The Statement and Its Logic
One of the principal goals of chemical synthesis is to convert the maximum amount of reactants into products while spending the least energy. The equilibrium constant $K_c$ tells us the position a reaction settles into, but it is independent of the initial concentrations. The practical question is the reverse one: if a system already at equilibrium is disturbed, in which direction does the net reaction proceed to restore balance? The answer is given by Le Chatelier's principle.
A change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner as to reduce or counteract the effect of the change. The principle is applicable to all physical and chemical equilibria.
The quantitative engine behind the principle is the reaction quotient $Q_c$. At equilibrium $Q_c = K_c$. Any stress that makes $Q_c < K_c$ drives the reaction forward; any stress that makes $Q_c > K_c$ drives it backward. The reaction always moves to bring $Q_c$ back to $K_c$. Keeping this $Q$-versus-$K$ comparison in mind removes almost all guesswork from shift questions — and it also explains the central trap of the topic, namely that a shift in position is not the same thing as a change in the value of $K$.
Effect of Concentration Change
When the concentration of a reactant or product is changed, the composition of the equilibrium mixture changes so as to minimise the effect of that change. Two statements cover every case: the stress of an added substance is relieved by net reaction that consumes it, while the stress of a removed substance is relieved by net reaction that replenishes it. Consider the gas-phase synthesis of hydrogen iodide:
$$\ce{H2(g) + I2(g) <=> 2HI(g)}$$
Adding $\ce{H2}$ raises the denominator of $Q_c = \ce{[HI]^2/([H2][I2])}$, so $Q_c$ falls below $K_c$ and the reaction moves forward, consuming the added $\ce{H2}$ and $\ce{I2}$ to make more $\ce{HI}$. The new equilibrium has more $\ce{HI}$ than before, and a hydrogen concentration that is lower than just after the addition but still higher than in the original mixture. Removing a product works the same way: it keeps $Q_c$ below $K_c$ and continually pulls the reaction forward — the basis of liquefying and bleeding off ammonia in industry, and of driving $\ce{CaCO3 -> CaO + CO2}$ to completion by sweeping $\ce{CO2}$ out of the kiln.
The thiocyanatoiron equilibrium: $\ce{Fe^3+(aq) + SCN^-(aq) <=> [Fe(SCN)]^2+(aq)}$, deep red.
Adding potassium thiocyanate deepens the red colour — the equilibrium shifts right. Adding oxalic acid, which traps $\ce{Fe^3+}$ as the stable $\ce{[Fe(C2O4)3]^3-}$, removes a reactant; the complex dissociates to replenish $\ce{Fe^3+}$ and the colour fades. Adding $\ce{HgCl2}$, which sequesters $\ce{SCN^-}$ as $\ce{[Hg(SCN)4]^2-}$, fades the colour for the same reason.
Effect of Pressure and Volume
A pressure change brought about by changing the volume only matters for gaseous reactions in which the total number of moles of gas differs between the two sides. Pure solids and pure liquids are ignored, because the volume and concentration of a condensed phase are essentially independent of pressure. The rule follows directly from the principle: increasing the pressure (by compressing the system) shifts the equilibrium towards the side with fewer moles of gas, because that direction lowers the pressure and counteracts the stress.
Take the methanation reaction, in which four moles of gas become two:
$$\ce{CO(g) + 3H2(g) <=> CH4(g) + H2O(g)}$$
Halving the volume doubles every partial pressure and concentration. Substituting double values into the reaction quotient gives $Q_c < K_c$ (the denominator, raised to the higher power, grows faster), so the reaction proceeds forward. Conversely, for $\ce{C(s) + CO2(g) <=> 2CO(g)}$, the forward direction increases the moles of gas from one to two, so raising the pressure pushes this equilibrium backward.
Effect of Inert Gas Addition
Adding an unreactive gas such as argon is the classic trap, and the answer depends entirely on what is held constant. The reaction quotient changes only if an added gas is itself a reactant or a product. An inert gas is neither, so its effect is governed purely by what it does to the partial pressures of the reacting species.
| Condition | What happens to partial pressures | Effect on equilibrium |
|---|---|---|
| Inert gas added at constant volume | Partial pressures and molar concentrations of the reacting species are unchanged | No shift. $Q_c = K_c$ still holds. |
| Inert gas added at constant pressure | Total volume increases; partial pressures of reacting gases fall (system effectively diluted) | Shifts towards the side with the greater number of gaseous moles. |
At constant volume the argon simply occupies the same box alongside the reacting molecules without altering how often they collide, so nothing moves. At constant pressure the system must expand to accommodate the argon, which spreads the reacting molecules out exactly as a volume increase would — and a volume increase favours the more-moles side.
Every shift prediction rests on the $Q$-versus-$K$ comparison. Revise how the constant itself is written in Equilibrium Constant ($K_c$ and $K_p$).
Effect of Temperature Change
Temperature is the one factor that is fundamentally different from the rest. When concentration, pressure or volume is altered, the equilibrium constant $K_c$ is untouched and only the composition adjusts to restore $Q_c = K_c$. When the temperature is changed, the value of $K_c$ itself changes. The direction of that change is fixed by the sign of $\Delta H$ for the reaction.
| Reaction type | Sign of $\Delta H$ | Effect of raising temperature |
|---|---|---|
| Exothermic | $\Delta H < 0$ | $K$ decreases; equilibrium shifts towards reactants (heat treated as a product) |
| Endothermic | $\Delta H > 0$ | $K$ increases; equilibrium shifts towards products (heat treated as a reactant) |
A useful device is to write the enthalpy term as if it were a chemical species. For an exothermic reaction, heat is a product, so raising the temperature is like adding product and the equilibrium retreats. The dimerisation of nitrogen dioxide demonstrates this cleanly:
$$\ce{2NO2(g) <=> N2O4(g)} \qquad \Delta H = -57.2~\text{kJ mol}^{-1}$$
$\ce{NO2}$ is brown and $\ce{N2O4}$ is colourless. In a freezing mixture the exothermic forward reaction is favoured and the brown colour fades; in hot water at 363 K the reverse reaction is favoured and the brown colour intensifies. The opposite signature appears in an endothermic equilibrium such as $\ce{[Co(H2O)6]^3+(aq) + 4Cl^-(aq) <=> [CoCl4]^2-(aq) + 6H2O(l)}$, where the blue tetrachloridocobaltate dominates at room temperature and the pink hexaaqua ion returns on cooling.
Effect of a Catalyst
A catalyst increases the rate of a reaction by providing a new pathway of lower activation energy. Crucially, it lowers the activation energy of the forward and reverse reactions by exactly the same amount, so both rates rise equally. The consequence is that a catalyst helps equilibrium to be reached sooner but does not move it: it does not change the equilibrium composition, the position, or the value of $K$. A catalyst does not appear in the balanced equation or in the equilibrium constant expression. The one caveat from NCERT is practical — if a reaction has an exceedingly small $K$, no catalyst can rescue the yield, because there is simply very little product to form at equilibrium.
Three confusions the examiner exploits every year
These three statements are the most-tested distinctions in the whole topic. Read them together so they cannot be swapped on you in a multi-statement question.
Only temperature changes $K$. Concentration, pressure, volume, inert gas and catalyst all shift the position but leave the numerical value of $K$ untouched.
Inert gas at constant volume does nothing. It changes neither partial pressures nor concentrations, so $Q_c = K_c$ still holds. A shift occurs only at constant pressure (towards more moles of gas).
A catalyst does not shift equilibrium. It speeds the forward and reverse reactions equally and only reduces the time to reach equilibrium. It never alters yield, position or $K$.
Master Shift Table
The whole of §6.8 collapses into a single reference grid. For each stress, read off the direction of the shift in position and — separately — whether the equilibrium constant $K$ changes. Note the deliberate split between the two right-hand columns: it is the discipline of keeping them apart that defeats the standard trap.
| Stress applied | Direction of shift in position | Change in $K$ |
|---|---|---|
| Increase concentration of a reactant | Forward (consumes added reactant) | No change |
| Increase concentration of a product / remove product | Backward / forward to replenish or consume | No change |
| Increase pressure (decrease volume) | Towards the side with fewer moles of gas | No change |
| Decrease pressure (increase volume) | Towards the side with more moles of gas | No change |
| Increase temperature — exothermic ($\Delta H < 0$) | Backward (towards reactants) | $K$ decreases |
| Increase temperature — endothermic ($\Delta H > 0$) | Forward (towards products) | $K$ increases |
| Add inert gas at constant volume | No shift | No change |
| Add inert gas at constant pressure | Towards the side with more moles of gas | No change |
| Add a catalyst | No shift (equilibrium reached faster) | No change |
Industrial Applications
The economic value of Le Chatelier's principle is that it tells a chemist how to bias an equilibrium towards product. Two NCERT case studies are perennial NEET material because each forces a compromise between yield (governed by equilibrium) and rate (governed by kinetics).
Haber process — synthesis of ammonia
$$\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \qquad \Delta H = -92.38~\text{kJ mol}^{-1}$$
The forward reaction is exothermic and converts four moles of gas into two. High pressure therefore favours ammonia because it pushes towards the fewer-mole side; about 200 atm is used. A low temperature would maximise the equilibrium yield of an exothermic reaction, but at low temperature the rate is hopelessly slow. The resolution is a moderate temperature of around 500 °C together with an iron catalyst, which delivers a satisfactory rate while keeping the equilibrium yield reasonable. Continuous liquefaction and removal of ammonia keeps $Q_c$ below $K_c$ and drives the reaction onward.
Contact process — manufacture of sulphuric acid
$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)} \qquad K_c = 1.7 \times 10^{26}$$
Here the enormous $K_c$ shows the equilibrium lies almost entirely towards $\ce{SO3}$, so yield is not the problem; the oxidation is simply very slow. Platinum or divanadium pentoxide ($\ce{V2O5}$) is used as a catalyst to raise the rate without disturbing the favourable equilibrium position — a clean illustration that a catalyst addresses kinetics, not the equilibrium itself.
Le Chatelier's principle in one screen
- A system at equilibrium responds to a stress so as to counteract that stress; predict the direction by comparing $Q$ with $K$.
- Adding a substance shifts the reaction to consume it; removing one shifts the reaction to replenish it.
- Increasing pressure (decreasing volume) shifts towards fewer moles of gas; solids and liquids are ignored.
- Only a temperature change alters $K$: $K$ falls for exothermic and rises for endothermic reactions as $T$ increases.
- Inert gas at constant volume — no shift; at constant pressure — shift towards more gaseous moles.
- A catalyst never shifts equilibrium or changes $K$; it only shortens the time to reach equilibrium.
- Haber: high pressure, ~500 °C, Fe catalyst, remove $\ce{NH3}$. Contact: $\ce{V2O5}$ or Pt to fix the slow rate of an already favourable equilibrium.