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Concept 01 / 08Equilibrium

Le Chatelier's Principle: Effect of Concentration, Pressure, Temperature and Catalysts

Le Chatelier's principle applied to changes in concentration, pressure or volume, temperature and addition of inert gas, the effect of a catalyst, why only temperature changes K, and industrial applications to the Haber and Contact processes.

Figure 1
Le Chatelier's principle. If a system at equilibrium is disturbed by a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to counteract the change. Concentration. Adding a reactant (or removing a product) makes Q<KQ < K; the reaction proceeds forward until Q=KQ = K again. Adding a product shifts it backward. KK itself does not change. In the figure, adding H2\mathrm{H_{2}} to an ammonia equilibrium consumes some of it together with N2\mathrm{N_{2}} and forms more NH3\mathrm{NH_{3}}. Removing a product continuously (for example CO2\mathrm{CO_{2}} escaping from a lime kiln, or water removed in esterification) drives a reaction to completion. Pressure (by changing volume). Compressing a gaseous equilibrium shifts it towards fewer moles of gas; expanding shifts it towards more moles. If Δng=0\Delta n_{g} = 0, pressure has no effect (H2+I2⇌2HI\mathrm{H_{2} + I_{2} \rightleftharpoons 2HI}). For N2+3H2⇌2NH3\mathrm{N_{2} + 3H_{2} \rightleftharpoons 2NH_{3}} (4→24 \rightarrow 2 moles), high pressure favours ammonia. Solids and liquids are hardly affected. Inert gas. Added at constant volume: partial pressures of the reacting gases are unchanged, so there is no shift. Added at constant total pressure: the volume increases, which acts like a decrease in pressure, and the equilibrium shifts towards more moles of gas. Temperature. Temperature is the only factor that changes KK. Raising the temperature favours the endothermic direction; lowering it favours the exothermic direction. For an exothermic forward reaction (ΔH<0\Delta H < 0), KK decreases as temperature rises; for an endothermic one, KK increases. Quantitatively (van 't Hoff equation): log⁡K2K1=ΔH∘2.303R(1T1−1T2)\log\frac{K_{2}}{K_{1}} = \frac{\Delta H^{\circ}}{2.303R}\left(\frac{1}{T_{1}} - \frac{1}{T_{2}}\right). Example: N2O4\mathrm{N_{2}O_{4}} (colourless) ⇌ 2NO2\mathrm{2NO_{2}} (brown) is endothermic; the mixture turns darker brown on heating. Catalyst. A catalyst speeds up the forward and backward reactions equally, by lowering the activation energy of both. Equilibrium is reached faster, but its position and KK are unchanged. Industrial applications. Haber process, N2+3H2⇌2NH3\mathrm{N_{2} + 3H_{2} \rightleftharpoons 2NH_{3}}, ΔH=−92.4 kJ mol−1\Delta H = -92.4\,\mathrm{kJ\,mol^{-1}}: high pressure (about 200 atm200\,\mathrm{atm}) favours ammonia; low temperature favours it too but makes the reaction too slow, so a compromise of about 700 K700\,\mathrm{K} with an iron catalyst (molybdenum promoter) is used, and ammonia is removed by liquefaction. Contact process, 2SO2+O2⇌2SO3\mathrm{2SO_{2} + O_{2} \rightleftharpoons 2SO_{3}}, exothermic: excess air, moderate temperature (about 720 K720\,\mathrm{K}), V2O5\mathrm{V_{2}O_{5}} catalyst, pressure of 11 to 2 atm2\,\mathrm{atm} (conversion is already high). Key Takeaways: - Concentration and pressure changes shift the position but not KK; temperature changes KK. - Heating favours the endothermic direction. - Compression favours fewer gas moles; no effect if Δng=0\Delta n_{g} = 0. - Inert gas at constant VV: no effect; at constant PP: shift towards more gas moles. - Catalysts do not shift equilibrium. - Pitfall: melting of ice (Vwater<ViceV_{\text{water}} < V_{\text{ice}}) is favoured by pressure, which is why ice melts under pressure; the principle applies to physical equilibria too. Brain-Teaser 1: For PCl5(g)⇌PCl3(g)+Cl2(g)\mathrm{PCl_{5}(g) \rightleftharpoons PCl_{3}(g) + Cl_{2}(g)}, ΔH>0\Delta H > 0, predict the effect of (a) raising temperature, (b) compressing the mixture, (c) adding Cl2\mathrm{Cl_{2}}, (d) adding argon at constant pressure. Solution: (a) Forward (endothermic), KK increases. (b) Backward, towards fewer moles (11 instead of 22). (c) Backward, KK unchanged. (d) Volume increases, so forward, towards more moles. Brain-Teaser 2: Why does the yield of ammonia fall at higher temperature, yet industrial plants do not operate at room temperature? Solution: The reaction is exothermic, so KK decreases with temperature and the equilibrium yield falls. But at low temperatures the rate is impractically slow even with a catalyst. About 700 K700\,\mathrm{K} is a compromise between yield and rate, compensated by high pressure and recycling of unreacted gases. Brain-Teaser 3: KpK_{p} of a reaction doubles when the temperature rises from 300 K300\,\mathrm{K} to 310 K310\,\mathrm{K}. Estimate ΔH∘\Delta H^{\circ}. Solution: log⁡2=ΔH∘2.303×8.314(1300−1310)\log 2 = \frac{\Delta H^{\circ}}{2.303 \times 8.314}\left(\frac{1}{300} - \frac{1}{310}\right). 1300−1310=1.075×10−4\frac{1}{300} - \frac{1}{310} = 1.075 \times 10^{-4}. ΔH∘=0.301×19.151.075×10−4=53.6 kJ mol−1\Delta H^{\circ} = \frac{0.301 \times 19.15}{1.075 \times 10^{-4}} = 53.6\,\mathrm{kJ\,mol^{-1}} (positive, since KK increased with TT).