1. The Nature of the Equilibrium At equilibrium the rate of the forward reaction and the rate of the reverse reaction are equal This implies that all reactions are reversible This is true (theoretically), but not especially practical in many instances An example: Combustion of a hydocarbon to the products CO 2 and H 2 O goes in the forward direction very readily. The reverse reaction, CO 2 and H 2 O back to the hydrocarbon, is, for all practical purposes, impossible The ratio of the concentrations of product(s) to reactant(s) can be expressed mathematically to describe what is referred to as the “position of equilibrium”

2. The “Position of Equilibrium” The position of equilibrium can be expressed mathematically as the “equilibrium constant”, abbreviated ‘K’ aA + bB  cC + dD

3. ‘K’‘K’ Here are the rules: K is always concentration product(s) over concentration reactant(s) Solid and liquid reactants/products are omitted from the ‘K’ expression When a reaction occurs in water, the quantity of water on the reactant side vs. what is left the product side after the reaction occurs is basically the same This would be the equivalent of placing the same number in both the numerator and denominator of a fraction. The numbers would cancel anyway Recall that we dealt with solutions in this way. We regarded the density and specific heat of solutions as identical to that of water, because the solutions before, during, and after the reaction were mostly water anyway.

4. More Rules Concentrations are raised to the stoichiometric coefficient ‘K’ is particular to the reaction itself and also the temperature at which the reaction occurs

5. So…Who Cares About ‘K’ Anyhow? Well, we do, because it is useful in several ways: We can determine if a reaction is reactant-favored or product-favored by the relative quantities of products vs. reactants in ‘K’ When a reaction is product favored K>1, because the numerator will be numerically greater than the denominator When a reaction is reactant favored K<1, because the denominator will be numerically greater than the numerator We can calculate the quantities of reactants/products that are present at equilibrium

6. A Word About ‘Q’ This still refers to the ratio of concentration of product(s) to concentration of reactant(s) BUT…the reaction doesn’t have to be at equilibrium The value of ‘Q’ changes over time (can be greater than ‘K’ or less than ‘K’) When Q = K…that is something VERY special…proceed to the next slide to see…

7. A Graphical Representation of ‘Q’ and ‘K’ H 2 + I 2  2 HI The shaded regions of the graphs are referred to as the “kinetic regions”. The kinetics we studied apply to the rates of the changes of the concentrations of reactant(s) and product(s). ‘Q’ would be used to express the ratio of product concentrations to reactant concentrations at any given point in this region. The value of ‘Q’ changes over time. The unshaded portions are the “equilibrium regions”. The ratio of concentrations of product(s) to reactant(s) is not changing. ‘K’ describes this ratio, which remains constant. In these equilibrium regions Q = K.

8. More About Graphs The reaction depicted in the graph on the left is product-favored, because the concentration of the product, HI, increases as the concentrations of the reactants, H 2 and I 2, are decreasing. The reaction depicted in the graph on the right is reactant-favored, because the concentration of the product, HI, is decreasing as the concentrations of the reactants, H 2 and I 2, are increasing. Both these graphs depict the reaction: H 2 + I 2  2 HI In the graph on the left the reaction proceeds left to right (toward the products). In the graph on the right the reaction proceeds from right to left (toward the reactants).

9. Equilibrium calculations We can predict the direction of a reaction by calculating the reaction quotient. Reaction quotient, Q For the reaction: aA + bB eE + fF Q has the same form as K c with one important difference. Q can be for any set of concentrations, not just at equilibrium. Q = [E] e [F] f [A] a [B] b

10. Reaction quotient Any set of concentrations can be given and a Q calculated. By comparing Q to the K c value, we can predict the direction for the reaction. Q < K c Q < K c Net forward reaction will occur. Q = K c Q = K c No change, at equilibrium. Q > K c Q > K c Net reverse reaction will occur.

11. Reaction quotient example For an earlier example H 2 (g) + I 2 (g) 2HI (g) K c was determined to be 54 at 425.4 o C. If we had a mixture that contained the following, predict the direction of the reaction. [H 2 ] = 4.25 x 10 -3 M [I 2 ]= 3.97 x 10 -1 M [HI]= 9.83 x 10 -2 M

12. Reaction quotient example Q = = = 5.73 Since Q is < K c, the system is not in equilibrium and will proceed in the forward direction. [ HI ] 2 [ H 2 ] [ I 2 ] (9.83 x 10 -2 ) 2 (4.25 x 10 -3 )(3.97 x 10 -1 )

13. Finding Equilibrium When Q<K reaction must proceed in the products direction (forward) in order to reach equilibrium When Q>K reaction must proceed in the reactants direction (reverse) in order to reach equilibrium

14. Chemical equilibrium Homogeneous equilibria Equilibria that involve only a single phase.Examples. All species in the gas phase H 2 (g) + I 2 (g) 2HI (g) All species are in solution. HC 2 H 2 O 2 (aq) H + (aq) + C 2 H 3 O 2 - (aq)

15. ICE The equilibrium constant can be calculated if certain information is given Some equilibrium problems can be solved by inspection, but others require a more step-by- step approach We will place the given information into an ICE table We will then use the information in the ICE table to calculate ‘K’

16. Determining equilibrium constants Equilibrium constants can be found by experiment. If you know the initial concentrations of all of the reactants, you only need to measure the concentration of a single species at equilibrium to determine the K c value. Let’s consider again the following equilibrium: H 2 (g) + I 2 (g) 2HI (g)

17. Determining equilibrium constants H 2 (g) + I 2 (g) 2HI (g) Assume that we started with the following initial concentrations at 425.4 o C. H 2 (g) 0.005 00 M I 2 (g) 0.012 50 M HI (g) 0.000 00 M At equilibrium, we determine that the concentration of iodine is 0.007 72 M

18. Determining equilibrium constants The equilibrium expression for our system is: K c = Based on the chemical equation, we know the equilibrium concentrations of each species. I 2 (g)= the measured amount = 0.007 72 M That means that 0.004 78 mol I 2 reacted to produce HI in 1.00 L of solution. [HI] 2 [H 2 ] [I 2 ]

19. Determining equilibrium constants Kc = = = 54 [HI] 2 [H 2 ] [I 2 ] (0.009 56) 2 (0.000 22)(0.007 72)

20. ICE-ier Notice the one that uses K p The set-up of ‘K p ’ works exactly the same way as K c Keep in mind (for later) how to use the ideal gas law and Dalton’s law to calculate partial pressures

21. Partial pressure equilibrium constants At constant temperature, the pressure of a gas is proportional to its molarity. PV = nRT Remember, for an ideal gas: PV = nRT and molarity is: M = mole / liter or n/V so: P = R T M Where R is the gas law constant and T is the absolute temperature (Kelvin) Recall also…P A V=n A RT, when looking at the partial pressure and number of moles of gas “A” in a mixture of gases (Dalton’s Law of Partial Pressures)

22. Partial pressure equilibrium constants For equilibria that involves gases, partial pressures can be used instead of concentrations. aA (g) + bB (g) eE (g) + fF (g) K p  = K p is used when the partial pressures are expressed in units of atmospheres. p E e p F f p A a p B b

23. Partial pressure equilibrium constants In general, K p = K c, instead K p = K c (RT)  n g  n g is the number of moles of gaseous products minus the number of moles of gaseous reactants.  n g = (e + f) - (a + b)

24. Partial pressure equilibrium constants For the following equilibrium, K c = 1.10 x 10 7 at 700. o C. What is the K p ? 2H 2 (g) + S 2 (g) 2H 2 S (g) K p = K c (RT)  n g T = 700 + 273 = 973 K R= 0.0821  n g = ( 2 ) - ( 2 + 1) = -1 atm L mol K

25. Partial pressure equilibrium constants K p  = K c (RT)  n g = 1.10 x 10 7 (0.0821 )(973 K) = 1.378 x10 5 atm L mol K []

26. Using K to Determine Equilibrium Concentrations If you can to determine K from equilibrium concentrations, logic would have it that you can determine equilibrium concentrations from K. Observe…

27. Calculating equilibrium concentrations If the stoichiometry and K c for a reaction is known, calculating the equilibrium concentrations of all species is possible. Commonly, the initial concentrations are known. One of the concentrations is expressed as the variable x. All others are then expressed in terms of x.

28. Equilibrium calculation example A sample of COCl 2 is allowed to decompose. The value of K c for the equilibrium COCl 2 (g) CO (g) + Cl 2 (g) is 2.2 x 10 -10 at 100 o C. If the initial concentration of COCl 2 is 0.095M, what will be the equilibrium concentrations for each of the species involved?

29. Equilibrium calculation example COCl 2 (g)  CO (g) + Cl 2 (g) Initial conc., M0.0950.0000.000 Change in conc.- X + X + X due to reaction Equilibrium Concentration, M(0.095 -X) X X K c == [ CO ] [ Cl 2 ] [ COCl 2 ] X 2 (0.095 - X)

30. Equilibrium calculation example X 2 (0.095 - X) K c = 2.2 x 10 -10 = Rearrangement gives: X 2 + 2.2 x 10 -10 X - 2.09 x 10 -11 = 0 This is a quadratic equation. Fortunately, there is a straightforward equation for their solution

31. Quadratic equations An equation of the form a X 2 + b X + c = 0 Can be solved by using the following x = Only the positive root is meaningful in equilibrium problems. (Obviously you cannot have negative concentrations!) -b + b 2 - 4ac 2a

32. Equilibrium calculation example -b + b 2 - 4ac 2a 2.2 x 10 -10 2.09 x 10 -11 1X 2 + 2.2 x 10 -10 X - 2.09 x 10 -11 = 0 b c a b c X = - 2.2 x 10 -10 + [(2.2 x 10 -10 ) 2 - (4)(1)(- 2.09 x 10 -11 )] 1/2 2 X = 9.1 x 10 -6 M

33. Summary of method of calculating equilibrium concentrations Write an equation for the equilibrium. Write an equilibrium constant expression. Express all unknown concentrations in terms of a single variable such as x. Substitute the equilibrium concentrations in terms of the single variable in the equilibrium constant expression. Solve for x. Use the value of x to calculate equilibrium concentrations.

34. Being Disruptive Chemical reactions don’t get to hang up a “Do Not Disturb” sign and there are LOTS of ways to disturb/disrupt equilibrium The following are various “stresses” that can be placed on a system at equilibrium: LeChatlier’s Principle states:

35. LeChatlier’s Principle A change in any of the factors determining an equilibrium will cause the system to adjust to reduce/minimize the effect of the change Also sometimes stated: If a stress is placed on a system at equilibrium, the position of the equilibrium shifts so as to minimize the effect of the stress

36. Predicting shifts in equilibria Equilibrium concentrations are based on: The specific equilibrium The starting concentrations Other factors such as: Temperature Pressure Reaction specific conditions Altering conditions will stress a system, resulting in an equilibrium shift.

37. Changes in concentration Changes in concentration do not change the value of the equilibrium constant at constant temperature. When a material is added to a system in equilibrium, the equilibrium will shift away from that side of the equation. When a material is removed from a system in equilibrium, the equilibrium will shift towards that side of the equation.

38. Changes in concentration HI H2H2 I2I2 Log concentration Time Example. Example. I 2 is added to an equilibrium mixture. The system will adjust all of the concentrations to reestablish a new equilibrium with the same K c.

39. Changes in concentration HI H2H2 I2I2 Log concentration Time Example. Example. Some H 2 is removed. Again, the system adjusts all of the concentrations to reestablish a new equilibrium with the same K c.

40. Changes in pressure Changing the pressure does not change the value of the equilibrium constant at constant temperature. Solids and liquids are not effected by pressure changes. Changing pressure by introducing an inert gas will not shift an equilibrium. Pressure changes only effect gases that are a portion of an equilibrium.

41. Changes in pressure In general, increasing the pressure by decreasing volume shifts equilibria towards the side that has the smaller number of moles of gas. H 2 (g) + I 2 (g) 2HI (g) N 2 O 4 (g) 2NO 2 (g) Unaffected by pressure Increased pressure, shift to left

42. Changes in temperature Changes in temperature usually change the value of the equilibrium constant. K c can either increase or decrease with increasing temperature. The direction and degree of change is dependent on the specific reaction. T, o C K p 6492.7 x 10 0 7606.3 x 10 1 8718.2 x 10 2 9826.8 x 10 3 T, o C K p 6492.7 x 10 0 7606.3 x 10 1 8718.2 x 10 2 9826.8 x 10 3 T, o C K p 2279.0 x 10 -2 4278.1 x 10 -5 6271.3 x 10 -6 8279.7 x 10 -8 T, o C K p 2279.0 x 10 -2 4278.1 x 10 -5 6271.3 x 10 -6 8279.7 x 10 -8 CH 4 (g) + H 2 O (g) 2H 2 (g) + CO (g) N 2 (g) + 3H 2 (g) 2NH 3 (g)