1. Dynamic Equilibrium …going back and forth… …at the same time… …at the same rate…

2. What happens in a reversible reaction? Consider: aW + bX ↔ cY + dZ Where W reacts with X to produce Y and Z a, b, c, and d are the coefficients of the balanced equation

3. What happens in a reversible reaction? Consider: aW + bX ↔ cY + dZ 1) W is mixed with X and begins to react quickly W and X are at maximum concentrations Y and Z are not present at the beginning

4. What happens in a reversible reaction? Consider: aW + bX ↔ cY + dZ 2) The “forward” reaction (W + X) continues, but is slowing 3) When enough product is formed, the “reverse” reaction begins W and X concentrations are decreasing Y and Z concentrations are increasing

5. What happens in a reversible reaction? Consider: aW + bX ↔ cY + dZ When the rate of the forward reaction is equal to the rate of the reverse reaction, the system has reached “dynamic equilibrium” Rate FWD = Rate REV

6. Dynamic Equilibrium The reaction does not stop! Products are still being formed Products are still combining to reform the reactants Nothing “appears” to be happening Concentrations stop changing Color changes cease, etc…

7. Graphing R  P time [conc.] Red = reactants Blue = products

8. Dynamic Equilibrium The only thing “equal” about equilibrium are the rates of the forward and reverse reactions Equilibrium might be reached when there is mostly products, mostly reactants, or maybe a 50/50 mix of both

9. Dynamic Equilibrium Note: the initial amounts of reactants and products do not matter Example: N 2 O 4(g) ↔ 2 NO 2(g) Start with only [N 2 O 4 ] ….. at eq., [N 2 O 4 ] = 0.83M and [NO 2 ] = 0.33M Start with only [NO 2 ]…. at eq., [N 2 O 4 ] = 0.83M and [NO 2 ] = 0.33M

10. Dynamic Equilibrium Quantitatively the situation at equilibrium can be expressed using the reactant and product concentrations in the “mass-action expression” Consider: aW + bX ↔ cY + dZ

11. Dynamic Equilibrium Consider: aW + bX ↔ cY + dZ At equilibrium, because the concentrations stop changing, Q becomes constant and is replaced by K – the equilibrium constant.

12. Dynamic Equilibrium Note : 1) [products] on top and [reactants] on bottom 2) Coefficients become exponents

13. Dynamic Equilibrium K = the “equilibrium constant” K is unitless K is temperature dependent as long as the temperature is constant, so is K for a reversible reaction

14. Dynamic Equilibrium Note : aW (aq) + bX (s) ↔ cY (aq) + dZ (l) Pure substances (solids, liquids) do not have a changeable [molarity] and so drop out of the equilibrium expression

15. K – the equilibrium constant The size of K tells us something about the equilibrium “position” i.e. what are the concentrations of the reactants and products at equilibrium? Because the products are in the numerator, the larger the K value, the more products that are present at equilibrium and the fewer reactants that remain.

16. K – the equilibrium constant General rule: K<10 -4 the equilibrium mixture is mostly reactants The reaction does not “proceed” very far forward in order to reach equilibrium The smaller K is, the fewer products formed

17. K – the equilibrium constant General rule: 10 -4 <K<10 4 the equilibrium mixture has substantial amounts of reactants and products Not necessarily a “50/50” mix, but reasonably similar amounts of reactants and products

18. K – the equilibrium constant General rule: K>10 4 the equilibrium mixture is mostly products The larger K gets, the more the forward reaction “goes to completion”

19. Dynamic Equilibrium …going back and forth… …at the same time… …at the same rate…

20. Dynamic Equilibrium …going back and forth… …at the same time… …at the same rate…

21. LeChatelier’s Principle If a system at equilibrium is disturbed it will respond in the direction that counteracts the disturbance and re- establishes equilibrium Disturbed(?) 1. add/remove a chemical 2. change temperature 3. change pressure (gases) 4. add/remove catalyst

22. 1. adding/removing a chemical If you add a chemical, the system tries to “remove” it This is done by reacting it away This uses up the chemicals on its “side” of the equation and making more of the chemicals on the other “side” Equilibrium is re-established (Q =K), but the individual concentrations are different

23. 1. adding/removing a chemical Consider: A + B ↔ C + D If you add more A… The system tries to remove it by reacting it away, which makes more products [C]  [D]  [B] ↓ It is said the equilibrium has “shifted to the right” or “shifted towards the products”

24. 1. adding/removing a chemical Consider: A + B ↔ C + D If you add more C… The system tries to remove it by reacting it away, which makes more reactants [A]  [B]  [D] ↓ It is said the equilibrium has “shifted to the left” or “shifted towards the reactants”

25. 1. adding/removing a chemical Consider: A + B ↔ C + D If you remove some B… The system tries to replace it by reacting to make more of it (and whatever else is on its side of the equation) [A]  [C] ↓ [D] ↓ It is said the equilibrium has “shifted to the left” or “shifted towards the reactants”

26. 1. adding/removing a chemical Consider: A + B ↔ C + D If you remove some D… The system tries to replace it by reacting to make more of it (and whatever else is on its side of the equation) [C]  [A], [B] ↓ The reaction is driven forward in this case, or towards the products

27. Just how does one “remove” a chemical? Note: [D] is the concentration of D Solids do not have a molarity because they are not dissolved into anything If one product in an aqueous system is a solid, the solid is called a “precipitate” and is not in the equilibrium mixture This “drives” the reaction forward Double replacements

28. Just how does one “remove” a chemical? Same for gases in an open container They can bubble out of the mixture (leave) Ex: opening a soda bottle H 2 CO 3(aq) ↔ H 2 O (l) + CO 2(g) Ex: Mg (s) + 2 HCl (aq) ↔ H 2(g) + MgCl 2(aq) If the container is open, the reaction just keeps going forward

29. 2. Changing the volume Remember Boyle’s Law Changing the volume of a container of gases changes their pressure as well Inverse relationship If V↓, P  If V , P↓

30. 2. Changing the volume If V↓, P  The equilibrium will shift to try to make the P↓ How is this done? Shift to whichever side has less gas Fewer moles of a gas Coefficients in the balanced equation Less gas means lower pressure

31. 2. Changing the volume Ex: N 2(g) + 3 H 2(g) ↔ 2 NH 3(g) 4 moles of gas in the reactants, 2 in products If V↓, P  …the system will try to make P↓ by shifting to the products (less gas) Every time the reaction proceeds forward, 4 moles of gas becomes 2…which means the P↓ Vice versa if V 

32. 3. Changing the temperature Consider : A + B ↔ C + D + Heat For this system… The forward reaction is exothermic The reverse reaction is endothermic Treat heat as if it were a substance being added or removed Add heat, equilibrium shifts away from the side with heat [A],[B]  [C],[D]↓ Remove heat, equilibrium shifts toward the side with heat [A],[B]↓ [C],[D] 

33. 4. Catalytic effect Adding or removing a catalyst has no effect on the value of K The activation energy is lowered for the forward and the reverse reaction, and they both speed up by the same amount, so Rate FWD still = Rate REV If not at equilibrium, it will be reached quicker is a catalyst is used.

34. LeChatelier’s Principle If a system at equilibrium is disturbed it will respond in the direction that counteracts the disturbance and re- establishes equilibrium Disturbed(?) 1. add/remove a chemical 2. change temperature 3. change pressure (gases) 4. add/remove catalyst

35. Now…lets try it out… Ex: PCl 3(g) + Cl 2(g) ↔ PCl 5(g) ΔH= -88kJ How will [Cl 2 ] at equilibrium be changed by… Adding some PCl 3 ? System will try to react the PCl 3 away More products are formed To do this, more Cl 2 is consumed [Cl 2 ] ↓

36. Now…lets try it out… Ex: PCl 3(g) + Cl 2(g) ↔ PCl 5(g) ΔH= -88kJ How will [Cl 2 ] at equilibrium be changed by… Adding some PCl 5 ? The system will try to react it away More reactants are formed [Cl 2 ] 

37. Now…lets try it out… Ex: PCl 3(g) + Cl 2(g) ↔ PCl 5(g) ΔH= -88kJ How will [Cl 2 ] at equilibrium be changed by… Increasing the temperature? Heat is added, the system shifts away from the side with heat ΔH = negative, so heat is a product System shifts towards the reactants [Cl 2 ] 

38. Now…lets try it out… Ex: PCl 3(g) + Cl 2(g) ↔ PCl 5(g) ΔH= -88kJ How will [Cl 2 ] at equilibrium be changed by… Decreasing the volume? If V↓, P  System shifts towards the side with less gas to make P↓ Product side has fewer moles of gas [Cl 2 ] ↓