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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 H2CO3(l) H2O(l) + CO2(g) 1 The Equilibrium Constant: K (Temperature Dependent) (mechanism independent) aA + bB dD + eE Kc [products] [reactants Kc raised to their coeff . ] raised to their coeff . [D] d a [E] [A] [B] e b Law of Mass Action: Values of Kc are constant for a RXN at a given temperature. Any equilibrium mixture of the above system at that temperature should give the SAME Kc value. 2 We also have Kp 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) Kp (P D ) (P E ) d e a b (P A ) (P B ) Since PV = nRT (ideal gas law) n V P RT moles concentrat Liters n ion and .. P RT V so pressure is proportional to concentration. K P K c RT n 0.0821L•atm mol•K moles of GAS temp in K 3 Problem: PCl3(g) + Cl2(g) PCl5(g) In a 5.00 L vessel (@ 230oC) an equilibrium mixture is found to contain: 0.0185 mol PCl3, 0.0158 mol PCl5 and 0.0870 mol of Cl2. Question: determine Kc and Kp Solution: Kc [PC l5 ] [PC l3 ][C l2 ] 0.0158 m ol 5.00 L 0.0185 m ol 0.0870 m ol 5.00 L 5.00 L Kc = 49.08 K P K c (RT) n 49.08(0.08 21 503) -1 one mol less gas on product side. 1.19 Question: what does the value of K mean? 4 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 H2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of Kc? CO(g) + 2H2(g) CH3OH(g) 0.3000 0 I. 0.1500 C. -x -2x +x x 0.3000 - 2x 0.1500 - x E. Since @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187 Therefore x = 0.0313 We can now solve for each of the other Eq. terms. 5 Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of Kc? I. C. CO(g) + 2H2(g) CH3OH(g) 0.3000 0 0.1500 -x -2x +x x 0.3000 - 2x 0.1500 - x E. Since @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187 Therefore x = 0.0313 We can now solve for each of the other Eq. terms. H2: 0.3000 - 2x = 0.2374 moles CH3OH: x = 0.0313 moles CO(g) + 2H2(g) CH3OH(g) Therefore E. 0.1187 0.2374 0.0313 6 Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of Kc? CO(g) + 2H2(g) CH3OH(g) 0.3000 0 0.1500 I. C. -x E. 0.1500 - x +x 0.3000 - 2x x CO(g) + 2H2(g) CH3OH(g) 0.1187 0.2374 0.0313 Therefore E. Now find Kc: -2x Kc [C H 3 O H] [C O ][H 2 ] 2 0.0313 1.500 0.1187 0.2374 1.500 1.500 2 Kc = 10.52 7 More I.C.E. Problem: @ 77oC, 2.00 mol of NOBr (nitrosyl bromide) placed in a 1.000L flask dissociates to the extent of 9.4%. Find Kc. C. 2NOBr(g) 2.00 -2x E. 2.00 - 2x I. 2NO(g) + 0 +2x 2x Br2(g) 0 +x x Now what? Since 9.4% dissociates, the Change in NOBr, -2x = 2.00(0.094) and.... |x| = 0.0940 so substituting the x value into the “E. term” gives: 2NOBr(g) E. 1.812 2NO(g) + 0.188 Br2(g) 0.0940 8 More I.C.E. Problem: @ 77oC, 2.00 mol of NOBr (nitrosyl bromide) placed in a 1.000L flask dissociates to the extent of 9.4%. Find Kc. I. C. E. 2NOBr(g) 2.00 -2x 2NO(g) + 0 +2x Br2(g) 0 +x 2.00 - 2x 2NOBr(g) 2x 2NO(g) + x Br2(g) E. 1.812 0.188 0.0940 2 Kc [NO] 2 [Br 2 ] [N OBr] 2 0.188 0.0940 1 1 1.812 1 2 What does the value of Kc tell us? = 1.01 x 10-3 9 Treatment of Pure Solids and Liquids (as solvents) in K expressions. S(s) + O2(g) SO2(g) would expect Kc = [SO2] [S(s)][O2] but since M is meaningless for solids, solids are dropped out. and... Kc [S O2 ] [O2 ] Experimentally Kc is found to be 4.2 x 1052 @ 25oC and independent of S. 10 AgCl(s) Ag+(aq) + Cl-(aq) Ksp = [Ag+(aq)][Cl-(aq)] = 1.8 x 10-10 @ 25oC This is an EQUILIBRIUM value independent of the amount of solid AgCl left sitting on the bottom of the container. 11 PURE LIQUIDS (SOLVENTS) NH3(g) + H2O(l) NH4+(aq) + OH-(aq) K [NH 4 ][O H ] 1.8 x 10 -5 [NH 3 ] Note: by convention the water is ignored. HCOOH(aq) + H2O(l) HCOO-(aq) + H3O+(aq) K [HC O O ][H 3 O ] 1.8 x 10 4 [HC O O H] 12 CaCO3(s) CaO(s) + CO2(g) What is the Kc expression? Kc = [CO2] What is the Kp expression? Kp = PCO2 13 Reversing Equations: Reactants become Products and Products become Reactants. What is the relationship between the two K values? 1. 2H2(g) + O2(g) 2H2O(g) K1 [H 2 O ] 2 2 3.4 x 10 81 [H 2 ] [O 2 ] 2. 2H2O(g) 2H2(g) + O2(g) 2 [H 2 ] [O 2 ] 82 K2 2.9x10 2 [H 2 O ] Relationship: K2 = 1/K1 = K1-1 14 Knet for summing RXN’s: If a RXN can be obtained from the sum of RXN’s, KRXN = K1K2 RXN 1: S(s) + O2 (g) SO2(g) K 1 [SO 2 ] [SO 3 ] 1/2 K net [SO 3 ] [O 2 ] K 1K 2 [O 2 ] x 2.6 x 10 [SO 2 ][O 2 ] Net RXN: S(s) + 3/2O2(g) SO3(g) [SO 2 ] 52 [O 2 ] K2 RXN 2: SO2(g) + 1/2O2(g) SO3(g) 4.2 x 10 [SO 3 ] 1/2 [SO 2 ][O 2 ] KRXN = K1K2 3/2 1.1 x 10 [SO 3 ] [O 2 ] 3/2 15 65 12 The RXN Quotient: Qc Consider a system that may not yet be @ Equilibrium. aA + bB dD + eE Qc [D] d a [E] [A] [B] If Qc = Kc ? If Qc < Kc ? If Qc > Kc ? e b @ Equilibrium ratio ratio Prod. React. Prod. is too small so RXN is too large so RXN React. 16 Example: PCl5(g) PCl3(g) + Cl2(g) @ 250oC Kc= 4.0 x 10-2 If: [Cl2] and [PCl3] = 0.30M and [PCl5] = 3.0M, is the system at Equilibrium? If not, which direction will it proceed? Qc = [PC l 3 ][C l 2 ] [0.30][0.3 0] -2 = 3.0 x 10 Find Qc and compare to Kc to decide. [PC l 5 ] [3.0] Qc < Kc (not @ Eq.) Which way must the RXN go to achieve Equilbrium? Remember ratio is prod./React more products makes the number bigger RXN goes 17 Calculations Using Kc: (1st case....Perfect Square) H2(g) + I2(g) 2HI(g) @ 699K Kc = 55.17 Experiment: 1.00 mol of each H2 and I2 in a 0.500 L flask. Find [ ] of products and reactants @ Equilibrium. [HI] C. H2(g) + I2(g) 2HI(g) [ ] [ ] [ ] 0 2.00 2.00 2x -x -x [H2 ][I2 ] E. 2.00 -x [2x] conc. in mol/L I. 2 Kc 2 Kc 2.00 - x 2x 2 [2x] [2.00- x][2.00- x] 2 = 55.17 [2.00- x] “perfect square” 18 Calculations Using Kc: (1st case....Perfect Square--continued) H2(g) + I2(g) 2HI(g) @ 699K H2(g) + I2(g) 2HI(g) [ ] [ ] [ ] 2x 2.00 -x 2.00 - x 2 Kc [HI] [H2 ][I2 ] E. 2 Kc Kc = 55.17 2 [2x] [2x] [2.00- x][2.00- x] 2 = 55.17 [2.00- x] “perfect square” 55.17 [2x] [2.00- x] 7.428(2.00 - x) = 2x 1.58 = x 19 Calculations Using Kc: (1st case....Perfect Square) H2(g) + I2(g) 2HI(g) @ 699K 2 Kc [HI] [H2 ][I2 ] E. Kc = 55.17 H2(g) + I2(g) 2HI(g) [ ] [ ] [ ] 2x 2.00 -x 2.00 - x 7.428(2.00 - x) = 2x 1.58 = x [H2] = 2.00 - 1.58 = 0.42M [I2] = 0.42M [HI] = 2(1.58) = 3.16M 20 Kc Problems with Quadratic Equation ax2 + bx +c = 0 x b 2 b 4ac 2a If equilibrium expression is not a perfect square must use quadratic equation. 21 Problem: H2(g) + I2(g) 2HI(g) @ 458 oC Kc = 49.7 Experiment: 1.00 mol H2, 2.00 mol I2 in a 1.00 L flask. Find: conc. of the equilibrium mixture. H2(g) + I2(g) 2HI(g) [ ] [ ] [ ] 2.00 0 I. 1.00 -x 2x -x C. 2x 2.00 - x E. 1.00 - x K 2 Kc [HI] [H2 ][I2 ] 2 K c 49.7 [2x] [1.00 - x][2.00 - x] 0.920x2 - 3.00x + 2.00 = 0 22 Problem: H2(g) + I2(g) 2HI(g) @ 458 oC Kc = 49.7 Experiment: 1.00 mol H2, 2.00 mol I2 in a 1.00 L flask. Find: conc. of the equilibrium mixture. 2 Kc [HI] [H2 ][I2 ] I. C. 2 K c 49.7 [2x] [1.00 - x][2.00 - x] E. H2(g) + [ ] 1.00 -x 1.00 - x I2(g) 2HI(g) [ ] [ ] 2.00 0 2x -x 2.00 - x 2x 0.920x2 - 3.00x + 2.00 = 0 2 solutions for x: 1.63 0.70 x = 2.33 or 0.93 will give positive solution for Eq. Conc. 23