Le Chatelier's Principle This principle is used along with reaction kinetics to help industry predict product formation and maximize $$$ by manipulating the yield of the desired product.

Include: Interpret concentration versus time graphs. OUTCOME QUESTION(S): C12-4-09 LeCHATELIER’S PRINCIPLE Use Le Chatelier’s principle to predict and explain shifts in equilibrium. Include: Interpret concentration versus time graphs. Describe some practical applications of Le Chatelier’s principle. Vocabulary & Concepts

Le Chatelier's Principle (1884) When a system at equilibrium is subjected to a stress, the system will adjust to relieve the stress and return to equilibrium. Remember: Kc value is constant, BEFORE the stress and AFTER the reaction adjusts stress: A change in conditions that causes a related change in collision rates that upsets the reaction equilibrium.

1. Concentration stress A + B C Kc = Kc = 1.35 Stress: a change in concentration by adding or removing products or reactants. Adjustment: change in collision rates and redistribution of particles. Increase [C]: Kc = [C] [A][B] A + B C We say “shifts left” Increased rate of reverse collisions Kc = 1.35 Reverse reaction is favoured Re-establish ratio; same Kc

A + B C A + B C Kc = Kc = 1.35 Kc = Kc = 1.35 Increase [B]: [C] [A][B] We say “shifts right” Increased rate of forward collisions Forward reaction is favoured Re-establish ratio; same Kc Kc = 1.35 Decrease (removing) [A]: Kc = [C] [A][B] A + B C We say “shifts left” Decreased rate of forward collisions Reverse reaction is favoured Re-establish ratio; same Kc Kc = 1.35

2 NO2 (g) N2O4 (g) Notice equilibrium re-established after adjusting to the stress smog smog Spike indicates that concentration was added or removed.

2 NH3 (g) N2(g) + 3H2(g) H2(g) + I2(g) 2 HI(g) Which way will the system shift IF the concentration of ammonia is increased? 2 NH3 (g) N2(g) + 3H2(g) Increased rate of forward collisions “shifts right” Which way with the system shift IF the concentration of iodine is decreased? H2(g) + I2(g) 2 HI(g) Decreased rate of forward collisions “shifts left”

*Equilibrium constant changes with new temperature – NEW Kc* 2. Temperature stress Stress: a change in temperature by adding or removing heat. Adjustment: change in collision rate and redistribution of particles. Put the energy into the equation and treat the change like a concentration question Exothermic: A  B (- ∆H ) Endothermic: A  B (+ ∆H) + HEAT HEAT + *Equilibrium constant changes with new temperature – NEW Kc*

+ A B heat + A B heat = Kc = Kc Increase temperature: [B] [A] Increased rate of reverse collisions Endothermic reaction is favoured New ratio created; new Kc We say “shifts left” Decrease temperature: + A B heat = [B] [A] Kc Decreased rate of reverse collisions Exothermic reaction is favoured New ratio created; new Kc We say “shifts right”

1 2 NO2 + N2O4 heat 2 ∆H = –58 kJ smog smog A curved change indicates it is NOT a concentration stress (it could be either temperature or pressure/volume) To clarify which stress: Use concentrations to find the value of the Kc before(1) and after(2) stress (2) 1 2 (1) (1) A change in Kc value indicates temp stress (2)

2 NH3 (g) N2(g) + 3H2(g) H2(g) + I2(g) 2 HI(g) Which way with the system shift IF the temperature is decreased for this exothermic reaction? 2 NH3 (g) N2(g) + 3H2(g) + HEAT Decreased rate of reverse collisions “shifts right” Which way with the system shift IF the temperature is increased for this endothermic reaction? H2(g) + I2(g) 2 HI(g) HEAT + Increased rate of forward collisions “shifts right”

Remember: pressure changes only affect gases 3. Pressure stress Stress: a change in pressure by increasing or decreasing the volume of the system. Remember: pressure changes only affect gases Adjustment: change in collision rate and redistribution of particles. Equilibrium expressions for pressure would be calculated using partial pressure values. Kp value would still be constant before and after pressure stress = {Productsp} {Reactantsp} Kp

A + 2 B C B A C B C = Kp Increased Pressure (↓V ): 1 + 2 : 1 {Cp} 1 + 2 : 1 C B = {Cp} {Ap}{Bp} Kp C Increased rate of forward collisions We say “shifts right” Forward reaction is favoured Re-establish ratio; same Kp The reaction will shift to the side with fewer molecules (reduce the number of molecules, reduce the pressure)

A + 2 B C B A B B A B C = Kp Decrease Pressure (↑V ): 1 + 2 : 1 {Cp} 1 + 2 : 1 = {Cp} {Ap}{Bp} Kp B B A B C Decreased rate of forward collisions We say “shifts left” Reverse reaction is favoured Re-establish ratio; same Kp The reaction will shift to the side with more molecules (increase the number of molecules, increase the pressure)

Same number on both sides – no shift will affect pressure Which way with the system shift IF the size of the container is cut in half ? 2 NH3 (g) N2(g) + 3H2(g) Increased rate of reverse collisions “shifts left” Which way with the system shift IF the pressure is decreased? H2(g) + I2(g) 2 HI(g) 1 + 1 : 2 No change in collisions rates “no shift” Same number on both sides – no shift will affect pressure

Factors that do not affect Equilibrium Systems 1. Catalysts: Lowers activation energy for both forward and reverse reaction equally. Same equilibrium positions established, just more quickly 2. Inert gases: Unreactive – therefore can not affect equilibrium of a concentration-based equation. Catalysts, inert gases, pure solids, pure liquids do NOT appear in the Equilibrium expression – so are not a stress if altered

Include: Interpret concentration versus time graphs. CAN YOU / HAVE YOU? C12-4-09 LeCHATELIER’S PRINCIPLE Use Le Chatelier’s principle to predict and explain shifts in equilibrium. Include: Interpret concentration versus time graphs. Describe some practical applications of Le Chatelier’s principle. Vocabulary & Concepts