2. Chapter 15: Chemical Equilibrium By: Ms. Buroker

3. What is Chemical Equilibrium? The state where the concentrations of all reactants and products remain constant with time. 1.) Reactions in closed vessels will reach equilibrium 2.) Reactions which are “far to the right” 3.) Reactions which are “far to the left”

4. Chemical Equilibrium: “Big K” kinetics: rate constant “little k” kinetics “little k” told us how fast a reaction proceeds and is used to indicate a possible mechanism. Eq. tells us to what extent a RXN proceeds to completion react.  prod. @ Eq: rate forward = rate reverse

5. The Equilibrium Condition 1.) Upon addition of reactants and/or products, they react until a constant amount of reactants and products are present = equilibrium. 2.) Equilibrium is dynamic since product is constantly made (forward reaction), but at the same rate it is consumed (reverse reaction).

6. The Equilibrium State Not all reactants are completely converted to product. Reaction equilibria deal with the extent of reaction. Arrows between reactants and products separate them and qualitatively indicate the extent of reaction. – Single arrow points to dominant side: H 2 (g) + O 2 (g)  H 2 O(g) – Double arrow indicates both reactants and products present after equilibrium obtained: N 2 O 4 (g)  2NO 2 (g). Equilibrium exists when rates of forward and reverse reaction are the same. E.g. When rate of N 2 O 4 decomposition equal the rate of formation of N 2 O 4, reaction at equilibrium N 2 O 4 (g)  2NO 2 (g).

7. The Equilibrium Constant The Link Between Chemical Equilibrium and Chemical Kinetics: Reactions with one elementary step: At equilibrium: R forward = R reverse E.g. Decomposition of N 2 O 4 : k f [N 2 O 4 ] = k r [NO 2 ] 2. or Equilibrium Constant

8. The Equilibrium Constant At equilibrium the ratio of concentrations equals a constant. A generalized form of this expression that describes the equilibrium condition: aA + bB + cC +...  mM + nN + oO.... Equilibrium Concentrations

9. Law of Mass Action: Values of K c are constant for a RXN at a given temperature. Any equilibrium mixture of the above system at that temperature should give the SAME K c value. Equilibrium Constants are Temperature Dependent and Mechanism Independent!! For the reverse reaction: K’ = 1/K If the original reaction is multiplied by a factor, then: K” = K n

10. What Does “K” Really Mean??? When K is much larger than 1 … A.) The reaction will consist mostly of products B.) The Equilibrium lies to the right When K is a very small value …. A.) The reaction will consist mainly of reactants B.) The Equilibrium condition is far left C.) The reaction does NOT occur to any significant event

11. Let’s Calculate a K Value … The following equilibrium concentrations were observed for the Haber process at 127 ○ C: [NH 3 ] = 3.1 x 10 -2 mol/L [N 2 ] = 8.5 x 10 -1 mol/L [H 2 ] = 3.1 x 10 -3 mol/L a.) Calculate the value of K at 127 ○ C for this reaction. Answer: 3.8 x 10 4 b.) Calculate K for the reverse reaction at 127 ○ C. Answer: 2.6 x 10 -5 c.) Calculate the value of the K for the reaction: 1/2N 2(g) + 3/2H 2(g) ↔ NH 3(g) Answer: 1.9 x 10 2

12. Okay … There is a term we refer to as the equilibrium position, which is a set of equilibrium concentrations. *** It’s important to note that this is different than the equilibrium constant, K. The equilibrium constant remains the same at a given temperature no matter what the concentration because it is a ratio. There is only one equilibrium constant for a particular system at a particular temperature, but there are an infinite number of equilibrium positions.

13. InitialEquilibriumInitialEquilibrium [SO 2 ] = 2.00M[SO2] = 1.50M[SO 2 ] = 0.500M[SO 2 ] = 0.590M [O 2 ] = 1.50M[O2] = 1.25M[O 2 ] = 0[O 2 ] = 0.0450M [SO 3 ] = 3.00M[SO3] = 3.50M[SO 3 ] = 0.350M[SO 3 ] = 0.260M Experiment 1 Experiment 2 The following results were collected for two experiments involving the reaction at 600 ○ C between gaseous sulfur dioxide and oxygen to for gaseous sulfur trioxide: Show that the equilibrium constant is the same in both cases. Step 1: Write a balanced chemical equation. 2SO 2(g) + O 2(g) ↔ 2SO 3(g) From the law of mass action:

14. Equilibrium Expressions Involving Pressures We also have K p which is sometimes used when dealing with gases with the “P” referring to the pressure of the gases. aA(g) + bB(g)  dD(g) + eE(g) Since: PV = nRT or then P = CRT Molar Concentration of Gas Note: Pressure is proportional to Concentration! For the General Reaction: aA(g) + bB(g)  dD(g) + eE(g) 0.0821Latm molK (d + e) – (a + b)

15. Let’s Try One … PCl 3 (g) + Cl 2 (g)  PCl 5 (g) In a 5.00 L vessel (@ 230 o C) an equilibrium mixture is found to contain: 0.0185 mol PCl 3, 0.0158 mol PCl 5 and 0.0870 mol of Cl 2. Determine K c and K p K c = 49.08 =

16. Heterogeneous Equilibria Equilibria involving more than one phase Ex: CaCO 3(s) ↔ CaO (s) + CO 2(g) Experimental results show that the position of a heterogeneous equilibria does not depend on the amounts of pure solids or liquids present. Concentrations of pure solids and liquids can not change; therefore:

17. CaCO 3 (s)  CaO(s) + CO 2 (g) What is the K c expression? Answer: K c = [CO 2 ] What is the K p expression? Answer: K p = P CO2

18. Using I.C.E. (Initial, Change, Equilibrium) Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and 0.300 mol of H 2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of K c ? CO(g) + 2H 2 (g)  CH 3 OH(g) I. C. E. 0.1500 0.3000 - X- 2X+ X 0.1187 If 0.1500 – X = 0.1187, Then X = 0.0313, So …. 0.23740.0313 = 10.52

19. More I.C.E. 2NOBr(g)  2NO(g) + Br 2 (g) I. C. E. 2.00 00 - 2X+ 2X+ X 2.00 – 2X + 2X +X Problem: @ 77 o C, 2.00 mol of NOBr (nitrosyl bromide) placed in a 1.000L flask dissociates to the extent of 9.4%. Find K c. Since 9.4% dissociates, the Change in NOBr, -2x = 2.00(0.094), So x = 0.094 1.8120.1880.094 = 1.01 x 10 -3

20. Applications of the Equilibrium Constant Reaction Quotient (Q) is determined using the law of mass action and initial concentrations instead of equilibrium concentrations. N 2(g) + 3H 2(g) ↔ 2NH 3(g) 1.) Q = K: The system is @ equilibrium … no shift will occur. 2.) Q > K: Shift to the left will occur 3.) Q < K: Shift to the right will occur Consider This … What about a system that may not yet be at equilibrium?

21. Equilibrium Quotient

22. 21 Example: PCl 5 (g)  PCl 3 (g) + Cl 2 (g) @ 250 o C K c = 4.0 x 10 - 2 If: [Cl 2 ] and [PCl 3 ] = 0.30M and [PCl 5 ] = 3.0M, is the system at Equilibrium? If not, which direction will it proceed? Q c = = 3.0 x 10 - 2 Q c < K c (not @ Eq.) Which way must the RXN go to achieve Equilbrium? Remember ratio is prod./React more products makes the number bigger RXN goes Find Q c and compare to K c to decide.

23. Le Chatelier’s Principle States: If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.

24. Le Chatelier Another Way … If a component (reactant or product) is added to a reaction system at equilibrium, the equilibrium concentration will shift in the direction that lowers the concentration of that component. If a compound is removed, the opposite effect occurs.

25. Le Chatelier Example #1 A closed container of ice and water is at equilibrium. Then, the temperature is raised. Ice + Energy  Water The system temporarily shifts to the _______ to restore equilibrium. right

26. Le Chatelier Example #2 N 2 O 4 (g) + Energy  2 NO 2 (g) The system temporarily shifts to the _______ to restore equilibrium. A closed container of N 2 O 4 and NO 2 is at equilibrium. NO 2 is added to the container. Left

27. A closed container of water and its vapor is at equilibrium. Vapor is removed from the system. water + Energy  vapor The system temporarily shifts to the _______ to restore equilibrium. Le Chatelier Example #3 right

28. Le Chatelier Example #4 A closed container of N 2 O 4 and NO 2 is at equilibrium. The pressure is increased. N 2 O 4 (g) + Energy  2 NO 2 (g) The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation. Left

29. Le Chatelier and Changes in Temp. Whether K c increases or decreases depends upon two factors,  T and  H rxn. 1.) An increase in temperature causes the equilibrium to shift towards the endothermic side. 2.) A decrease in temperature causes the equilibrium to shift in the opposite direction. Changes in concentrations or total pressure cause shifts in equilibrium without changing the value of the equilibrium constant. In contrast, changes in temperature cause the value of K c to change.

30. Le Chatelier and Changes in Temp. Whether K c increases or decreases depends upon two factors,  T and  H rxn. 1.) An increase in temperature causes the equilibrium to shift towards the endothermic side. 2.) A decrease in temperature causes the equilibrium to shift in the opposite direction. Reactants + heat ↔ Products Endothermic Reactants ↔ Products + heat Exothermic An increase in heat will cause the reaction to shift to the right. An increase in heat will cause the reaction to shift to the left. KcKc KcKc

31. Le Chatelier and Catalysts 1.) It has no effect on the equilibrium. 2.) It speeds up attainment of equilibrium. K c related to the  H and not E a