1. N 2 (g) + 3 H 2 (g) --> 2 NH 3 (g) Chemical Equilibrium

2. Many reactions do not go to completion - under the given conditions it is possible that not all of the reactants are consumed. Instead the extent of the reaction is determined by the equilibrium point. A + B --> C + D As the concentrations of C and D increase, C and D could react to form A and B - the REVERSE reaction. A + B C + D N 2 (g) + 3H 2 (g) 2NH 3 (g)

3. The forward and reverse directions “oppose” one another. At some point in time, the rate of the forward reaction will equal the rate of the reverse reaction - this point corresponds to EQUILIBRIUM. Hence, when equilibrium has been reached, the concentration of A, B, C and D stay constant, as long as the conditions are held the same. At the equilibrium point: A and B combine to form C and D; C and D combine to form A and B; but both occur at the same rate. There is no NET change in the concentrations of A, B, C & D.

4. Equilibrium When opposing forces acting on a system are equal in magnitude, the system is said to be in a state of equilibrium. A dynamic equilibrium is one at which changes to the system do occur at the microscopic level, but at the macroscopic level these changes are not observed. In general: Processes not at equilibrium will act or react to reach equilibrium.

5. AB

6. NH 3 (g) H 2 (g) + N 2 (g)

7. Characteristics of equilibrium 1) The attainment of equilibrium is spontaneous; i.e. it is a natural tendency 2) At equilibrium there is no macroscopic evidence of any changes in the system 3) A dynamic balance is established between opposing forces 4) Equilibrium is reached from either direction

8. The Equilibrium Expression the ratio: (concentration C) c (concentration D) d (concentration A) a (concentration B) b = K (constant) K is called the EQUILIBRIUM CONSTANT Note: K has a fixed value for a particular reaction and varies with temperature a A + b B c C + d D For a general reaction:

9. If all reactants and products are gases, the relationship between the partial pressures of all gases at equilibrium is: P C c P D d = K P A a P B b If all reactants and products are in solution, the relationship between the concentrations of all species at equilibrium is: [C] c [D] d = K [A] a [B] b Where [X] is the concentration (example molarity) of species X at equilibrium Homogenous reactions; reactants and products in the same phase

10. Heterogeneous reaction: reactants and products are not in the same phase a A(aq) + b B(aq) c C(aq) + d D(g) [C] c P D d = K [A] a [B] b

11. K is a dimensionless quantity. 2A(g) B(g) K = P B /P 2 A is actually K= P B /P ref (P A /P ref ) 2 P ref is set to 1 atm K is dimensionless For solutions, if concentration is M; [A] ref = 1M

12. HCl(aq) H + (aq) + Cl - (aq) [H + ] [Cl - ] = K [HCl] CH 3 COOH(aq) H + (aq) + CH 3 COO - (aq) [H + ] [CH 3 COO - ] = K [CH 3 COOH] K ~ 10 7 at 25 o C K ~ 10 -5 at 25 o C

13. The reaction of SO 2 (g) and O 2 (g) forming SO 3 (g) 2 SO 2 (g)+ O 2 (g) 2SO 3 (g) P 2 SO 3 = K P SO 2 2 PO2PO2 Equilibrium can be reached for different partial pressures of SO 2, O 2, and SO 3, depending on the starting conditions, but at 25 o C, the value of K is the same.

14. The Magnitude of the Equilibrium Constant The magnitude of the equilibrium constant reveals the extent to which the reaction will proceed in the desired direction. a A + b B c C + d D Reactions that have K values > 1 are favored in the direction written; i.e. forward direction. Reactions that have K values < 1 are favored in the reverse direction Reactions for which K is near I have substantial amounts of both reactants and products when equilibrium is established.

16. Applying the Equilibrium Expression to Gas Phase Reactions PCl 5 (g) PCl 3 (g) + Cl 2 (g)K = 2.15 What are the equilibrium partial pressures of all three gases in a closed container containing only PCl 5 at 0.100 atm and held at 250 o C? According to the ideal gas laws, the partial pressures of gases is proportional to the number of moles of each gas, as long as the volume and temperature are kept fixed. The stoichiometry of this reaction is 1 : 1 : 1

17. PCl 5 (g) PCl 3 (g) + Cl 2 (g) Initial P (atm) Change in P (atm) Equilibrium P (atm) -xx x 0.1000 0 0.100-xx x If the partial pressure of PCl 5 decreases by x at equilibrium, the partial pressures of PCl 3 and Cl 2 increases by x at equilibrium. At equilibrium: P PCl 3 P Cl 2 = K P PCl 5

18. This is a quadratic equation of the form ax 2 + bx + c = 0 and the solution of this equation is of the form = 2.15 (0.100-x) (x) x 2 = 2.15 (0.100-x) x 2 + 2.15x - 0.215 = 0 x = (-b ± √ b 2 - 4ac) 2a

19. Using this expression and solving for x, the roots of the equation are x = 0.0957 and -2.25 atm. At equilibrium, the partial pressures of Cl 2 and PCl 3 are 0.0957 atm, and that of PCl 5 is (0.100 - 0.0957) = 0.004atm

20. 4 NO 2 (g) N 2 O(g) + 3 O 2 (g) The three gases are introduced into a container at partial pressures of 3.6 atm NO 2, 5.1 atm N 2 O and 8.0atm O 2 and react to reach equilibrium at a fixed temperature. The equilibrium partial pressure of NO 2 is measured to be 2.4 atm. Calculate the equilibrium constant of the reaction at this temperature, assuming that no competing reactions occur.

21. At equilibrium, the partial pressure of NO 2 is 2.4 atm 3.6 - 4x = 2.4 => x = 0.3 atm Hence P N 2 O at equilibrium = 5.7 atm; P O 2 = 8.9 atm 4 NO 2 (g) 2 N 2 O(g) + 3 O 2 (g) Initial P (atm) Change in P (atm) Equilibrium P (atm) - x2x/4 3x/4 3.65.1 8.0 2.45.1+2x 8.0+3x K = (5.7) 2 (8.9) 3 (2.4) 4 = 6.9 x 10 2 Change in P (atm)- 4x2x 3x

22. In applying the equilibrium expression the following must be considered. 1) The equilibrium constant for a reverse reaction is the reciprocal of the equilibrium constant for the corresponding forward reaction. 2 H 2 (g) + O 2 (g) 2H 2 O(g) = K 1 PH2O PH2O 2 P H 2 P O 2 2 2H 2 O(g) 2 H 2 (g) + O 2 (g) PH2O PH2O 2 = K 2 P H 2 P O 2 2 K 1 = K 2

23. 2) When the coefficients in a balanced chemical equation are multiplied by a constant factor, the corresponding equilibrium constant is raised to the power equal to that factor. = K 2 PH2O PH2O P H 2 P O 2 1/2 H 2 (g) + O 2 (g) H 2 O(g) 1 2 K 2 = √K 1 2 H 2 (g) + O 2 (g) 2H 2 O(g) = K 1 PH2O PH2O 2 P H 2 P O 2 2

24. 3) When chemical equations are added or subtracted to obtain a net equation, the corresponding equilibrium constants are multiplied or divided to obtain the equilibrium constant of the net equation. 2 BrCl(g) Cl 2 (g) + Br 2 (g) = K 1 = 0.45 at 25 o C P BrCl 2 P Cl 2 P Br 2 Br 2 (g) + I 2 (g) 2 IBr(g) = K 2 = 0.051 at 25 o C 2 P Br 2 P I 2 P IBr

25. Adding the two chemical equations gives: 2 BrCl(g) + Br 2 (g) + I 2 (g) 2 IBr(g) + Cl 2 (g) + Br 2 (g) 2 BrCl(g) + I 2 (g) 2 IBr(g) + Cl 2 (g) = K 3 P BrCl 2 P Cl 2 P IBr 2 PI2PI2 Looking at the expressions for K 1, K 2 and K 3 K 1 K 2 = K 3 Hence, K 3 = 0.023 at 25 o C

26. For the general reaction: P C c P D d = Q P A a P B b a A(g) + b B(g) c C(g) + d D(g) Define the reaction quotient, Q: where P is the partial pressure of a species at any point in time. Reaction Quotient

27. If Q = K, the reaction is at equilibrium If Q ≠ K, the reaction is not at equilibrium. If Q > K, reaction proceeds from right to left If Q < K, reaction proceeds from left to right a A(g) + b B(g) c C(g) + d D(g)

29. The Ideal Gas Equation and Chemical Equilibrium For gaseous reactants or products, the concentration may be in moles/liter. The concentrations of the gases in moles/liter must be converted to partial pressures. Concentration of a species A in moles/lit = [A] [A] = n V PAPA R T = This equation can be used to convert concentration of a gas in moles/lit to partial pressure of the gas.

30. where P ref is the reference pressure = 1 atm and ensures that K is unitless. a A(g) + b B(g) c C(g) + d D(g) Hence, for a general gas phase reaction [C] c [D] d = K [A] a [B] b R T P ref () (a+b-c-d)

31. The equilibrium constant for the reaction CH 4 (g) + H 2 O(g) CO(g) + 3 H 2 (g) Equals 0.172 at 900K. The concentrations of H 2 (g), CO(g), and H 2 O(g) in an equilibrium mixture of gases all equal 0.00642 mol/L. Calculate the concentration of CH 4 (g) in the mixture, assuming that this is the only reaction taking place. [CO] [H 2 ] 3 = K [CH 4 ] [H 2 O] R T P ref () -2 [0.00642 mol/L] [0.00642 mol/L ] 3 = 0.172 [CH 4 ] [0.00642 mol/L] [(0.08206 L atm mol -1 ) 900 K] -2 [CH 4 ] = 0.00839 mol/L

32. What happens if a system at equilibrium undergoes a change in conditions? The tendency of a system to achieve equilibrium is spontaneous. Once a system is at equilibrium it will remain at equilibrium. However, if conditions change which are different from the equilibrium condition (for example the temperature changes) the system will respond to this change in a way to achieve equilibrium again. Note: the concentrations of species when equilibrium is re- established need not be the same as the ones established at the previous equilibrium.

33. N 2 (g) + 3H 2 (g) 2NH 3 (g)

34. LeChatelier’s Principle indicates how a system will respond if a system at equilibrium experiences a change: If a stress is applied to a system at equilibrium, the system tends to react so that the stress is minimized P C c P D d = Q P A a P B b a A(g) + b B(g) c C(g) + d D(g) A) Changing the concentration of a reactant or product The reaction quotient determines the direction of the reaction. If the system is at equilibrium, Q = K If to a system at equilibrium a small amount of reactant is added, Q decreases and is < K. Hence the reaction will proceed to the right to reduce the concentration of the added reactant.

35. If a product is removed from the equilibrium mixture, Q also decreases, and the reaction once again proceeds to the right to increase the concentration of the product. B) Changing the Volume Decreasing the volume, increases the total pressure of the reaction mixture. The reaction will then proceed in the direction which reduces the total pressure. 2 NO 2 (g) N 2 O 4 (g) If the volume is decreased, the above reaction will move to the right, to decrease the total number of molecules and hence the pressure.

36. C) Changing Temperature The effect of changing the temperature of a system at equilibrium depends on whether the reaction proceeds by absorbing energy (endothermic) or by releasing energy (exothermic). An endothermic reaction lowers the temperature of the system and an exothermic reaction raises the temperature of the system. If a reaction is exothermic, raising the temperature causes the equilibrium to shift to the left. a A(g) + b B(g) c C(g) + d D(g)+ heat Forward reaction is exothermic; reverse reaction is endothermic.

37. For reactions where the equilibrium concentration of a product is low, product yields can be increased by adjusting conditions that encourage a reaction to proceed in the direction of the products. 2 NO 2 (g) N 2 O 4 (g)

38. Heterogeneous Reactions CaCO 3 (s) + SO 3 (g) CaSO 3 (s) + CO 2 (g) [CaSO 3 ] P CO 2 = K [CaCO 3 ] P SO 3 In all equilibrium expressions, the concentrations of all pure solids and liquids are set to 1. P CO 2 = K P SO 3 Hence,

39. In general, to write the equilibrium expression for a reaction 1) Concentration of gases are expressed as partial pressures 2) Concentration of dissolved species in solution are expressed as moles/liter 3) Concentrations of pure solids and liquids are set to 1 (for a solvent taking part in a reaction, its concentration is also set to 1 providing the solution is dilute)

40. Problem: The reaction between Ni(s) and CO(g) to form Ni(CO)4(g) is as follows: Ni(s) + 4CO(g) Ni(CO) 4 (g) A quantity of Ni(s) is added to a vessel containing CO at a partial pressure of 1.282 atm and 354 K. At the equilibrium point of this reaction, the partial pressure of CO is 0.709 atm. Calculate the equilibrium constant of this reaction at 354 K. Initial P (atm) Change in P (atm) Equilibrium P (atm) 0.5730.143 1.282 0 Ni(s) + 4CO(g) Ni(CO) 4 (g) 0.7090.143 P NiCO 4 K= P CO 4

41. P NiCO 4 K= P CO 4 0.143 K= (0.709) 4 = 0.567

42. Applications of Chemical Equilibria Why is CO lethal? Hemoglobin (Hb) is the main chemical component of red blood cells, carrying oxygen from the lungs to the body tissue, transporting oxygen from a region of high concentration to low concentration. This transportation is accomplished by the formation of the oxygen-hemoglobin complex, oxyhemoglobin (HBO 2 ) Hb(aq) + O 2 (g) HbO 2 (aq) K= [HbO 2 ] [Hb]PO2PO2

43. In the absence of Hb, the amount of O 2 in blood is low. Because of the formation of the HbO 2 complex, the amount of O 2 in blood is increased by a factor of 70. Hb(aq) + O 2 (g) HbO 2 (aq) LeChatelier’s principle predicts that in regions of high O 2 partial pressure, the Hb-HbO 2 equilibrium is shifted to the right, which is the case in the lungs In regions of low O 2 partial pressure, the equilibrium shifts to the left, resulting in a breakup of the HbO 2 complex, releasing O 2 to the body’s tissues.

44. What happens if a person breathes CO? Hb(aq) + CO(g) HbCO (aq) K(CO)= [HbCO] [Hb]P CO K(CO) > (KO 2 ) Which means that CO binds more tightly to Hb compared to O 2

45. When Hb is exposed to both O 2 and CO, there is competition for the Hb, and the following reaction takes place: HbO 2 (aq) + CO(g) HbCO(aq) + O 2 (g) The K for this reaction is: K= [HbO 2 ] [HbCO] PO2PO2 P CO = K(CO) K(O 2 ) K PO2PO2 P CO = [HbO 2 ] [HbCO]

46. The relative amounts of HbCO and HbO 2 therefore depend on the partial pressures of CO and O 2. HbO 2 (aq) + CO(g) HbCO(aq) + O 2 (g) Since K(CO) > K(O 2 ), K for the competition reaction is >1 In fact at 38 o C, the value of K is 210 and so the position of the equilibrium strongly favors the formation of the HbCO complex, with CO displacing O 2 from the HbO 2 complex, resulting in asphyxiation. However, the process is reversible - from LeChatelier’s principle, a large partial pressure of O 2 will result in the equilibrium of the competition reaction to favor the HbO 2 complex.

47. Extraction and Separation A solute dissolved in a solvent A can be extracted from this solvent by using another solvent B. The solute must dissolve in both solvents A & B and solvent B must be immiscible with the solvent A. For example, CCl 4 and H 2 O are immiscible. I 2 (s) dissolves in both solvents.

48. If to a solution of I 2 (aq) some CCl 4 is added and then the flask containing the I 2 (aq) and CCl 4 is shaken, some of the I 2 in the water layer will be extracted into the CCl 4 layer I 2 (aq) I 2 (CCl 4 ) The following equilibrium is then established: K= [I 2 ] CCl 4 [I 2 ] H 2 O H 2 O layer CCl 4 layer

49. Chromatography - extraction of small quantities of solute