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Published byMoses Walsh Modified over 9 years ago
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Le Chatelier’s Principle A reaction at equilibrium, when “stressed,” will react to relieve the stress. (If you mess with it, it will work to return to its equilibrium ratio.)
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Le Chatelier’s Principle “Stress” includes changes to: concentration pressure temperature
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Le Chatelier’s Principle Questions will tell you HOW equilibrium was stressed. You answer with what the reaction does to FIX itself. The choices are: It will shift LEFT, RIGHT, or nothing will happen.
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Le Chatelier: Concentration If the concentration of a reactant is increased the equilibrium will shift in the direction that uses the reactants, so that the reactant concentration decreases. (The forward reaction is favored– more products will be made as a result.)
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Le Chatelier: Concentration 2SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g) If the SO 2 or O 2 concentration was increased: (i.e., we added more reactants) Equilibrium will shift right to decrease the concentration of reactants. (Some of the reactant we added gets used up and more SO 3 (product) is produced.)
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Le Chatelier: Concentration “Equilibrium will shift right” Let’s define that: When a new equilibrium is reached (when the rate of forward and reverse reactions become equal again), there will be more product than there was before. The equilibrium RATIO did not change.
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Le Chatelier: Concentration The forward reaction is also favored if the concentration of the product is decreased. If product is removed, equilibrium shifts to the right in order to: Decrease the amount of reactant (by using some up) Increase the amount of product (by making some more)
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Le Chatelier: Concentration Equilibrium shifts to the right when Concentration of reactants increases Concentration of products decreases The reverse is also true: Equilibrium shifts to the LEFT when Concentration of reactants decreases Concentration of products increases
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Le Chatelier: Concentration 2SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g) Adding more SO 2 (g) shifts it to the right The question might say “increasing [SO 2 ]” which implies that we’re measuring concentration. The question might say “increasing P SO2 ” which implies that we’re measuring concentration of a GAS. (Both mean the same thing.)
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Le Chatelier: Concentration Be careful! Solids (s) and liquids (l) cannot become more “concentrated.” There are only two options. You have some (concentration = 100%) Or you don’t (concentration = 0%) 2SO 2 (g) + O 2 (l) ⇌ 2 SO 3 (g) Adding more SO 2 (g) shifts it to the right But Adding more O 2 (l) makes no difference! (because it is liquid)
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Le Chatelier: Concentration Increasing or decreasing the amount of solids (s) or liquids (l) DOES NOT AFFECT equilibrium That means the answer is “no change” Increasing or decreasing the amount of gases (g) or aqueous solutions (aq) DOES AFFECT equilibrium. That means the answer is either “shifts left” or “shifts right.”
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Le Chatelier: Pressure Changing the pressure of the system This is indicated either by “pressure of the system increased/decreased” OR “volume of the container is increased/decreased” Making a container larger makes pressure DECREASE. Pressure and volume are opposites!
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Le Chatelier: Pressure Changing the pressure of the system If pressure is increased the equilibrium will shift to favor a decrease in pressure. If the pressure is decreased the equilibrium will shift to favor an increase in pressure.
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Le Chatelier: Pressure The reaction will “shift to favor a decrease/increase in pressure” 2SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g) Count the moles of gas on each side of the equilibrium arrow. Reactants: 3 moles of gas Products: 2 moles of gas (Coefficients MATTER!)
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Le Chatelier: Pressure 2SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g) (3 moles of gas > 2 moles of gas) Shifting left will increase pressure Shifting right will decrease pressure (If there aren’t any gases, it’s a tie, and nothing happens.)
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Le Chatelier: Pressure 2SO 2 (g) + O 2 (l) ⇌ 2 SO 3 (g) (2 moles of gas = 2 moles of gas) It’s a tie. Nothing happens.
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Le Chatelier: Temperature Changing the temperature of the system If the temperature is increased the equilibrium will shift to favor the reaction which will absorb heat. (The endothermic reaction is favored.)
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Le Chatelier: Temperature Changing the temperature of the system If the temperature is decreased the equilibrium will shift to favor the reaction which will release heat. (The exothermic reaction is favored.)
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Le Chatelier: Temperature N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g) ; ΔH= −92 kJ The reaction is exothermic. Energy is a product. An increase in temperature shifts this reaction to the left. A decrease in temperature shifts this reaction to the right.
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Exothermic vs Endothermic ΔH= N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g); ΔH= −92 kJ 2 NH 3 (g) ⇌ N 2 (g) + 3 H 2 (g); ΔH= 92 kJ Energy as part of the reaction: N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g) + energy N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g) + 92 kJ 2 NH 3 (g) + energy ⇌ N 2 (g) + 3 H 2 (g) 2 NH 3 (g) + 92 kJ ⇌ N 2 (g) + 3 H 2 (g)
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