Lecture Notes on Chemical Equilibrium - General Chemistry | CHEM 142, Study notes of Chemistry

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Chapter 6 Chemical Equilibrium Chapter 6: Chemical Equilibrium 6.1 The Equilibrium Condition 6.2 The Equilibrium Constant 6.3 Equilibrium Expressions Involving Pressures 6.4 The Concept of Activity 6.5 Heterogeneous Equilibria 6.6 Applications of the Equilibrium Constant 6.7 Solving Equilibrium Problems 6.8 LeChatelier’s Principle 6.9 Equilibria Involving Real Gases Concentration vs. Time CO(g) + H2O(g) = CO2(g) + H2(g) 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 The equilibrium we are going to consider ts: CoCl4 (aq) + 6H2O(l) <> blue [Co(H0)¢}aq) + 4CI (aq) r s per 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. Characteristics of the Equilibrium Expression • The equilibrium expression for a reaction written in reverse is the reciprocal of that for the original expression. • If the original reaction is multiplied by a factor n, the new equilibrium constant is the original raised to the power n. • The apparent units for K are determined by the powers of the concentration terms. 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/L21 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) (Apparent) Units for K and KP • At first sight, it would seem that the units for K would be in concentrations raised to a reaction-specific power, and those for KP in pressure units raised to a reaction-specific power. But these are ‘apparent’ units. • For theoretical reasons, we will refer each concentration or pressure to a reference state, which always causes the units of concentration or pressure to cancel. Thus, K and KP are expressed without units. Calculations Using 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. Example 6.2 (P 202) - I For the synthesis of ammonia at 500oC, the equilibrium constant is 6.0 x 10-2 L2/mol2. Predict the direction in which the system will shift to reach equilibrium in each of the following cases. a) [NH3]0 = 1.0 x 10-3 M; [N2]0= 1.0 x 10-5 M; [H2]0=2.0 x 10-3 M b) [NH3]0 = 2.00 x 10-4 M; [N2]0= 1.50 x 10-5 M; [H2]0= 3.54 x 10-1 M c) [NH3]0 = 1.0 x 10-4 M; [N2]0= 5.0 M; [H2]0= 1.0 x 10-2 M Solution a) First we calculate the reaction quotient Q: Q = = = 1.3 x 107 L2/mol2 [NH3]02 [N2]0[H2]03 (1.0 x 10-3 mol/L)2 (1.0 x 10-5 mol/L)(2.0 x 10-3 mol/L)3 Since K = 6.0 x 10-2 L2/mol2, Q is much greater than K. For the system to attain equilibrium, the concentrations of the products must be decreased and the concentrations of the reactants increased. The system will shift to the left: 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 Initial Concentrations and K –I Problem: Given the that the reaction to form HF from molecular hydrogen and fluorine has a reaction quotient of 115 at a certain temperature. If 3.000 mol of each component is added to a 1.500 L flask, calculate the equilibrium concentrations of each species. H2 (g) + F2 (g) 2 HF(g) Plan: Calculate the concentrations of each component, and then figure the changes, and solve the equilibrium equation to find the resultant concentrations. Solution: K = = 115[HF] 2 [H2] [F2] [H2] = = 2.000 M 3.000 mol 3.000 mol 3.000 mol [F2] = = 2.000 M [HF] = = 2.000 M 1.500 L 1.500 L 1.500 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 –III x = 1.56 M = [CH4] Therefore: [H2S] = 8.00 M + 2x = 8.00 M + 2(1.56 M) = 11.12 M [CS2] = 4.00 M - x = 4.00 M - 1.56 M = 2.44 M [H2] = 8.00 M - 4x = 8.00 M - 4(1.56 M) = 1.76 M [CH4] = 1.56 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 Effect of a Change in Concentration–I Given an equilibrium equation such as : CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) If one adds ammonia to the reaction mixture at equilibrium, it will force the reaction to go to the right producing more product. Likewise, if one takes ammonia from the equilibrium mixture, it will force the reaction back to produce more reactants by recombining H2 and HCN to give more of the initial reactants, CH4 and NH3. CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) Add NH3 Forces equilibrium to produce more product. Forces the reaction equilibrium to go back to the left and produce more of the reactants. Remove NH3 The Effect of a Change in Concentration–II CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) If to this same equilibrium mixture one decides to add one of the products to the equilibrium mixture, it will force the equilibrium back toward the reactant side and increase the concentrations of reactants. Likewise, if one takes away some of the hydrogen or hydrogen cyanide from the product side, it will force the equilibrium to replace it. CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) CH4 (g) + NH3 (g) HCN(g) + 3 H2 (g) Add H2Forces equilibrium to go toward the reactant direction. Remove HCNForces equilibrium to make more produce and replace the lost HCN. The Effect of a Change in Concentration Consider the following reaction: 2 H2S(g) + O2(g) = 2 S(s) + 2 H2O(g) What happens to: (a) [H2O] if O2 is added? (b) [H2S] if O2 is added? (c) [O2] if H2S is removed? (d) [H2S] if S is added? Figure 6.9: Brown NO2(g) and colorless N2O4(g) at equilibrium in a syringe Source: Ken O’Donoghue 2 NO2 (g) N2O4 (g) Brown Colorless 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. ΔH 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. ΔH is positive. • Addition of heat to an endothermic reaction shifts the equilibrium to the right. • The value of K increases in consequence. The Effect of a Change in Temperature on the Position of Equilibrium How does an increase in temperature affect the equilibrium concentration of the indicated substance and K for the following reactions: (a) CaO(s) + H2O (l) = Ca(OH)2 (aq) ΔH0 =-82 kJ (b) (a) CaCO3 (s) = + CaO(s) + CO2 (g) ΔH0 = 178 kJ (c) SO2 (g) = S(s) + O2(g) ΔH0 = 297 kJ Predicting the Effect of a Change in Concentration on the Position of the Equilibrium Problem: Carbon will react with water to yield carbon monoxide and and hydrogen, in a reaction called the water gas reaction that was used to convert coal into a fuel that can be used by industry. C(s) + H2O (g) CO(g) + H2 (g) What happens to: (a) [CO] if C is added? (c) [H2O] if H2 is added? (b) [CO] if H2O is added? (d) [H2O] if CO is removed? Plan: We either write the reaction quotient to see how equilibrium will be effected, or look at the equation, and predict the change in direction of the reaction, and the effect of the material desired. Solution: (a) No change, as carbon is a solid, and not involved in the equilibrium, as long as some carbon is present to allow the reaction. (b) The reaction moves to the product side, and [CO] increases. (c) The reaction moves to the reactant side, and [H2O] increases. (d) The reaction moves to the product side, and [H2O] decreases. Predicting the Effect of Temperature and Pressure Problem: How would you change the volume (pressure) or temperature in the following reactions to increase the chemical yield of the products? (a) 2 SO2 (g) + O2 (g) 2 SO3 (g); H0 = 197 kJ (b) CO(g) + 2 H2 (g) CH3OH(g); H0 = -90.7 kJ (c) C(s) + CO2 (g) 2 CO(g); H0 = 172.5 kJ (d) N2(g) + 3 H2(g) 2 NH3(g); H0 = -91.8 kJ Plan: For the impact of volume (pressure), we examine the reaction for the side with the most gaseous molecules formed. For temperature, we see if the reaction is exothermic, or endothermic. An increase in volume (pressure) will force a reaction toward fewer gas molecules. Solution: To get a higher yield of the products you should: (a) Increase the pressure, and increase the temperature. (b) Increase the pressure, and decrease the temperature. (c) A pressure change will not change the yield, an increase in the temperature will increase the product yield. (d) Increase the pressure, and decrease the temperature.