# Factors that Affect Equilibrium > Le Chatelier's principle, and how changing concentration, pressure, volume and temperature, or adding a catalyst, shifts an equilibrium. Includes reading graphs and collision theory. From HSC Chemistry on WhizzNotes, free study notes for NSW students: https://whizznotes.com/chemistry/factors-affecting-equilibrium/. Updated 3 November 2026. Year 11 and Year 12 boxes go deeper than the Year 10 text around them. ### Story time: The seesaw in the playground Two children sit on a seesaw, perfectly balanced. A third child jumps on one end. The seesaw tips, and the children shuffle along until it balances again, but in a new position. In 1884, the French chemist Henri Le Chatelier noticed that chemical equilibria behave the same way. Push on a balanced reaction and it shifts to push back, until it settles at a new balance. His rule lets chemists make more ammonia for fertiliser, more sulfuric acid, and even explains why your blood stays at the right pH. **In short:** Le Chatelier's principle says that when an equilibrium is disturbed, it shifts to partly undo the change. Adding a reactant shifts it towards products, higher pressure shifts it to the side with fewer gas moles, heating shifts it in the endothermic direction, and a catalyst causes no shift. ## Le Chatelier's principle > If a system at equilibrium is disturbed, the system adjusts to **partly undo** the change, and a new equilibrium is reached. "Shift right" (or "shift forwards") means more products are made. "Shift left" means more reactants are made. The system only **partly** undoes the change. It never gets all the way back to where it started. ## Changing concentration - **Add** a reactant: the system uses some of it up, so it **shifts right**. - **Remove** a product: the system makes more of it, so it **shifts right**. - **Add** a product, or **remove** a reactant: it **shifts left**. ### Worked example: Iron thiocyanate $$\mathrm{Fe^{3+}(aq)} + \mathrm{SCN^-(aq)} \rightleftharpoons \mathrm{FeSCN^{2+}(aq)}$$ Pale yellow on the left, blood red on the right. - Add more $\mathrm{Fe^{3+}}$: shifts right, solution turns **redder**. - Add $\mathrm{Ag^+}$, which removes $\mathrm{SCN^-}$ as a solid: shifts left, solution turns **paler**. Ions can be removed by making a **precipitate** or by reacting them. Adding $\mathrm{OH^-}$ removes $\mathrm{H^+}$ (they make water). This is a common trick in exam questions. For example, in $\mathrm{Cr_2O_7^{2-}} + \mathrm{H_2O} \rightleftharpoons \mathrm{2CrO_4^{2-}} + \mathrm{2H^+}$ (orange to yellow), adding $\mathrm{NaOH}$ removes $\mathrm{H^+}$, so it shifts right and turns **yellow**. **Adding water to a solution** dilutes every ion. The system shifts to the side with **more dissolved particles**. For the iron thiocyanate equilibrium (2 ions on the left, 1 on the right), it shifts left and the red fades. **Solids and pure liquids** do not appear in the equilibrium expression. Adding more solid carbon or more liquid water (as the solvent) does **not** shift the equilibrium. ## Changing pressure or volume (gases) Squash a gas mixture into a **smaller volume** and its pressure goes up. The system shifts to the side with **fewer gas moles**, which lowers the pressure. $$\mathrm{N_2(g)} + \mathrm{3H_2(g)} \rightleftharpoons \mathrm{2NH_3(g)}$$ Left side: 4 moles of gas. Right side: 2 moles. Higher pressure shifts right and makes more ammonia. - Increase the **volume**: pressure drops, so it shifts to the side with **more** gas moles. - If both sides have the **same** number of gas moles (like $\mathrm{H_2} + \mathrm{I_2} \rightleftharpoons \mathrm{2HI}$), pressure has **no effect** on the position. - Adding an **inert gas** at constant volume does nothing, because the concentrations of the reacting gases do not change. ### Worked example: Brown gas in a syringe $\mathrm{2NO_2(g)} \rightleftharpoons \mathrm{N_2O_4(g)}$: brown on the left, colourless on the right. Push the plunger in. The gas first looks **darker** (the same brown gas is packed tighter). Then the system shifts right to fewer moles, and the colour **fades a little**. It ends darker than at the start, but lighter than straight after the push. Inject more colourless $\mathrm{N_2O_4}$ instead, and the system shifts **left** to use some of it up, so the brown colour slowly gets **darker**. ## Changing temperature Treat heat like a chemical in the equation. - **Exothermic** forward reaction ($\Delta H < 0$): heat is a "product". Heating shifts **left**. Cooling shifts **right**. - **Endothermic** forward reaction ($\Delta H > 0$): heat is a "reactant". Heating shifts **right**. Temperature is the **only** change that alters the value of the equilibrium constant $K$ (next chapter). ### Worked example: Cobalt colours $$\mathrm{[Co(H_2O)_6]^{2+}(aq)} + \mathrm{4Cl^-(aq)} \rightleftharpoons \mathrm{[CoCl_4]^{2-}(aq)} + \mathrm{6H_2O(l)} \quad \Delta H > 0$$ Pink on the left, blue on the right. Warm it and it turns **blue** (the endothermic way). Cool it in ice and it turns **pink**. Add concentrated $\mathrm{HCl}$ (more $\mathrm{Cl^-}$) and it turns blue. ## Adding a catalyst A catalyst speeds up the forward and reverse reactions **equally**. The equilibrium is reached **faster**, but the position and amounts do **not** change. ## Summary table | Change | Shift | Value of $K$ | |---|---|---| | Add reactant | Right | Same | | Remove product | Right | Same | | Increase pressure (smaller volume) | To fewer gas moles | Same | | Increase temperature, exothermic | Left | Decreases | | Increase temperature, endothermic | Right | Increases | | Add catalyst | No shift | Same | ## Reading concentration-time graphs These graphs often appear in exams. Look at **how** the lines change at the moment of the disturbance: - **One line jumps** up or down suddenly, then the others change smoothly: a chemical was **added or removed**. - **All lines jump** at once (by the same factor): the **volume changed**. - **No line jumps**, but all start changing smoothly: the **temperature changed**. ## Industrial chemistry: the Haber process Ammonia is made by $\mathrm{N_2} + \mathrm{3H_2} \rightleftharpoons \mathrm{2NH_3}$, $\Delta H = -92$ kJ. - **High pressure** (about 200 atm) shifts right. Very high pressure is costly and dangerous. - **Low temperature** gives the best yield, but the reaction would be far too slow. A **compromise** of about 450 °C is used. - An **iron catalyst** speeds it up. - Ammonia is **removed** as it forms, so the reaction keeps shifting right. **Mind map** ```text Le Chatelier Concentration Add reactant: right Remove product: right Solids: no effect Pressure Fewer gas moles side Equal moles: no effect Inert gas: no effect Temperature Exo: heat shifts left Endo: heat shifts right Changes K Catalyst Faster, no shift Graphs One jumps: added All jump: volume Smooth: temperature ``` ### Year 11: Collision theory explains the shift Collision theory explains **why** the system shifts. Add more reactant. Reactant particles now collide more often, so the **forward rate** jumps up. The forward reaction is now faster than the reverse, so products build up. As products build up, the reverse rate rises, until the two rates are equal again. Raise the temperature. Particles move faster and more collisions have the activation energy, so **both** rates rise. But the endothermic direction has the higher activation energy, so its rate rises **more**. The system shifts in the endothermic direction. On a **rate-time graph**, a concentration change makes **one** rate line jump. A temperature or volume change makes **both** lines jump, by different amounts. ### Year 12: Equilibria in the body and the ocean **Blood.** Carbon dioxide dissolves in blood: $$\mathrm{CO_2(g)} + \mathrm{H_2O(l)} \rightleftharpoons \mathrm{H_2CO_3(aq)} \rightleftharpoons \mathrm{H^+(aq)} + \mathrm{HCO_3^-(aq)}$$ Breathing too fast (hyperventilating) removes $\mathrm{CO_2}$. The equilibria shift **left**, $[\mathrm{H^+}]$ falls, and blood pH rises. **Ocean acidification.** More $\mathrm{CO_2}$ in the air shifts the same equilibria **right**, making more $\mathrm{H^+}$. The $\mathrm{H^+}$ reacts with carbonate ions, $\mathrm{H^+} + \mathrm{CO_3^{2-}} \rightarrow \mathrm{HCO_3^-}$, so there is less carbonate for corals and shellfish to build their shells. **Volume change and Q.** For $\mathrm{N_2O_4} \rightleftharpoons \mathrm{2NO_2}$, $K = \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]}$. Doubling the volume halves every concentration, so $Q = \dfrac{(x/2)^2}{y/2} = \dfrac{K}{2}$. Since $Q < K$, the system shifts right, towards more gas moles, as Le Chatelier predicts. ## Quick summary - Le Chatelier: a disturbed equilibrium shifts to **partly undo** the change. - Add reactant or remove product: shift right. Solids and pure liquids: no effect. - Higher pressure (smaller volume): shift to the side with fewer gas moles. - Heating shifts in the **endothermic** direction and changes $K$. - A catalyst gives no shift; equilibrium is just reached faster. - On graphs: one line jumps = chemical added; all jump = volume change; smooth change = temperature. ## Quiz Try these questions from the chapter quiz. 1. Le Chatelier's principle says a disturbed equilibrium will: - A. Stop reacting - B. Shift to partly undo the change - C. Shift to make the change bigger - D. Return exactly to where it started - Answer: B. Shift to partly undo the change. The system pushes back against the change, but only partly. 2. Adding more reactant to an equilibrium mixture shifts it: - A. To the right - B. To the left - C. Not at all - D. Right, then back to exactly the start - Answer: A. To the right. The system uses up some of the added reactant. 3. For $\mathrm{N_2(g)} + \mathrm{3H_2(g)} \rightleftharpoons \mathrm{2NH_3(g)}$, increasing the pressure: - A. Shifts right - B. Shifts left - C. Has no effect - D. Stops the reaction - Answer: A. Shifts right. The right side has fewer gas moles (2 compared with 4). The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/chemistry/factors-affecting-equilibrium/#quiz