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Le Chatelier's Principle. Use Le Chatelier’s Principle to explain the effects on the position of a system at equilibrium from:  Changing concentration.

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Presentation on theme: "Le Chatelier's Principle. Use Le Chatelier’s Principle to explain the effects on the position of a system at equilibrium from:  Changing concentration."— Presentation transcript:

1 Le Chatelier's Principle

2 Use Le Chatelier’s Principle to explain the effects on the position of a system at equilibrium from:  Changing concentration  Changing temperature  Changing pressure  Adding a catalyst

3 Le Chatelier's Principle (1884) When a system at equilibrium is subjected to a stress, the system will adjust so as to relieve the stress. Remember: K c value is constant. BEFORE the stress, and AFTER the reaction adjusts to the stress.

4 Types of Stress

5 1. Concentration stress Any change in concentration of products or reactants from equilibrium values. [increase] System will react by consuming the excess. [decrease] System will react by producing to replace the loss.

6 More C means increased rate of reverse reaction. K c = [C] [A][B] CBA+ K c = 1.35 Increase [C] - the reaction shifts left to use it up. Excess C used up until ratio of product to reactant concentrations is equal to K c once again.

7 K c = [C] [A][B] BC A+ K c = 1.35 Forward reaction rate increases (favored) and K c is reestablished. Removing a particle is like decreasing [ ]. Reaction will shift to replace the loss. Increase [B] - the reaction shifts right to use it up.

8 2 NO 2 (g) N 2 O 4 (g) car exhaust smog Huge spike indicates that [ ] was changed by adding more particles.

9 2 NO 2 (g) N 2 O 4 (g) car exhaust smog A huge spike indicates that [ ] was changed by removing particles.

10 2 NO 2 (g) N 2 O 4 (g) car exhaust smog

11 Temperature

12 The system relieves the stress by either replacing lost heat or consuming added heat. Reaction reestablishes new eqlbm (with new [ ]s) at new temperature – BUT also changes the K c. Consider heat a component of the system: Exothermic A  B (- ∆H ) Endothermic A  B (+ ∆H) HEAT + + HEAT 2. Temperature stress

13 Temperature increase / add heat Reaction shifts left to use up the heat. Endothermic reaction (reverse) favored. Temperature decrease / removing heat Reaction shifts right to produce more heat. Exothermic reaction (forward) favored. +heat AB + A B K c = [B] [A] K c = [B] [A]

14 ∆H = -58 kJ 2 NO 2 (g) N 2 O 4 (g) car exhaust smog

15 ∆H = -58 kJ 2 NO 2 (g) N 2 O 4 (g) car exhaust smog

16 Volume/Pressure

17 Changing the pressure of a system only affects those equilibria with gaseous reactants and/or products. 3. Volume stress Shifts to compensate pressure changes will effect all concentrations – BUT, K c value will return as equilibrium reestablishes. A + 2 B  C

18 A + 2 B C Volume decrease – (↑P ) A B B C Reaction shifts to decrease total number of gas particles – reduces pressure. C

19 A + 2 B C Volume increase– (↓P ) A B B C Reaction shifts to increase total number of gas particles – increase pressure. B B A

20 1. IF the size of the container is cut in half? Increases pressure  reduce pressure by reducing the number of molecules in the container. Reverse reaction favoured. The equilibrium shifts left. 2 NH 3(g) N 2(g) + 3 H 2(g)

21 2. IF the reaction chamber is increased in volume? Increasing volume, reduces pressure  increase pressure by increasing number of molecules in container. Reverse reaction is favoured. The equilibrium shifts left. 2 NO 2 (g) N 2 O 4 (g) car exhaust smog

22 Therefore, in response to pressure changes, the equilibrium position remains unchanged. H 2(g) + I 2(g) 2 HI (g) 3. Increase container OR increase the pressure? : 2 Pressure changes have NO effect on this eqlbm – same # of particles regardless of shift (stoich).

23 Factors (stresses) that do not affect Equilibrium Systems

24 Catalysts Lowers activation energy for both forward and reverse reaction equally. Equilibrium established more quickly, but position and ratios of concentrations will remain the same. K value remains the same.

25 Inert Gases (noble gases) Unreactive – are not part of a reaction, therefore can not affect [ ], pressure or volume of a equilibrium system. Catalysts, inert gases, pure solids or pure liquids do NOT appear in the mass action expression - so they cannot have an effect if altered.

26 Le Chatelier's AND life

27 Haemoglobin protein used to transport O 2 from lungs to body tissue. Lungs - [O 2 ] is high - forward reaction favored Haemoglobin binds to the excess O 2. Tissue - [CO 2 ] is high and [O 2 ] is low - reverse reaction favored. Hb releases O 2. Hb (aq) + O 2 (g)  HbO 2 (aq) Haemoglobin Production and Altitude

28 Hb (aq) + O 2 (g)  HbO 2 (aq) High altitudes - [O 2 ] is very low - reverse reaction favored. Hb release O 2, fewer Hb bind oxygen. Result in exaggerated lack of oxygen to the tissues, resulting in headache, nausea and fatigue. Over time, body adjusts by producing more haemoglobin molecules. Increases [Hb] in the blood stream shifts equilibrium right - more O 2 bound and transported to the tissue.

29 Appliance - NO energy - forward reaction favored Energy release to run appliance. Outlet (recharge) - high energy - reverse favored Reforming the reactants, storing the energy for use. Rechargable Batteries Lead-acid PbO 2 + Pb + 4 H SO 4 2-  2 PbSO H 2 O + energy Nickel-cadmium Cd + 2 NiO(OH) + 2 H 2 O  2 Ni(OH) + Cd(OH) 2 + energy Electrical energy (like heat) is written in the reaction.

30 THE HABER PROCESS

31 N 2(g) + 3H 2(g) 2NH 3(g) ΔH = kJ mol -1  high pressure  medium temperature - catayst  remove ammonia  high reactant concentrations


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