shifts in the direction that tends to reduce the effect of that change. Underneath the qualitative wording sits a precise quantitative trigger you have already met — the reaction quotient $Q$. At equilibrium $Q = K$; any disturbance momentarily pushes $Q$ away from $K$, and the system reacts in whichever direction restores the equality.

where: $[\text{products}]$ and $[\text{reactants}]$ are the instantaneous concentrations (or partial pressures) at the moment of disturbance, raised to their stoichiometric powers $p$ and $r$, and $K$ is the equilibrium constant at the prevailing temperature. Every one of the five classic disturbances below is really just a different way of nudging $Q$ off $K$.

Concentration. Adding more of a reactant (or removing a product) makes the denominator of $Q$ effectively larger (or the numerator smaller), so $Q$ drops below $K$ and equilibrium shifts forward, consuming some of what was added. Adding more product (or removing reactant) does the reverse, shifting backward. This is why removing a product as it forms — distilling it off, or letting a gas escape — drives a reaction further toward completion than the same mixture in a sealed vessel ever could.

Pressure and volume (gaseous equilibria only). Shrinking the volume raises every partial pressure, and $Q$'s response depends on how many gas molecules sit on each side. If $\Delta n_g = n_{\text{gas, products}} - n_{\text{gas, reactants}}$ is negative, compressing the mixture shifts equilibrium toward the side with fewer gas moles, partially relieving the pressure rise; expanding the volume shifts it toward the side with more moles. When $\Delta n_g = 0$, compression changes every partial pressure by the same factor, $Q$ is untouched, and the position of equilibrium is unaffected.

Inert gas. Added at constant volume, an inert gas raises the total pressure but leaves every reacting gas's own partial pressure — and hence $Q$ — exactly where it was, so equilibrium is undisturbed. Added at constant total pressure, the vessel must expand to accommodate it, diluting the reacting gases exactly as a genuine volume increase would, so equilibrium shifts toward the side with more gas moles (if $\Delta n_g \neq 0$).

Temperature. Heat behaves like a reactant or product in an exothermic or endothermic reaction respectively, so raising temperature always shifts equilibrium in the endothermic direction — the direction that absorbs the added heat. For an exothermic forward reaction ($\Delta H < 0$), heating shifts equilibrium backward and $K$ decreases; for an endothermic forward reaction ($\Delta H > 0$), heating shifts it forward and $K$ increases. Temperature is the only one of these five factors that changes the numerical value of $K$; concentration, pressure and volume changes shift the equilibrium position at a fixed temperature but leave $K$ itself unchanged.

Catalyst. A catalyst lowers the activation energy of the forward and reverse steps equally, speeding up how quickly equilibrium is reached without shifting its position or altering $K$ — because it accelerates both directions by the same factor, the ratio of forward to reverse rates at equilibrium, which fixes $K$, is untouched.

Le Chatelier's principle is not an independent law of nature — it is a shortcut for the $Q$-versus-$K$ reasoning already used to predict reaction direction. At equilibrium, $Q = K$ exactly. Any disturbance is, at its core, a sudden change to the concentrations, partial pressures, or the value of $K$ itself, and the system responds by running forward or backward until $Q$ catches back up to (the new) $K$. For concentration, pressure, and volume changes, only the instantaneous values inside $Q$ move — $K$ itself is fixed at constant temperature — so the system reacts in whichever direction restores $Q = K$, and this restoring shift is, by construction, the direction that partially cancels the imposed change. For temperature changes, it is $K$ itself that moves (larger $K$ for the endothermic direction as heat is supplied), so the system must shift toward the new $K$, and this is the same as saying the reaction runs in the heat-absorbing direction. A catalyst changes neither $Q$ (at the instant of disturbance, nothing about concentrations changed) nor $K$ (its position depends only on the rate-constant ratio, which is unaffected by a species that accelerates both directions equally), so there is nothing for the system to shift toward — only the speed of getting to the same equilibrium changes. This single $Q$-vs-$K$ lens is why the principle applies identically to physical equilibria (melting, dissolution, vaporisation) as well as chemical ones: any process with a $Q$ and a $K$ obeys it.

Effect of pressure and temperature on the Haber process. For $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ ($\Delta H < 0$, exothermic), predict the effect of (a) increasing the pressure by decreasing the volume, and (b) increasing the temperature, on the yield of ammonia, and (c) state why an industrial plant still adds a catalyst even though it does not shift the equilibrium.

Step 1. Count gas moles on each side: reactants $= 1 + 3 = 4$ mol gas; products $= 2$ mol gas, so $\Delta n_g = 2 - 4 = -2$ — the product side has fewer moles of gas.

Step 2. (a) Compressing the vessel raises every partial pressure, which momentarily makes $Q$ smaller than the new effective $K$-comparison for the fewer-mole side (equivalently: raising pressure favours the side that reduces total moles, since fewer moles means a smaller pressure contribution). Since products (2 mol) are fewer than reactants (4 mol), equilibrium shifts forward, increasing $\text{NH}_3$ yield. This is exactly why the industrial Haber process is run at high pressure, typically 150–250 atm.

Step 3. (b) The forward reaction is exothermic ($\Delta H < 0$), so heat behaves like a product. Raising temperature shifts equilibrium backward (toward the endothermic reverse direction), decreasing $\text{NH}_3$ yield and lowering $K$ — even though the higher temperature increases the rate at which equilibrium is reached.

Step 4. (c) A catalyst (finely divided iron with promoters) does not change the equilibrium yield calculated in Steps 2–3 at all — it speeds up both the forward and reverse rates equally. Its industrial role is purely kinetic: it lets the plant reach a usefully high $\text{NH}_3$ yield in a practical time at a moderate compromise temperature (about 450 °C), rather than needing an impractically low temperature (which would maximise equilibrium yield per Step 3 but make the reaction too slow to be economic). Answer: high pressure and a moderate, catalyst-assisted temperature together maximise the practical ammonia yield.

• Applying "shift toward fewer moles" to a constant-volume inert-gas addition. That rule needs partial pressures to actually change; adding inert gas at constant volume leaves every reacting gas's partial pressure — and $Q$ — untouched.
• Believing a catalyst shifts equilibrium or raises the equilibrium yield. A catalyst only shortens the time to reach equilibrium; forward and reverse rates rise equally, so the equilibrium position and $K$ stay exactly the same.
• Forgetting that pure solids and liquids don't appear in $Q$ or $K$. Adding or removing a pure solid or liquid reactant/product (as long as some remains) does not shift equilibrium, since its activity is constant and never enters $Q$.
• Assuming temperature changes only affect rate, not $K$. Temperature is the one variable here that genuinely changes the numerical value of $K$ — raising it for endothermic reactions, lowering it for exothermic ones — unlike concentration, pressure, or volume changes, which move the position but leave $K$ fixed.

Ask the tutor: Want to try a mixed set of Le Chatelier predictions — including a solubility or ionisation equilibrium from later in this chapter?