Chemical Equilibrium - General Chemistry - Slides | CHEM 142, Study notes of Chemistry

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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 OMIT: Equilibria Involving Real Gases Next HW due Weds. at usual time, on Ch. 6 1 Molecular Picture of Establishment of Equilibrium CO(g) + H2O(g) = CO2(g) + H2(g) INITIAL 7 CO(g) + 7 H2O(g) + 0 CO2(g) + 0 H2(g) AFTER IT STOPS CHANGING 2 CO(g) + 2 H2O(g) + 5 CO2(g) + 5 H2(g) 2 Characteristics of Chemical Equilibrium States • Reaching equilibrium requires reactions to occur. • Once reached, they show no macroscopic evidence of further change. • Reached through dynamic balance of forward and reverse reaction rates. 5 Figure 5.21: The collision rate of gas particles defines the maximum blue-pink reaction rate! Z = Collision rate (of one pink with blues)= # collisions with blues per s = 4 [Nblue/V] d2 (πRT/M)1/2 = 4 [blues] d2 (πRT/M)1/2 6 Equilibrium arises through dynamic balance between forward and back reactions Forward rate = k1[NO2][NO2]=k1[NO2]2 Back rate = k-1[NO3][NO] k1 and k-1 reflect probabilities that one collision leads to reaction Reverse: NO3(g) + NO(g) → 2NO2(g) k -1 7 Reaching Equilibrium on the Macroscopic and Molecular Level N2O4 (g) 2 NO2 (g) Colorless Brown K = [NO2]2 / [N2O4] Units are mol/L. 10 Equilibrium from Different Starting Points CO(g) + 2 H2(g) ↔ CH3OH(g) [ ] [ ][ ] 3 2 2 has same value at equil. for all 3 starting points CH OH K CO H = 11 N2(g) + 3 H2(g) 2 NH3(g) Fe K = [NH3] 2 [N2][H2]3 12 Small K N2 (g) + O2 (g) 2 NO(g) K = 1 x 10 –30 Essentially only reactants at eqbm. (1015 x products) Intermediate K 2 BrCl(g) Br2 (g) + Cl2 (g) K = 5 Comparable amounts of products and reactants at eqbm. Large K 2 CO(g) + O2 (g) 2 CO2 (g) K = 2.2 x 1022 Essentially only products at eqbm. Equilibrium Constants can have a wide range of values 15 CS2(g) + 3 O2(g) CO2(g) + 2 SO2(g) • The equilibrium expression for a reaction written in reverse is the reciprocal of that for the original expression. CO2(g) + 2 SO2(g) CS2(g) + 3 O2(g) • If the original reaction is multiplied by a factor n, the new equilibrium constant is the original raised to the power n. 2 CS2(g) + 6 O2(g) = 2 CO2(g) + 4 SO2(g) [ ][ ] [ ][ ]322 2 22 1 OCS SOCO =K [ ][ ] [ ][ ] 1222 3 22 2 1 SOCO OCS K K == [ ] [ ] [ ] [ ] ( ) 2 4 22 2 3 12 6 2 2 CO SO CS O K K= = 16 Example: Calculation of the Equilibrium Constant from equilibrium amounts At 454 K, the following reaction takes place: 3 Al2Cl6(g) 2 Al3Cl9(g) At this temperature, the equilibrium concentration of Al2Cl6(g) is 1.00 M and the equilibrium concentration of Al3Cl9(g) is 1.02 x 10-2 M. Compute the equilibrium constant at 454 K. Strategy: Substitute values into K [ ] [ ] ( ) ( ) 14 3 22 3 62 2 93 1004.1 00.1 1002.1 ClAl ClAl −−− ×=×== M M MK 17 Like Example 6.1 (P 201) - II b) Calculate the value of the equilibrium constant at 927oC for: H2O(g) + CH4 (g) = CO(g) + 3 H2 (g) K = = [CO] [H2]3 [CH4][H2O] Easier: This K is just the reciprocal of K from part a: 1 Ka K = 1 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 Easier: K = (K from part a)1/3 20 Like Example 6.1 (P 201) - 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) 1 Ka 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) Easier: K = (Ka)1/3 = (0.0393L2/mol2)1/3 Easier: This K is just the reciprocal of K from part a: 21 Determining Equilibrium Concentrations from K Example: Methane can be made by reacting carbon disulfide with hydrogen gas. K for this reaction is 27.8 (L/mol)2 at 900°C. 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? Strategy: (1)Calculate the equilibrium concentrations from the moles given and the volume of the container. (2)Use the value of K to solve for the concentration of methane. (3) Calculate the number of moles of methane from M and V. 22 Expressing K with Pressure Units For gases, PV=nRT can be rearranged to give: P = RTn V or: =n P V RT Since = Molar concentration of gasn V For an equilibrium between gaseous compounds, there is a relationship between the equilibrium constants K and KP. Example: 2 NO (g) + O2 (g) 2 NO2 (g) Kp = P 2NO2 P 2NO PO2 [NO2]2 [NO]2 [O2] K = and Using concentrations Using pressures 25 How is KP related to K? Answer: Through the use of the ideal gas law. RTCRTRTnVP AAA =⎟ ⎠ ⎞ ⎜ ⎝ ⎛== V nor P AA ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) c d c d C D C D P a b a b A BA B c d c d C D c d a b n a b a b A B P P C RT C RT K C RT C RTP P C C RT K RT K RT RTC C + + − + Δ + × × = = × × = × = = Where Δn = c+d-(a+b) Note that K =KP when Δn = 0 26 For reaction 2NO(g) + O2(g) 2NO2(g) Sum of coefficients (+ for products, – for reactants) Δn = 2 – 2 – 1 = – 1 [ ] [ ] [ ] 2 2 2 2 NO NO O K = 2 2 2 NO 2 NO O P P K P P = 1( )P KK K RT RT −= = 27 CaCO3 (s) CaO (s) + CO2 (g) KP = PCO2 (the 2 solids do NOT appear in KP!) The position of an equilibrium involving a pure solid or liquid does not depend upon the amounts of pure solid or liquids present. The activity of a pure solid or liquid is always equal to 1 if present, so its “concentration” does not appear in K expression. For pure solids and liquids, activity = 1.0 30 Equilibrium involving pure solids or liquids: set their activity = 1 Example: NH4NO2(s) = N2(g) + 2 H2O(g) The equilibrium constant for this reaction would normally be expressed as: [ ][ ] [ ] 2 2 2 4 2 N H O ' NH NO K = 4 2 However, a pure solid or liquid retains the same activity during the reaction. Thus we set the activity of NH NO ( ) to one. s [ ][ ] ( )( )2 2 22 2 2 N H ON H O and P PpK K= = 1.0000 1.0000 31 The Reaction Quotient, Q [ ] [ ] [ ] [ ] Consider the reaction: a A( ) B( ) C( ) D( ) The reaction quotient, Q is defined as where the subscripts indicate momentary concentrations at some time t before (or after) equi c d t t a b t t g b g c g dm g C D Q A B t + = + = librium has established. Q has same form as K, but the concentrations are the actual rather than the equilibrium concentrations 32 Reaction Direction and the Relative Sizes of Q and K “Excess reactants initially” “Excess products initially” 35 Which Direction? • If Q < K, the system will shift to the right by converting reactants to products. • If Q = K, the system is at equilibrium; the concentrations will not change. • If Q > K, the system will shift to the left by converting products back to reactants. 36 Example: Predict Reaction Direction using Q For the following reaction: CH4(g) + 2 H2S(g) = CS2(g) + 4 H2(g) 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 960oC. At this temperature, K = 0.036. In which direction will the reaction proceed in order to reach equilibrium? Strategy: 1) Calculate the actual concentrations 2) Calculate Q. 3) Compare Q and K 37 Example: Calculating Equilibrium pressures and concentrations from K and initial conditions. Consider the equilibrium: CO(g) + H2O(g) = CO2(g) + H2(g) 0.250 mol CO and 0.250 mol H2O are placed in a 125 mL flask at 900 K. What is the composition of the equilibrium mixture if K = 1.56? The original reactant concentrations are: [CO]0 = [H2O]0 = 0.250 mol/ 0.125 L = 2.00 M Q = 0.Therefore, Q < K, so reactants are consumed and products made. 40 CO(g) + H2O(g) = CO2(g) + H2(g) Construct the reaction table: 002.002.00Init. [i] H2(g)CO2(g)H2O(g)CO(g)Conc. (M) 41 +x+x-x-xChange = x times coefficient = neg. for reactants 002.002.00Init. [i] H2(g)CO2(g)H2O(g)CO(g)Conc. (M) CO(g) + H2O(g) = CO2(g) + H2(g) 42 [ ] [ ] [ ] [ ] 2 2 2 Calculating equilibrium concentrations: CO H O 2.00 2.00 M - 1.11 M 0.89 M CO H 1.11 M x x = = − = = = = = [ ][ ] [ ][ ] 2 2 2 Check results: CO H 1.11 1.11 1.6 1.56 CO H O 0.89 0.89 K ×= = = ≈ × 45 Solving Equilibrium Problems • Write the balanced equation for the reaction. • Write the equilibrium expression K. • List the initial concentrations, [i]initial. • Calculate Q and determine the direction of shift to equilibrium. • Define change in [i] to reach equilibrium: Change = x times the stoich. coefficient • Express equilibrium concentrations in terms of x. • Substitute the equilibrium concentrations into the equilibrium expression for K. • Solve for x and calculate [i]eqbm for all i. • Check the solution by calculating K and making sure it is identical to the original K. 46 Like Example 6.3 (P203-5) - I Calculate the equilibrium concentrations when Hydrogen Chloride gas is made from Hydrogen gas and Chlorine gas. Initially 4.000 mol of H2, and 4.000 mol of Cl2, are added to 2.000 mol of gaseous HCl in a 2.000 liter flask. H2 (g) + Cl2 (g) 2 HCl(g) K = 2.76 x 102 = [Cl2] = [H2] = 4.000mol/2.000L = 2.000M [HCl] = 2.000 mol/2.000L = 1.000M Initial Concentration Change Equilibrium Conc. (mol/L) (mol/L) (mol/L) [H2]o = 2.000M -x [H2] = [Cl2]o = 2.000M [Cl2] = [HCl]o = 1.000M [HCl] = [HCl]2 [H2] [Cl2] 47 Like Example 6.3 (P203-5) - II K = 2.76 x 102 = = = [HCl]2 [H2] [Cl2] (1.000 + 2x)2 (2.000 –x)(2.000 – x) (1.000 + 2x)2 (2.000 – x)2 Take the square root of each side: 16.61 = (1.000 + 2x) (2.000 – x) 33.22 – 16.61x = 1.000 + 2x Therefore: [H2] = 0.269 M 32.22 = 18.61x [Cl2] = 0.269 M x = 1.731 [HCl] = 4.462 M Check: = = 276 OK! [HCl]2 [H2] [Cl2] (4.462)2 (0.269)(0.269) 50 Example: Calculate equilibrium partial pressures using exact solution of Quadratic Equation -4 2 2 The reaction between nitrogen and oxygen to form nitric acid proceeds according to the reaction: N (g) + O (g) 2NO(g) K 4.10 10 at 2000. K= × 2 20.500 moles of N and 0.860 moles of O are put into a 2.00 L vessel initially. Calculate the partial pressure of all the species at equilibrium. 51 2 2 Initial concentration of N = 0.500 mol 2.00 L = 0.250 M, Initial concentration of O = 0.860 mol 2.00 L = 0.430 M, Initial concentration of NO 0 Q =0; therefore the reaction proceeds to the right Construct = the reaction table: 00.4300.250Initial NO(g)Molarity 2N (g) 2O (g) 2 2N (g) + O (g) 2NO(g) 52 [ ] [ ][ ] ( ) ( )( ) 2 2 4 2 2 Substituting the equilibrium concentrations from the table into the equilibrium expression: NO 2 4.10 10 N O 0.250 0.430 x K x x −= = = × − − 2 2 This is a quadratic equation of the general form 0 where the (two) roots can be obtained from the quadratic formula: 4x 2 ax bx c b b ac a± + + = − ± − = 2 4 5 This expression simplifies to 4.00 2.78 10 4.41 10 0x x− −+ × − × = 55 [ ] -3 -3 We obtain two possible solutions x 3.35 10 and x 3.28 10 Since only the positive root leads to all positive concentrations, we ignore the negative root. Calculating equilibrium concentrations: NO = − × = × [ ] [ ] [ ] [ ][ ] ( ) -3 2 2 22 -3 4 4 2 2 2 6.56 10 N 0.250 0.247 O 0.430 0.427 Check: 6.56 10NO 4.09 10 4.10 10 N O 0.247 0.427 x M M x M M x M K − − = = × = − = = − = × = = = × ≈ × × 56 Example: Solving equilibrium problems with simplifying assumptions Phosgene decomposes into CO and Cl2 when heated according to the equation. Calculate the concentration of all species at equilibrium if 5.00 moles of phosgene is placed into a 10.0 L flask -4 2 2COCl (g) CO(g) + Cl (g) K = 8.3 10 at 360 C× [ ]2 5.00 molesCOCl (g) 0.500 10.0 L M= = 57 xx0.500 - xEquil. +x+x-xChange 000.500Init. Molarity -4 2 2COCl (g) CO(g) + Cl (g) K = 8.3 10 at 360 C× 2COCl (g) CO(g) 2Cl (g) 60 [ ][ ] [ ] 2 42 2 2 2 2 4 2 2 2 2 CO Cl 8.3 10 COCl 0.500 For K to be so small, 0.500 and so 0.500 0.500 8.3 10 2.037 10 2.04 10 0.500 0.500 Exact solution is 1.996 10 2.00 10 Approximation gives error o xK x x x x x x x x x − − − − − − = = = × − − >> − ≈ ≈ = × = × → × − = × → × f 2% 61 Calculating K from Concentration Data–I Problem: Hydrogen iodide decomposes at moderate temperatures by the reaction below: When 4.00 mol HI was placed in a 5.00 L vessel at 458°C, the equilibrium mixture was found to contain 0.442 mol I2. What is the value of Kc ? Plan: First we calculate the molar concentrations, and then put them into the equilibrium expression to find it’s value. Solution: To calculate the concentrations of HI and I2, we divide the amounts of these compounds by the volume of the vessel. 2 HI(g) H2 (g) + I2 (g) Starting conc. of HI = = 0.800 M4.00 mol 5.00 L Equilibrium conc. of I2 = = 0.0884 M 0.442 mol 5.00 L Conc. (M) 2HI(g) H2 (g) I2 (g) Starting 0.800 0 0 Change Equilibrium 0.0884 62 Calculating K from Concentration Data–II [HI] = M = (0.800 - 2 x 0.0884) M = 0.623 M [H2] = x = 0.0884 M = [I2] Kc = = = 0.0201 [H2] [I2] [HI]2 ( 0.0884)(0.0884) (0.623)2 Therefore the equilibrium constant for the decomposition of Hydrogen Iodide at 458°C is only 0.0201 meaning that the decomposition does not proceed very far under these temperature conditions. We were given the initial concentrations, and that of one at equilibrium, and found the others that were needed to calculate the equilibrium constant. 65 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 66 Le Chatelier’s Principle If a change in conditions (a ‘stress’ such as change in P, T, or concentration) is imposed on a system at equilibrium, the equilibrium position will shift in a direction that tends to reduce that change in conditions. Henri Le Chatelier, 1884 67 The Effect of a Change in Concentration • If a gaseous reactant or product is added to a system at equilibrium, the system will shift in a direction to to reduce the concentration of the added component. • If a gaseous reactant or product is removed from a system at equilibrium, the system will shift in a direction to to increase the concentration of the removed component. [ ] [ ][ ] [ ][ ] [ ] 2 3 2 1 2 1 3 31 2 1 2 2NO (g) NO(g) + NO (g) NO NO NO NO NO NO k k k k − − = = 70 Examples: 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? The reaction proceeds to the right so H2O increases. (b) [H2S] if O2 is added? Some H2S reacts with the added O2 to move the reaction to the right, so [H2S] decreases. 71 2 H2S(g) + O2(g) 2 S(s) + 2 H2O(g) (c) [O2] if H2S is removed? The reaction proceeds to the left to re-form H2S, more O2 is formed as well, O2 increases. (d) [H2S] if S(s) is added? S is a solid, so its activity does not change. Thus, [H2S] is unchanged. 72 Examples: The Effect of a Change in Pressure How would you change the total pressure or volume in the following reactions to increase the yield of the products: (a) CaCO3(s) = CaO(s) + CO2 (g) The only gas is the product CO2. To move the reaction to the right increase the volume. (b) S(s) + 3 F2 (g) = SF6 (g) With 3 moles of gas on the left and only one on the right, we increase the pressure to form more SF6. (c) Cl2(g) + I2(g) = 2 ICl (g) The number of moles of gas is the same on both sides of the equation, so a change in pressure or volume will have no effect. 75 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 76 The Effect of a Change in Temperature Only temperature changes will alter the equilibrium constant, and that is why we always specify the temperature when giving the value of Kc. The best way to look at temperature effects is to realize that energy is a component of the equation, the same as a reactant, or product. For example, if you have an exothermic reaction, heat (energy) is on the product side of the equation, but if it is an endothermic reaction, it will be on the reactant side of the equation. O2 (g) + 2 H2 (g) 2 H2O(g) + Energy = Exothermic Electrical energy + 2 H2O(g) 2 H2 (g) + O2 (g) = Endothermic A temperature increase favors the endothermic direction and a temperature decrease favors the exothermic direction. 77 Endothermic Reactions • Absorb energy upon reaction. Treat energy as a reactant. Use Le Chatelier’s principle. • Increasing T adds energy to system. The equilibrium shifts towards the products. (System absorbs energy) • The value of K increases in consequence. 80 Example: 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) + energy Increasing T adds thermal energy. This shifts the system to the left, where it absorbs energy. [Ca(OH)2] and K decrease. (b) CaCO3 (s) + energy = + CaO(s) + CO2 (g) Increasing T adds thermal energy. This shifts the system to the right, where it absorbs energy.[CO2] and K increase. (c) SO2 (g) + energy = S(s) + O2(g) Increasing T adds thermal energy. This shifts the system to the right, where it absorbs energy.[SO2] will decrease and K increases. 81 The equilibrium we are going to consider is: CoC, (aq) + 6H,O(l) + blue [Co(H,0),]* (aq) + 4CI- (aq)