Factors Affecting Equilibria

Squeeze a sealed flask of an equilibrium mixture, warm it, or pour in a little more of one reactant, and the mixture does not simply sit there — it shifts, quietly re-arranging concentrations until forward and reverse rates match again at a new balance point. That predictable "push-back" is Le Chatelier's principle, one of the most quoted rules in chemistry, and it turns qualitative common sense (add more, get more product) into a tool sharp enough to explain why the Haber process runs at high pressure and moderate heat rather than at the temperature that looks fastest. This section covers Unit VI, Equilibrium of CBSE/NCERT Class 11 Chemistry: stating Le Chatelier's principle and applying it to concentration, pressure/volume, temperature, inert-gas, and catalyst changes for both chemical and physical equilibria, building on the $K_c$ , $K_p$ , and $Q$ machinery from the previous node. The reasoning behind every shift is the comparison of the reaction quotient $Q$ against the equilibrium constant $K$ — the same $Q$ -versus- $K$ test used earlier to decide which way an unbalanced system moves.

The core insight

A system at equilibrium responds to any imposed change in concentration, pressure, volume, or temperature by shifting in whichever direction partially cancels that change, because the disturbance first pushes the reaction quotient $Q$ away from $K$ and the system reacts until $Q = K$ is restored.

The core idea

Le Chatelier's principle, precisely stated. If a system at equilibrium is subjected to a change in concentration, pressure, volume, or temperature, the equilibrium shifts in the direction that tends to counteract (partially oppose) the effect of that change. Every specific case below is this one sentence applied to a different knob.

Concentration. Adding more reactant, or removing product, drives the reaction forward; adding more product, or removing reactant, drives it backward. The mechanism is the $Q$ -versus- $K$ test: adding a reactant lowers $[\text{products}]/[\text{reactants}]$ , so $Q < K$ , and the forward reaction runs until $Q = K$ again.

Pressure and volume (gas-phase reactions). Compressing the container (raising pressure) shifts equilibrium toward the side with fewer moles of gas; expanding it shifts equilibrium toward the side with more moles of gas. If the two sides have equal gas moles, $\Delta n_g = 0$ , pressure and volume changes have no effect at all.

Inert gas. At constant volume, adding an inert gas changes total pressure but leaves every reacting species' partial pressure and concentration untouched, so equilibrium is unaffected. At constant pressure, adding an inert gas forces the container to expand, diluting the reacting species — equivalent to lowering pressure, so equilibrium shifts toward the side with more gas moles.

Temperature. This is the only listed factor that changes the value of $K$ itself, not merely the equilibrium position. For an exothermic forward reaction ( $\Delta H < 0$ ), heating shifts equilibrium backward and lowers $K$ ; for an endothermic forward reaction ( $\Delta H > 0$ ), heating shifts equilibrium forward and raises $K$ . Treating heat as a pseudo-product (exothermic) or pseudo-reactant (endothermic) makes the direction easy to recall.

Catalyst. A catalyst accelerates the forward and reverse reactions by the identical factor, so the system reaches equilibrium faster without moving the equilibrium position or changing $K$ .

Physical equilibria. The principle applies just as well outside chemical reactions. Compressing an ice–water equilibrium favours the liquid, which occupies less volume than the same mass of ice — this is why pressure lowers ice's melting point. Heating a liquid–vapour equilibrium favours the vapour, since evaporation absorbs heat, raising the equilibrium vapour pressure.

Q = \frac{[\text{products}]^{\text{stoich}}}{[\text{reactants}]^{\text{stoich}}}, \qquad Q < K \Rightarrow \text{shift forward}, \quad Q > K \Rightarrow \text{shift backward}. \qquad (1)

where: $Q$ is the reaction quotient, computed with the same expression as $K$ but using the concentrations (or pressures) present the instant after the disturbance, and $K$ is the fixed equilibrium constant at the given temperature. Every one of the five factors above is just a way of forcing $Q$ away from $K$ , except temperature, which moves $K$ itself.

Figure

A balance beam labelled Q equals K tipping toward products when a reactant is added and tipping back level once equilibrium is restored

Derivation AHow compressing a gas mixture picks the side with fewer moles

The goal is to show, using only the definition of concentration and simple algebra, why compressing a gas-phase equilibrium always shifts it toward the side with fewer moles of gas — not as a memorised rule but as a direct consequence of how $Q_c$ responds when every concentration is squeezed by the same factor, so that a single sign of $\Delta n_g$ decides whether $Q_c$ rises above or falls below $K_c$ the instant the volume changes.

Step 1. Write the general gas equilibrium $aA + bB \rightleftharpoons cC + dD$ with reaction quotient $Q_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$ , and define $\Delta n_g = (c+d) - (a+b)$ , the change in gas moles from reactants to products. At the instant before any disturbance, $Q_c = K_c$ .

Step 2. Suppose the container volume is suddenly scaled from $V_0$ to $V_{\text{new}} = xV_0$ , with $x < 1$ for a compression. Because concentration is moles divided by volume, every species' concentration is instantly multiplied by the same factor $1/x$ (moles unchanged, volume smaller), so $[C]_{\text{new}} = [C]_{\text{old}}/x$ and likewise for every other species.

Step 3. Substitute the scaled concentrations into $Q_c$ . The numerator picks up a factor of $(1/x)^{c+d}$ and the denominator a factor of $(1/x)^{a+b}$ , so $Q_{c,\text{new}} = Q_{c,\text{old}}\left(\dfrac{1}{x}\right)^{(c+d)-(a+b)} = K_c\left(\dfrac{1}{x}\right)^{\Delta n_g}$ , using $Q_{c,\text{old}} = K_c$ from Step 1.

Step 4. Compare $Q_{c,\text{new}}$ to $K_c$ . For a compression, $x < 1$ so $1/x > 1$ . If $\Delta n_g < 0$ (fewer gas moles on the product side), $(1/x)^{\Delta n_g} < 1$ , so $Q_{c,\text{new}} < K_c$ and the reaction runs forward, toward products — the side with fewer moles. If $\Delta n_g > 0$ , the same algebra gives $Q_{c,\text{new}} > K_c$ , so the reaction runs backward, again toward the side with fewer moles. If $\Delta n_g = 0$ , the exponent vanishes and $Q_{c,\text{new}} = K_c$ exactly — no shift, matching the stated exception. Verified: for any $\Delta n_g \ne 0$ and any compression $x<1$ , $(1/x)^{\Delta n_g}$ always lands $Q_{c,\text{new}}$ on the side of $K_c$ that favours fewer gas moles, and $\Delta n_g = 0$ gives $Q_{c,\text{new}} = K_c$ identically.

Worked example

Problem: For $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ , the equilibrium concentrations in a $2.0\ \text{L}$ vessel at a given temperature are $[\text{N}_2] = 0.50\ \text{mol L}^{-1}$ , $[\text{H}_2] = 1.5\ \text{mol L}^{-1}$ , $[\text{NH}_3] = 0.20\ \text{mol L}^{-1}$ . The vessel is suddenly compressed to $1.0\ \text{L}$ (so $x = 0.5$ ) at constant temperature. Using the Step 3 result, find $Q_c$ immediately after compression and confirm the predicted direction of shift.

Step 1. Identify $K_c$ before disturbance: since the system starts at equilibrium, $K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \dfrac{(0.20)^2}{(0.50)(1.5)^3} = \dfrac{0.040}{1.6875} = 0.0237$ .

Step 2. Find $\Delta n_g$ : product side has $2$ mol gas, reactant side has $1+3=4$ mol gas, so $\Delta n_g = 2-4 = -2$ .

Step 3. Apply the derivation's result with $x = 0.5$ : $Q_{c,\text{new}} = K_c\left(\dfrac{1}{0.5}\right)^{-2} = K_c \times 2^{-2} = K_c/4 = 0.0237/4 = 0.00593$ .

Step 4. Compare: $Q_{c,\text{new}} = 0.00593 < K_c = 0.0237$ , so the reaction runs forward, consuming $\text{N}_2$ and $\text{H}_2$ to form more $\text{NH}_3$ , exactly as Le Chatelier's principle predicts for the side with fewer gas moles ( $2$ versus $4$ ). Compression drives the ammonia equilibrium forward, and $Q_c$ falls to one-quarter of $K_c$ the instant the volume halves, before the system reacts back toward $Q_c = K_c$ .

Competitive stretch

If instead the vessel were expanded to $4.0\ \text{L}$ ( $x = 2$ ) at constant temperature, $Q_{c,\text{new}} = K_c(1/2)^{-2} = K_c \times 4 = 0.0949$ , which is greater than $K_c$ , so the reaction shifts backward, decomposing some $\text{NH}_3$ back to $\text{N}_2$ and $\text{H}_2$ — consistent with expansion favouring the side with more gas moles.

Common traps

Believing a catalyst changes equilibrium yield — it only changes how fast equilibrium is reached; the position and $K$ are unaffected.
Applying pressure/volume reasoning when $\Delta n_g = 0$ — with equal gas moles on both sides, pressure and volume changes cause no shift at all.
Forgetting that only temperature changes the actual value of $K$ — concentration, pressure, and volume changes shift the equilibrium position but leave $K$ unchanged.
Mishandling inert-gas addition — no effect at constant volume, but a real shift at constant pressure, because the container must expand.

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Why doesn't adding a catalyst increase the equilibrium yield if it speeds up the reaction?

A catalyst provides an alternative pathway with lower activation energy, but that lower barrier applies equally to the forward and reverse reactions. Both rates increase by the same factor, so the ratio of rate constants — which fixes $K$ — never changes. The system reaches the same equilibrium position it would have reached anyway, just faster. Yield is set entirely by $K$ , temperature, and the imposed conditions, never by the catalyst.

If I add an inert gas like argon to an equilibrium mixture, why does it matter whether volume or pressure is held constant?

At constant volume, adding argon raises the total pressure, but every reacting species still occupies the same volume with the same moles, so its own concentration and partial pressure are unchanged — $Q_c$ and $Q_p$ stay exactly where they were, and there is no shift. At constant pressure, the container must expand to make room for the extra argon, which dilutes every reacting species — this is mathematically identical to lowering the pressure by expansion, so equilibrium shifts toward the side with more gas moles.

Why does the Haber process use high pressure but only a moderate temperature, if high temperature also speeds up the reaction?

Ammonia synthesis is exothermic, so Le Chatelier's principle says high temperature actually shifts equilibrium backward, lowering both $K$ and the equilibrium yield of $\text{NH}_3$ — even though it does speed up the rate of reaching whatever equilibrium is reached. High pressure, by contrast, shifts equilibrium forward (toward the side with fewer gas moles) and raises the rate too, so pressure is pushed as high as engineering allows. Temperature is kept only moderate, and a catalyst is added, to reach an acceptable rate without sacrificing too much yield — a compromise between rate and equilibrium position, not a single-factor optimisation.

Factors Affecting Equilibria

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads