I. Basis of Equilibrium. A. Q and equilibrium. EQUILIBRIA 1. Consider the general reaction bb + cc dd + ee a. Αs time elapses, [B] and [C] decrease causing the rate of the forward reaction to decrease. Conversely, as the concentrations of product build up, the rate of the reverse reaction increases. When the forward and reverse reaction rates are equal. When this occurs, equilibrium is established. b. At equilibrium: Rate forward Rate reverse. The concentrations of reactants and products will settle down to their equilibrium values and will no longer change with time.. The extent of a reaction can be followed using the proper reaction quotient, Q For a general reaction of the type, bb + cc dd + ee d a E e Q a D a b c B a C b. Where the a s are the activities, which are the effective concentrations. For calculations we make the following approximations for the activities: For gases, substitute their partial pressures in atm. For solutes, substitute their molar concentrations. For pure solids and liquids, substitute unity (1). c.since at equilibrium the concentrations of reactants and products reach constant values, Q will equal a constant, K, called the equilibrium constant. Q eq K d. For gaseous reactions, a X P X therefore for bb(g) + cc(g) At equilibrium, Q P K P d P E e dd(g) + ee(g) Q P P D P b c Where P s are the instantaneous partial pressures B P C d P E e K P P D P b c Where P s are the equilibrium partial pressures B P C The symbols Q P and K P indicate that the partial pressures are being used for activities 3. Q and the spontaneous reaction direction. a. If Q < K, the reaction is spontaneous in the forward direction. The reactant concentrations will decrease and products increase. 1
b. If Q > K, the reverse direction is the spontaneous one. The reactant concentrations will increase and products decrease. c. In all cases, the reaction proceeds in one direction or the other until Q K. 3. Some examples of equilibrium constant expressions: a. Gaseous reactions. N (g) + 3H (g) NH 3 (g) K K P P NH3 P N P 3 1.x10 6 at 700 C H Where the partial pressures are their equilibrium values b. Heterogeneous equilibria. C(s) + H O(g) CO(g) + H (g) K P COP H P H O Note that the activity of the pure solid, C(s), equals 1. In dealing with heterogeneous equilibria, the pure liquid or solid does not appear in the K, or Q, expression, as long as some is present, the amount is not important. c. Weak acid dissociation. HNO (aq) + H O(l) H 3 O + (aq) + NO (aq) K K a [H 3O + ][NO ] [HNO ] 7.x10 4 at 5 C The symbol K a is used for the acid dissociation constant. Note that in aqueous solutions, the activity of water1. d. Water self-dissociation. H O(l) H 3 O + (aq) + OH (aq) K K w [H 3 O + ][OH ] 1.0x10 14 at 5 This expression must be satisfied in any aqueous solution. e. Consider the reaction: Zn(s) + H + (aq) Zn + (aq) + H (g) Q a a Zn + (aq) H (g ) [Zn+ ]P H a Zn(s) a H [H + ] + (aq ) B. K C vs. K P 1. For gaseous reactions, K P can be obtained directly from thermodynamics. However, in some calculations it is more convenient to express concentrations in the unit of molarity. When molarity is used, the symbol for the equilibrium constant is K C.
K C [D]d [E] e [B] b c where [ ] stands for molarity [C]. Although K P and K C are not necessarily equal, if one is known the other can be calculated. K P P d e DP E P b c and K B P C [D]d [E] e If the gases behave ideally, P C [B] b [C] c i n irt V [i]rt Therefore, P A [A]RT, etc. Substituting into the K P expression, K P [D]d (RT) d [E] e (RT) e [B] b (RT) b [C] c (RT) c [D]d [E] e [B] b [C] c (RT)(d +e)-(b+ c) K C (RT)!n where Δn the change in the number of moles of gas. The value of R is 0.0806 L atm/mol K. Note that RT would have the dimension of atm/molarity 3. Examples of K P /K C interconversions a.consider the reaction N (g) + 3H (g) NH 3 (g) at 700 C. K P P NH 3 3 1.x10-6 atm - P N P H K P has a practical unit of atm -. K C can be calculated from K C [NH 3 ] K P (RT)! "n 1.x10-6 atm - (0.0806)(973)atm/M 7.7x10-3 M - [N ][H ] 3 b. Consider the reaction H (g) + I (g) HI(g) at 430 C. P HI K P PH P I [HI] 54.3 and K C 54.3 [H ][I ] Note that when Δn 0, K P K C. 3
C. Calculations using K C. 1. Concentration Changes and Equilibrium for CO(g) + H (g) CH 3 OH(g) at 500 K K C [CH 3 OH] [CO][H ] 14.5 INITIAL CONDITIONS EQUILIBRIUM CONDITIONS [CO] [H ] [CH 3 OH] Init Q C [CO] [H ] [CH 3 OH] Eq Q C 0.100 0.100 0.000 0.000 0.0911 0.08 0.0089 14.5 0.000 0.000 0.100 0.0753 0.151 0.047 14.5 0.000 0.000 0.409 0.16 0.34 0.47 14.5 0.100 0.100 0.100 100 0.138 0.176 0.06 14.5 0.075 0.00 0.116 38.7 0.100 0.50 0.091 14.5 4.440 8.30 10.000 0.033 0.850 1.050 13.590 14.5 Note that irrespective of where one starts out, at equilibrium, Q C Eq K C 14.5. Numerical examples Consider the reaction H (g) + I (g) HI(g) at 430 C. [HI] K C 54.3 [H ][I ] a. Suppose.000 moles of H and 3.000 moles of I are mixed in a 5.000 L container at 430 C and the reaction takes place. Calculate the equilibrium concentrations of H, I, and HI. Initial conditions: [H ].000 mol 5.000 L 0.400 M; [I ] 3.000 mol 5.000 L 0.600 M [HI] 0. Q C (init) 0 < K C. H and I are consumed and HI is formed. Let x the decrease in [H ] required to reach equilibrium. H + I HI Eqiul. Concs. 0.400 - x 0.600 - x x [HI] (x) 4x K 54.3 C [H ][I ] (0.400 - x)(0.600 - x) 0.40 - x + x! 50.3x - 54.3x + 13.03 0. This is in the standard form ax +bx+c0 and x can be determined from the quadratic formula 4
x - b ± b - 4ac a x 54.3 ± (54.3) - 4(50.3)(13.03) 54.3 ± 18.07 0.719, 0.360 (50.3) 100.6 Because of the quadratic equation, two values of x are solutions. However only x 0.360 is physically significant (since x the decrease in [H ], it is not possible for x to be greater than the initial H concentration, that is you cannot have a negative concentration). the equilibrium concentrations are: [H ] 0.400-0.360 0.040 M; [I ] 0.600-0.360 0.40 M; [HI] (0.360) 0.70 M b. Suppose 1.000 mole of H, 1.500 moles of I and 10.000 moles of HI were equilibrated in a 10.000 L contained at 430 C. Calculate the equilibrium concentrations of all species. Initial: [H ] 1.000 mol/10.000 L 0.100 M; [I ] 1.500 mol/10.000 L 0.150 M; [HI] 10.000 mol/10.000 L 1.000 M. Q C (init) 1.000 /(0.100)(0.150) 66.8 > K C [HI] decreases and [H ] and [I ] increases Let x the increase in [H ]. H + I HI Equil. Concs. 0.100 + x 0.150 + x 1.000 - x [HI] (1.000 - x) K C 54.3 [H ][I ] (0.100 + x)(0.150 + x) 1.000-4.000x + 4x 0.015 + 0.50x + x You verify that the only physically significant solution is x 0.010. The equilibrium concentrations are: [H ] 0.100 + 0.010 0.110 M; [I ] 0.150 + 0.010 0.160 M; [HI] 1.000 - (0.010) 0.980 M. II. Equilibrium Shifts Consider the reaction N (g) + 3H (g) NH 3 (g) ΔH - 45.9 kj/mol A. Consider how the equilibrium point will be affected by certain changes in experimental conditions. 1. If the equilibrium shifts towards the product side, the number of moles of product (NH 3 ) will increase and the moles of H and/or N will decrease. A shift towards the reactant side will increase the moles of reactants and decrease the moles of product. 5
. In rationalizing the shifts, we will use LeChatelier's Principle and a comparison of Q C and K C. a. LeChatelier's Principle Whenever a stress (change in conditions) is placed on a system in equilibrium, the equilibrium will shift so as to counteract the stress. b. At equilibrium, Q C K C. Any change that causes Q C K C will initiate a shift in the equilibrium to restore the condition Q C K C. Q C [NH 3 ] [N ][H ] 3 K C at equilibrium. B. See how the number of moles of N will be affected by the following changes. 1. Some H is added. a. Moles of N will decrease, equilibrium is shifted towards products. b. When H is added, Q C decreases but K C remains the same, therefore, a net reaction producing more NH 3 must occur until Q C is again equal to K C... The pressure is increased by decreasing the volume. a. Moles of N will decrease. b. LeChatelier's Principle: If P is increased by decreasing the volume, the equilibrium will shift towards the side of the equation having the fewer gas molecules. c. At equilibrium, Q C K C. This can be written Q C n N V n NH3 V n H 3 V n NH 3 n NH3 3 n N n H x V K C If V is decreased then the ratio 3 must increase so that K C is restored. n N n H Therefore, the number of moles of NH 3 must increase while the moles of N and H decrease. d. Note that, if the pressure was increased by adding an inert gas, the equilibrium would not shift, since that does not change the concentrations of the reactants or of the products. 6
3. The temperature is increased. a. The moles of N will increase. b. Recall that K for an exothermic reaction decreases as the temperature increases, while K for an endothermic reaction increases with increasing temperature. c. Since the reaction is exothermic, heat can be thought of as a product and the reaction can be written as N (g) + 3 H (g) NH 3 (g) + Heat Using LeChatlier s principle, if heat is put in the system (the temperature is increased), the equilibrium will shift in favor of the reactants. 4. A catalyst is added. Moles of N will not change. A catalyst decreases the time required to reach equilibrium but does not change the value of K. 7
GASEOUS EQUILIBRIUM PROBLEMS 1. At a particular temperature the reaction PCl 5 (g) PCl 3 (g) + Cl (g) has a value K C 0.51. Calculate the equilibrium concentrations of PCl 5, PCl 3, and Cl when 3.000 moles of PCl 5 are introduced into a 5.000 L container at this temperature.. At 5 C the equilibrium constant, K C, for the reaction N O 4 (g) NO (g) is equal to 0.11. Suppose that.00 moles of N O 4 and 1.50 moles of NO were introduced into a 5.00 L container at 5 C. Calculate their equilibrium concentrations. 3. At 5 C the value of K C for the reaction NOCl(g) NO(g) + Cl (g) is equal to 0.080. Calculate K P for the reaction at 5 C. 4. Consider the endothermic reaction N (g) + O (g) NO(g) in a state of equilibrium. Indicate how the number of moles of O will be affected by the following changes a. N is added. b. The temperature is decreased. c. The pressure is decreased by increasing the volume. d. A catalyst is added. 5. Which of the changes listed in Problem 4 will lead to a change in the value of K? 6. Consider the reaction PbS(s) + 3O (g) PbO(s) + SO (g). a. Write the expression for K P for the reaction. b. How would the moles of O be affected if a small amount of PbO were added to the system at equilibrium? c. How would the number of moles of O be affected if the pressure was increased by decreasing the volume? 7. Write expressions for the equilibrium constants, Kc, for each of the following: a) CO (g) + O (g) CO (g) b) 4NH 3 (g) + 7O (g) 4NO (g) + 6H O (g) c) K O (s) K O (s) + O (g) 8
d) Ni (s) + 4CO (g) Ni(CO) 4 (g) e) C (s) + CO (g) CO (g) 8. For each of the reactions in question (7), write the equation showing the relation between K p and K c. 9. At 150 C, K c for the reaction, NO (g) N (g) + O (g), is 0. The value of H for this reaction is 18 kj. a) Suppose that.0 moles of NO, 5.0 moles of N and 89.0 moles of O are mixed in a 4.0 L container at 150 C. Calculate the value of Q c when the substances are first mixed. In which direction will the reading go to reach equilibrium? b) Calculate at 150 C, the concentration of all species present at equilibrium if i. 1.0 moles of NO is placed in a 4.0 L ii..0 moles of N and 3.0 moles of O are placed in a 0.50 L container. iii. 1.0 moles of NO, 4.0 moles of N, and 5.0 moles of O are placed in a 10 L container. iv. 1.0 moles of NO,.0 moles of N and 3.0 moles of O are placed in a.0 L container. c) Suppose the above reaction is at equilibrium and the following changes are made. State how the number of moles or N present at equilibrium would be affected by each of the changes and how the equilibrium constant K c would change. i. More O (g) is added. ii. The pressure is decreased by increasing the volume. iii. The pressure is increased by adding an inert gas. decreased. v. A catalyst is added. d) What is the value of K p for this reaction at 150 C? iv. The temperature is 10 Consider the reaction C (s) + CO (g) CO (g). The value of H for this reaction is 17 kj. State how the number of moles of CO present at equilibrium will be affected by the following changes in conditions. a) The temperature is increased. b) The pressure is increased by decreasing the volume. c) More C (s) is added to the system (assume the volume does not change appreciably). 9
11. PCl 5 (g) PCl 3 (g) + Cl (g). If, at a particular temperature.4 moles of PCl 5 is placed in a 5.0 L container and, when equilibrium is attained, 0.40 moles of the PCl 5 remains. a) Calculate the equilibrium constant K c for the reaction. b) Suppose 5.0 moles of PCl 5, 4.0 moles of PCl 3 and 6.0 moles of Cl are placed in a.0 L container at the particular temperature and equilibrium is attained. Calculate the equilibrium concentrations of all species present. 10
Answers to selected questions: 1. [PCl 3 ] [Cl ] 0.354M, [PCl 5 ] 0.46M. [NO ] 0.M, [N O 4 ] 0.44M 3. K P 1.96 atm 4. a. decrease c. no change b. increase d. no change 5. b 6. a. K P P SO P 3 O b. no change c. decrease 7. a. K C [CO ] [CO] [O ] 8. b. K P K C (RT) -1 d. K P K C (RT) -3 c. K C [O ] e. K C [CO] [CO ] 9. a. 111, towards NO b. i. [NO] 0.06M, [N ] [O ] 0.11M; ii. [NO] 0.98M, [N ] 3.51M, [O ] 5.51M; iii. [NO] 0.10M, [N ] 0.40M, [O ] 0.50M; iv. [NO] 0.30M, [N ] 1.10M, [O ] 1.60M c. i. decrease; ii. no change; iii. no change; iv. decrease; v. no change d. 0 10. a. decrease b. increase c. no change 11. a..0 b. [PCl 5 ].65M, [PCl 3 ] 1.85M, [Cl ].85M 11