Equilibrium Constants: Calculations and Applications, Study notes of Chemistry

Download Equilibrium Constants: Calculations and Applications and more Study notes Chemistry in PDF only on Docsity! Chapter 6 Chemical Equilibrium Chemical Equilibrium Until Now we have assumed that a chemical reaction goes to completion as written. For example, we might suppose that when H2O is introduced into a flask with CO and sealed, it will all convert to H2 and CO2: H2O(g) + CO (g) H2(g) + CO2(g) However, we note that the reaction does not go to completion, but rather forms a certain, predictable amount of products and does not proceed further. This new stable state of the system which includes both reactants and products is called the equilibrium state. Reaction rates Kinetics of Approach to Equilibrium - Forward rate Reverse rate - Time The State of Equilibrium At equilibrium: ratefwd = raterev ratefwd = kfwd[N2O4] raterev = krev[NO2]2 For the Nitrogen dioxide - dinitrogen tetroxide equilibrium: N2O4 (g, colorless) = 2 NO2 (g, brown) kfwd[N2O4] = krev[NO2]2 kfwd [NO2]2 krev [N2O4] = = Keq 1) kfwd< krev N2 (g) + O2 (g) 2 NO(g) K = 1 x 10 -30 2) kfwd> krev 2 CO(g) + O2 (g) 2 CO2 (g) K = 2.2 x 1022 3) kfwd= krev 2 BrCl(g) Br2 (g) + Cl2 (g) K = 5 Reaching Equilibrium on the Macroscopic and Molecular Level N2O4 (g) 2 NO2 (g) Colorless Brown Equilibrium Condition Equilibrium Constant in Terms of Concentrations and Pressures Characteristics of True Chemical Equilibria • They show no macroscopic evidence of change. • They are reached through spontaneous processes. • A dynamic balance of forward and reverse processes exists within them. • They are the same regardless of the direction from which they are approached. Concentration The Haber Process N.(g) + 3 H.(g) == 2 NH,(g) Like Example 6.1 (P 195) - I The following equilibrium concentrations were observed for the Reaction between CO and H2 to form CH4 and H2O at 927oC. CO(g) + 3 H2 (g) = CH4 (g) + H2O(g) [CO] = 0.613 mol/L [CH4] = 0.387 mol/L [H2] = 1.839 mol/L [H2O] = 0.387 mol/L a) Calculate the value of K at 927oC for this reaction. b) Calculate the value of the equilibrium constant at 927oC for: H2O(g) + CH4 (g) = CO(g) + 3 H2 (g) c) Calculate the value of the equilibrium constant at 927oC for: 1/3 CO(g) + H2 (g) = 1/3 CH4 (g) + 1/3 H2O(g) Solution: a) Given the equation above: K = = = 0.0393 L2/mol2[CO] [H2]3 [CH4] [H2O] (0.387 mol/L) (0.387 mol/L) (0.613 mol/L) (1.839 mol/L)3 Like Example 6.1 (P 195) - II b) Calculate the value of the equilibrium constant at 927oC for: H2O(g) + CH4 (g) = CO(g) + 3 H2 (g) K = = = 25.45 mol2/L2 [CO] [H2]3 [CH4][H2O] (0.613 mol/L) (1.839 mol/L)3 (0.387 mol/L) (0.387 mol/L) This is the reciprocal of K: 1 K = = 25.45 mol2/L2 1 0.0393 L2/mol2 c) Calculate the value of the equilibrium constant at 927oC for: 1/3 CO(g) + H2 (g) = 1/3 CH4 (g) + 1/3 H2O(g) K = = [CO]1/3 [H2] [H2O]1/3[CH4]1/3 (0.387mol/L)1/3 (0.387 mol/L)1/3 (0.613 mol/L)1/3 (1.839 mol/L) K = = 0.340 L2/3/mol2/3 = (0.0393L2/mol2)1/3 (0.729) (0.729) (0.850)(1.839) Equilibrium Expressions Involving Pressures For a reaction of the type jA + kB = lC+mD It is sometimes convenient to write the equilibrium expression in terms of partial pressures, e.g. The the Ps indicate the partial pressures of the species in equilibrium and KP is a constant called the equilibrium constant in terms of partial pressures. ( )( ) ( )( )kBjA m D l C P PP PP K = Calculations using the Equilibrium Constant What We Can Learn About a Reaction from Its Equilibrium Constant 1. The tendency of the reaction to occur: – A value of K > 1 favors products – A value of K < 1 favors reactants – However, the value of K says nothing about the speed of the reaction. 2. Whether a given set of concentrations represents an equilibrium condition. 3. The equilibrium position that will be reached for a given set of initial conditions. The Reaction Quotient, Q [ ] [ ] [ ] [ ] ions.concentrat initial indicate zero of subscripts thewhere as defined is Q quotient, reaction The )( D)(C ) B( )A( :reaction heConsider t 00 00 kj ml BA DC Q gmglgkgj = +=+ Example 6.2 (P 202) - II b) We calculate the value of Q: [NH3]02 [N2]0[H2]03 Q = = = 6.01 x 10-2 L2/mol2 (2.00 x 10-4 mol/L)2 (1.50 x 10-5 mol/L) (3.54 x 10-1 mol/L)3 In this case Q = K, so the system is at equilibrium. No shift will occur. c) The value of Q is: [NH3]02 [N2]0[H2]03 Q = = = 2.0 x 10-3 L2/mol2 (1.0 x 10-4 mol/L)2 (5.0 mol/L) (1.0 x 10-2 mol/L)3 Here Q is less than K, so the system will shift to the right, attaining equilibrium by increasing the concentration of the product and decreasing the concentrations of the reactants. More Ammonia! Solving Equilibrium Problems • Write the balanced equation for the reaction. • Write the equilibrium expression. • List the initial concentrations. • Calculate Q and determine the direction of shift to equilibrium. • Define the change needed to reach equilibrium and define the equilibrium concentrations. • Substitute the equilibrium concentrations into the equilibrium expression and solve for the unknown. • Check the solution by calculating K and making sure it is identical to the original K. Determining Equilibrium Concentrations from K–I Problem: One laboratory method of making methane is from carbon disulfide reacting with hydrogen gas, and K this reaction at 900°C is 27.8. CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2 S(g) At equilibrium the reaction mixture in a 4.70 L flask contains 0.250 mol CS2, 1.10 mol of H2, and 0.45 mol of H2S, how much methane was formed? Plan: Write the reaction quotient, and calculate the equilibrium concentrations from the moles given and the volume of the container. Use the reaction quotient and solve for the concentration of methane. Solution: CS2 (g) + 4 H2 (g) CH4 (g) + 2 H2 S(g) K = = 27.8 [CH4] [H2S]2 [CS2] [H2]4 [CS2] = 0.250 mol 4.70 L [CS2] = 0.05319 mol/L Determining Equilibrium Concentrations from Initial Concentrations and K–II Concentration (M) H2 F2 HF Initial 2.000 2.000 2.000 Change -x -x +2x Final 2.000-x 2.000-x 2.000+2x K = = 115 = =[HF] 2 [H2][F2] (2.000 + 2x)2 (2.000 - x)(2.000 - x) (2.000 + 2x)2 (2.000 - x)2 Taking the square root of each side we get: (115)1/2 = =10.7238(2.000 + 2x) (2.000 - x) x = 1.528 [H2] = 2.000 - 1.528 = 0.472 M [F2] = 2.000 - 1.528 = 0.472 M [HF] = 2.000 + 2(1.528) = 5.056 M K = = [HF] 2 [H2][F2] (5.056 M)2 (0.472 M)(0.472 M) K = 115check: Using the Quadratic Formula to Solve for the Unknown Given the Reaction between CO and H2O: Concentration (M) CO(g) + H2O(g) CO2(g) + H2(g) Initial 2.00 1.00 0 0 Change -x -x +x +x Equilibrium 2.00-x 1.00-x x x Qc = = = = 1.56 [CO2][H2] [CO][H2O] (x) (x) (2.00-x)(1.00-x) x2 x2 - 3.00x + 2.00 We rearrange the equation: 0.56x2 - 4.68x + 3.12 = 0 ax2 + bx + c = 0 quadratic equation: x = - b + b 2 - 4ac 2a x = = 7.6 M and 0.73 M 4.68 + (-4.68)2 - 4(0.56)(3.12) 2(0.56) [CO] = 1.27 M [H2O] = 0.27 M [CO2] = 0.73 M [H2] = 0.73 M Predicting Reaction Direction and Calculating Equilibrium Concentrations –I Problem: Two components of natural gas can react according to the following chemical equation: CH4(g) + 2 H2S(g) CS2(g) + 4 H2(g) In an experiment, 1.00 mol CH4, 1.00 mol CS2, 2.00 mol H2S, and 2.00 mol H2 are mixed in a 250 mL vessel at 960°C. At this temperature, K = 0.036. (a) In which direction will the reaction go? (b) If [CH4] = 5.56 M at equilibrium, what are the concentrations of the other substances? Plan: The find the direction, we calculate Qc using the calculated concentrations from the data given, and compare it with Kc. (b) Based upon (a), we determine the sign of each component for the reaction table and then use the given [CH4] at equilibrium to determine the others. Solution: [CH4] = = 4.00 M 1.00 mol 0.250 L [H2S] = 8.00 M, [CS2] = 4.00 M and [H2 ] = 8.00 M Le Chatelier’s Principle “If a change in conditions (a “stress”) is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to reduce that change in conditions.” (1884) A + B C + D + Energy For example: In the reaction above, if more A or B is added you will force the reaction to produce more product, if they are removed, it will force the equilibrium to form more reactants. If C or D is added you will force the reaction to form more reactants, if they are Removed from the reaction mixture, it will force the equilibrium to Form more products. If it is heated, you will get more reactants, and if cooled, more products. Henri Louis Le Chatelier Source: Photo Researchers The Haber Process N2(g) + 3 H2(g) 2 NH3(g) + Energy The Effect of a Change in Pressure How would you change the total pressure of each of the following reactions to increase the yield of the products: (a) CaCO3 (s) = CaO(s) + CO2 (g) (b) S(s) + 3 F2 (g) = SF6 (g) (c) Cl2(g) + I2(g) = 2 ICl (g) Exothermic Reactions • Releases heat upon reaction. ∆H is negative. • Addition of heat to an exothermic reaction shifts the equilibrium to the left. • The value of K decreases in consequence. Endothermic Reactions • Absorbs heat upon reaction. ∆H is positive. • Addition of heat to an endothermic reaction shifts the equilibrium to the right. • The value of K increases in consequence. Shifting the N2O4(g) and 2NO2(g) equilibrium by changing the temperature al be