1. Factors that Affect Equilibrium Concentrations!

2. 2 Le Chatalier’s Principle The first person to study and comment on factors that change equilibrium concentrations was Henry Louis Le Chatalier. Le Chatalier’s Principle states that if a stress is applied to a system initially at equilibrium, the reaction will shift in the direction that minimizes the stress and brings the system back into a NEW equilibrium.

3. 3 Le Chatalier’s Principle He discovered that three factors can be changed to affect the equilibrium of a reaction 1) The concentrations of the chemicals involved. 2) The pressure and/or volume of the system. 3) The temperature of the system. This will help us in determining how a reaction will proceed as it attempts to reach equilibrium.

4. 4 3 H 2 + N 2  2 NH 3 Changes in Concentration The system is at equilibrium. How do I know this? *Concentrations remain the same. Why are they constant? *Rates of reaction are the same. What would happen if we pumped in more N 2 gas? Would the system stay in equilibrium? What would happen? How would the reaction shift?

5. 5 3 H 2 + N 2  2 NH 3 Changes in Concentration An increase in the nitrogen would make the reactants side increase. To relieve the “stress” more products would form. * Shift to the right

6. 6 In the previous slide when the reaction was at equilibrium equilibrium constant was: But when we added more nitrogen (a “stress”), the ratio changed because the concentrations changed! (Reaction Quotient = Q) KcKc QcQc KcKc

7. 7 The reaction quotient is less than the equilibrium constant, meaning the reaction must move from left to right to re-establish equilibrium. Q < K  0.0988 < 0.296  More Products! The system reacted to minimize the stress by reducing the amount of N 2 and H 2 in order to reach a new equilibrium!

8. 8 If we increase the concentration of a reactant, the reaction proceeds to the right, creating more products and returns to equilibrium. If we increase the concentration of a product, the reaction proceeds to the left creating more reactants and returns to equilibrium. CO (g) + H 2 O (g)  CO 2 (g) + H 2 (g) How will the reaction shift if we: a) Adding CO b) Adding CO 2 c) Removing H 2 O

9. 9 Effect of changes in pressure and volume Pressure and Volume are always related together when discussing the shifting of a reaction. A change in one causes the other, but what is their relationship? Pressure and Volume are inversely related. Meaning an increase in Volume causes a decrease in Pressure.

10. 10 N 2 (g) + 3 H 2 (g)  2 NH 3 (g) Say we decrease the volume of the ammonia gas in the above reaction. The stress on the equilibrium is an increase in pressure. Le Chatalier’s Principle tells us the system will shift so that the overall pressure will decrease. How would it achieve that, which way will it go? The reaction will favor the direction that has the fewest moles of gases. The fewer molecules, the fewer the collisions, thus lowering the pressure. Which way will that be in this reaction? The reaction will shift to the right producing more Ammonia!

11. 11 Pressure Change Through a Change in Volume An increase in the pressure, (caused by decreasing the volume) causes the reaction to shift to the side with fewer total moles of gas. A decrease in the pressure, (caused by increasing the volume) causes the reaction to shift to the side with greater total moles of gas.

12. 12 2 SO 2 (g) + O 2 (g)  2 SO 3 (g) Based on the above reaction, predict change in concentrations of the reactants and products if the volume changes from 10.0 L to 1.0 L. What happens to the Volume? What happens to the Pressure? Which way will the reaction shift?

13. 13 2 SO 2 (g) + O 2 (g)  2 SO 3 (g)

14. 14 One Thing to Note: If the total number of moles of gas on the reactants side EQUALS the moles on the product side then what would be true? H 2 (g) + l 2 (g)  2 HI(g) When the total moles are the same on both sides, then a pressure change due to volume change will not affect the equilibrium system. The concentrations will remain the same.

15. 15 Other things to note We ALWAYS talked about pressure changes in terms of volume change. If we increase the pressure by adding inert gas to the system, no equilibrium shift will be seen because the partial pressures of the gases in the equilibrium system have not changed! In other words, there is no actual stress being placed on the equilibrium system

16. 16 Practice Problem How do the concentration of the products change when each of the following reactions are subject to an increase in pressure by decreasing the volume? a) CO (g) + H 2 O (g)  CO 2 (g) + H 2 (g) b) 2 CO (g)  C (s) + O 2 (g) c) N 2 O 4 (g)  2 NO 2 (g)

17. 17 Changes in temperature and equilibrium The reaction for the formation of ammonia. N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  H  = -92.2 kJ. When this reaction takes place it gives off heat energy. That’s what  H  represents. This is called an EXOTHERMIC Reaction.

18. 18 Changes in temperature and equilibrium For an EXOTHERMIC reaction, Heat is a product. A + B  C + D + HEAT Treat heat like any other compound. If we increase the temperature, we will increase the “concentration” of this product. How will the reaction respond? An increase in temperature will drive the reaction towards the left to balance out this stress. For EXOTHERMIC, an increase in temperature will favor the REACTANTS.

19. 19 Changes in temperature and equilibrium The reaction for the production of nitrogen monoxide: N 2 (g) + O 2 (g)  2 NO (g)  H  = 180.5 kJ When  H is positive, it means that heat is required to start the reaction and is a reactant. This is called ENDOTHERMIC!

20. 20 Changes in temperature and equilibrium For an ENDOTHERMIC reaction, Heat is a reactant. A + B + HEAT  C + D Treat heat like any other compound. If we increase the temperature, we will increase the “concentration” of this reactant?. How will the reaction respond? An increase in temperature will drive the reaction towards the right to balance out this stress. For ENDOTHERMIC, an increase in temperature will favor the PRODUCTS.

21. 21 Changes in temperature and equilibrium The exact opposite would be true of a decrease in temperature. EXOTHERMIC: A + B  C + D + HEAT A decrease in the “concentration” of heat will cause the reaction to favor the products. ENDOTHERMIC: A + B + HEAT  C + D A decrease in the “concentration” of heat will cause the reaction to favor the reactants.

22. 22 Problem When air is heated at very high temperatures in an engine, the air pollutant nitric oxide is produced by the reaction N 2 (g) + O 2 (g)  2 NO (g)  H  = 180.5 kJ How does the concentration of NO vary with an increase in temperature? What does the  H value tell us about the reaction? Does that mean temperature is a reactant or a product? ENDOTHERMIC, Heat is a reactant, so increase temp and the product [NO] will increase. Decrease the temperature and [NO] will decrease.

23. 23 Problem Calcination of limestone (decomposition of calcium carbonate) occurs through the following reaction: CaCO 3 (s)  CO 2 (g) + CaO (s) After we establish this equilibrium system at a given temperature, what will be the effect on equilibrium of a) Increasing the volume of the container b) Increasing the pressure c) Decreasing [CO 2 ] d) If the reaction is determined to be exothermic, a decrease in temperature.

24. 24 Catalysis and equilibrium Since both the forward and reverse reactions pass through the same transition state, a catalyst reduces the activation energy for both the forward and reverse reactions, by the same amount. This increases the rates of both the forward and reverse reactions by the same amount!

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26. 26 Catalysis and equilibrium Another way to think of it is that a catalyst does not appear in the overall balanced equation for a reaction and therefore it won’t appear in the equilibrium constant equation meaning no change in the equilibrium constant will be seen when you add a catalyst.