1. Equilibrium

2.  In equilibrium, the concentrations of the chemical species are constant, not necessarily equal.  Equilibrium constants:  K = equilibrium constant  K a = weak acid  K b =weak base  K w = water  K p = pressure  K c = molar concentrations

3.  The state where the concentrations of all reactants and products remain constant with time.  In the past when you did stoichiometry problems it was assumed that the reactions went to completion. However, there are many reactions that stop far short of completion. An example is the dimerization (two molecules joining together) of nitrogen dioxide. NO 2(g) + NO 2(g)  N 2 O 4(g)NO 2(g) + NO 2(g)  N 2 O 4(g) (1:56) What can you deduce from your observations of this chemical reaction? How could you change the above equation to more closely resemble what you observed in this experiment?

4.  The law of mass action is a general description of the equilibrium condition:  aA + bB  cC + dD  [C] c [D] d  K = [A] a [B] b  There are no units for K  When K>1, products are favored at equilibrium.  When K< 1, reactants are favored at equilibrium.  K is a constant as long as temperature doesn’t change.

5.  Rules: Pure solids do not appear in expressions (M does not change)  Pure liquids do not appear in expressions.  Only solutions and gases will be in the equilibrium expressions because their concentrations do change.  Write the equilibrium constant for the following reaction: (NMSI pg2)  4NH 3(g) + 7O 2(g) ↔ 4NO 2(g) + 6H 2 O (g) (NMSI page 2)  [NO 2 ] 4 [H 2 O] 6  K = [NH 3 ] 4 [O 2 ] 7

6.  Write the expressions for K and K p for the following processes:  A. The decomposition of solid phosphorus pentachloride to liquid phosphorus trichloride and chlorine gas.  PCl 5(s)  PCl 3(l) + Cl 2(g)  K = [Cl 2 ] K p = [Cl 2 ] CuSO 4 · 5H2O (s)  CuSO 4(s) + 5 H 2 O (g) K = [H 2 O] 5 K p =[H 2 O] 5

7.  The following equilibrium concentrations were observed for the Haber process at 127 o C: [NH 3 ] = 3.1 x 10 -2 M [N 2 ] = 8.5 x 10 -1 M [H 2 ] = 3.1 x 10 -3 M Calculate the value of K at 127 o C for this reaction. 3H 2(g) + N 2(g) ↔ 2NH 3(g) [3.1 x 10 -2 ] 2 K = [3.1 x 10 - 3 ] 3 [8.5 x 10 -1 ] = 3.8 x 10 4

8.  b. Calculate the value of the equilibrium constant at 127 o C for the balanced chemical equation: 2NH 3(g) ↔ 3H 2(g) + N 2(g) [3.1 x 10 -3 ] 3 [8.5 x 10 -1 ] K ’ = [3.1 x 10 - 2 ] 2 K ’ = 1/K from the previous solution K ’ = 1/3.8 x 10 4 = 2.6 x 10 -5

9. c. Calculate the value of the equilibrium constant at 127 o C for the reaction given by the equation: ½ N 2(g) + 3/2 H 2(g)  NH 3(g) Compare to the first equation and this is all to the ½ power. So K ’’ = K 1/2 K ’’ = (3.8 x 10 4 ) 1/2 = 1.9 x 10 2

10.  The equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. K’ = 1/K  When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression raised to the nth power.  Thus K new = (K original ) n  K values are written without units.

11.  Equilibrium constants have the same value at a given temperature regardless of the amounts of gases that are mixed together initially but the equilibrium concentrations will not always be the same.  Each set of equilibrium concentrations is called an equilibrium position.  There is only one equilibrium constant at a given temperature but infinite numbers of equilibrium positions.

12.  The following results were collected for two experiments involving the reaction at 600 o C between gaseous sulfur dioxide and oxygen to form gaseous sulfur trioxide.  Show that the equilibrium constant is the same in both cases. Experiment OneExperiment Two InitialEquilibriumInitialEquilibrium [SO 2 ] 0 = 2.00M[SO 2 ] = 1.50 M[SO 2 ] 0 = 0.500 M[SO 2 ] = 0.590 M [O 2 ] 0 = 1.50 M[O 2 ] = 1.25 M[O 2 ] 0 = 0[O 2 ] = 0.0450 M [SO 3 ] 0 = 3.00 M[SO 3 ] = 3.50 M[SO 3 ] 0 = 0.350 M[SO 3 ] = 0.260 M

13. [3.50] 2  K = [1.50] 2 [1.25] = 4.36 [0.260] 2  K = [0.590] 2 [0.0450] = 4.32  These values are the same within experimental error.

14.  Equilibrium partial pressures can be found by: [P product ] n  K p = [P reac ] l [P reac ] m  Reaction: 2NO (g) + Cl 2(g) ↔ 2NOCl (g)  Pressures at 25 o C:  P NOCl = 1.2 atm  P NO = 5.0 x 10 -2 atm  P Cl2 = 3.0 x 10 -1 atm  Calculate the value of K p at 25 o C  1.9 x 10 3

15.  K c and K p are not interchangeable.  In the equation above:  Δ n = total number of moles of gas produced – total moles of gas reacted.  R = 0.086 L · atm/mol · K  T = Kelvin  Using the value of K p from the previous problem, calculate K at 25 o C for 2NO (g) + Cl 2(g) ↔ 2NOCl (g) K c = 1.9 x 10 3 ÷ (.08206 L atm/mol K · 298K) -1 = 4.6x10 4 K c and K p are equal if the number of moles of gas produced = number of moles gas produced. What does the value of K mean for both this and the previous problem with regard to reactants vs. products?? Both are greater than one meaning that the formation of product is favored over reactant.

16.  Q c has the appearance of K but the concentrations are not necessarily at equilibrium.  If Q < K, the system is not at equilibrium : Reactants  products to make Q = K at equil  Products are favored  If Q = K, the system is at equilibrium  If Q > K, the system is not at equilibrium: Products  reactants to make Q = K at equil.  Reactants are favored.

17.  [NH 3 ] 0 = 1.0 x 10 -3 M; [N 2 ] 0 = 1.0 x 10 -5 M; [H 2 ] 0 = 2.0 x 10 -3 M  [NH 3 ] 0 = 2.0 x 10 -4 M; [N 2 ] 0 = 1.5 x 10 -5 M; [H 2 ] 0 = 3.54 x 10 -1 M  [NH 3 ] 0 = 1.0 x 10 -4 M; [N 2 ] 0 = 5.0 M; [H 2 ] 0 = 1.0 x 10 -2 M  1. Q >K; shift left (Q = 1.25 x 10 7 )2. Q = 6.0 x 10 -2 no shift  3. Q < K; shift right (Q =.002 )

18. N 2 O 4(g) ↔ 2NO 2(g) Consider an experiment in which gaseous N 2 O 4 was placed in a flask and allowed to reach equilibrium at a temperature where K p = 0.133. At equilibrium, the pressure of N 2 O 4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO 2. K p = [NO 2 ] 2 0.133 = [NO 2 ] 2 = 0.600 atm [N 2 O 4 ] 2.71 atm

19.  Set up “RICE” TABLE:  R = write a balanced equation for the predominant reacting species  I = fill in the initial concentrations  C = determine the change that is taking place in terms of x  E = express the equilibrium concentrations in terms of x

20.  At a certain temperature a 1.00 L flask initially contained 0.298 mol PCl 3(g) and 8.70 x 10 -3 mol PCl 5(g). After the system had reached equilibrium, 2.00 x 10 -3 mol Cl 2(g) was found in the flask. Gaseous PCl 5 decomposes according to the reaction: PCl 5(g) ↔ PCl 3 (g) + Cl 2(g)  Calculate the equilibrium concentrations of all species and the value of K.  K c = [PCl 3 ] [Cl 2 ]  [PCl 5 ]  Set up a RICE table and then solve for K  K = 8.96 x 10 -2 R (reaction)PCl 5 ↔ PCl 3 + Cl 2 I (initial conc).00870 M.298 M? C (change)- x+ x ExEx.00870 -.0020.298 +.0020.0020 Equilibrium.00670.300.0020

21.  Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is 5.10. Calculate the equilibrium concentrations of all species if 1.000 mol of each component is mixed in a 1.000 L flask.  Balanced chemical equation:  CO (g) + H 2 O (g) ↔ CO 2(g) + H 2(g)  Build a RICE TABLE  Calculate Q and compare to K.  Since Q = 1 and K = 5.10 that means that the products are favored. Reaction CO (g) + H 2 O (g) ↔ CO 2(g) + H 2(g) Initial []1.0M Change Equilibrium x answer

22. Reaction CO (g) + H 2 O (g) ↔ CO 2(g) + H 2(g) Initial1.0 Change-x-x-x-x+ x Equil x 1 – x 1 + x Equil  Using K = 5.10, solve for x  K = (1 + x) 2 √5.10 = 1+x ÷ 1-x (1 – x) 2 = 2.26 – 2.26x = 1 + x x =.387 Fill in the rest of the table

23.  Assume that the reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant of 1.15 x 10 2 at a certain temperature. In a particular experiment, 3.000 mol of each component was added to a 1,500 L flask; 3.000 mol/1,500L = 2.0 M. Calculate the equilibrium concentrations of all species. H 2(g) + F 2(g) ↔ 2HF (g) Solve for Q. Q = [2] 2 ÷ [2][2] = 1. So Q ends up being <K and the forward reaction is favored. K = (2 + 2x) 2 ÷ (2-x)(2-x) x = 1.53 Reaction H 2(g) + F 2(g) ↔ 2HF (g) Initial2.0 M Change- x + 2x ExEx 2-x 2 + 2x E

24.  Assume that gaseous hydrogen iodide is synthesized from hydrogen gas and iodine vapor at a temperature where the equilibrium constant is 1.00 x 10 2. Suppose HI at 5.000 x 10 - atm, H 2 at 1.000 x 10 -2 atm and I 2 at 5.000 x 10 -3 atm are mixed in a 5.000 L flask. Calculate the equilibrium pressures for each.  First step? Calculate Q  K = 100 Q = 5000  K< Q so the reaction is reversed—there is more reactant than product.  That means that the reactants become + and product becomes -

25. ReactionH 2(g) + I 2(g) ↔ 2HI (g) Initial Pressure.01.005.5 Change+ x - 2 x Equilibrium x.01 + x.005 + x.5 - 2x Equilibrium

26.  100 = (.5 - 2x) 2 ÷ (.01 + x)(.005 + x)  100 =.25 – 2x + 4x 2 ÷.00005 +.015 x + x 2 Cross multiply by 100.25 – 2x + 4x 2 =.005 + 1.5x + 100x 2 96x 2 + 3.5x -.245 = 0 a b c Using quadratic formula, x = 0.0355 now finish up the table.

28.  Shifts occur to reestablish equilibrium.  Pure solids do not affect equilibrium positions.  Adding or removing a reagent causes the equilibrium to shift to reestablish K.  Shifts occur to reestablish equilibrium positions. K is generally [products]÷[reactants]  Increasing pressure causes equilibrium to shift to the side containing the fewest moles of gas. The converse is true.  Adding a catalyst causes no shift and has NO EFFECT on K. It just causes equilibrium to be reestablished faster.  Changing the temperature is like adding or removing a reactant or product.  Endothermic: A + B + heat ↔ C + D  Exothermic: A + B ↔ C + D + heat

29.  If a component (reactant or product) is added to a reaction system at equilibrium, the equilibrium will shift in the direction that lowers the concentration of that component.  Arsenic can be extracted from its ores by first reacting the ore with oxygen to form solid As 4 O 6, which is then reduced using carbon. As 4 O 6(s) + 6C (s) ↔ As 4(g) + 6CO (g) Predict the direction of the shift of the equilibrium position in response to each of the following changes in conditions.

30. a. Addition of carbon monoxide b. The shift will be away from the substance whose concentration is increased. The equilibrium position will shift to the left when CO is added. c. Addition or removal of carbon or tetraarsenic hexoxide. d. The amount of a pure solid has NO EFFECT on equilibrium position, so there is no shift. e. Removal of gaseous arsenic. f. If gaseous arsenic is removed, the equilibrium position will shift to the right to form more products. This is used quite a bit in industrial processes to increase yield.

31.  There are 3 ways to change pressure involving gases:  Add or remove a gaseous reactant or product.  Add an inert gas (one not involved in the process)  Change the volume of the container.

32.  Adding or removing a gaseous component will shift the equilibrium in the direction that lowers the concentration of that component.  Adding an inert gas that is not part of the process makes no change.  When the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume. This is done by decreasing the total number of gas molecules in the system. Volume is proportional to number of moles. So the shift will be to the side with the least number of moles of gas.  N 2(g) +3H 2(g) ↔ 2NH 3(g) shifts to the right  Increasing the volume is opposite, when the volume is increased, the system will shift to increase its volume. An increase in volume in NH 3 will produce a shift to the left to increase the total number of gas molecules present.

33.  Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced. a. The preparation of liquid phosphorus trichloride by the reaction: P 4(s) + 6Cl 2(g) ↔ 4PCl 3(l) Since P 4 and PCl 3 are a pure solid and a pure liquid, only the change in volume of the chlorine gas will affect the equilibrium. The volume is decreased, so the position of the equilibrium will shift to the right, since the reactant side contains six gaseous molecules and the product side has none.

34. b. The preparation of gaseous phosphorus pentachloride according to the equation: PCl 3(g) + Cl 2(g) ↔ PCl 5(g) Decreasing the volume will shift the given reaction to the right, since the product side contains only one gaseous molecule while the reactant side has two.

35. c. The reaction of phosphorus trichloride with ammonia: PCl 3(g) + 3NH 3(g) ↔ P(NH 2 ) 3(g) + 3HCl (g)  Both sides of the balanced reaction equation have four gaseous molecules. A change in volume will have no effect on the equilibrium position. There is no shift in this case.

36.  Predict how the value of K changes as the temperature is increased.  N 2(g) + O 2(g) ↔ 2NO (g) Δ H= 181 kJ  This is an endothermic reaction indicated by the positive value for Δ H. Energy can be viewed as a reactant and K increases as the temperature increases. The shift is to the right.

37.  2SO 2(g) + O 2(g) ↔ 2SO 3(g) Δ H = -198 kJ  This is an exothermic reaction (energy can be regarded as a product). As the temperature is increased, the value of K decreases (equilibrium shifts to the left).

38.  58 kJ + N 2 O 4(g) ↔ 2NO 2(g)  Predict the equilibrium shifts Shift Right Left Right None Left Right Left Change Addition of N 2 O 4(g) Addition of NO 2(g) Removal of N 2 O 4(g) Removal of NO 2(g) Addition of He Decrease container volume Increase container volume Increase temperature Decrease temperature

39. Thermodynamically favorable electrochemical cells have a positive voltage (E o ) while thermodynamically favorable reactions have a negative value of Δ G (free energy). If the equilibrium constant for such a reaction is greater than one, the products are favored and one would expect to see a positive E o and a negative Δ G. If K is less than one, the opposite would be expected.