Chap 15B - Chem 210 / Wang / Chapter 15B Equilibrium...

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Unformatted text preview: Chem 210 / Wang / Chapter 15B Equilibrium Calculations & LeChatelier's Principle Chapter 15B Chemical Equilibrium 15.1 The Concept of Equilibrium 15.2 The Equilibrium Constant; Kc and Kp 15.3 Interpreting and Working with Equilibrium Constants 15.4 Heterogeneous Equilibria 15.5 Calculating Equilibrium Constants 15.6 Application of Equilibrium Constants 15.7 LeChatelier’s Principle Equilibrium Calculations Question (a) K = 55.5 for: H2(g) + I2(g) ⇔ 2 HI(g) at 425°C (a) If 0.050 mol of H2 and 0.050 mol of I2 are placed in a 1.00 L vessel and heated to 425°C and allowed to reach equilibrium, what are the equilibrium values in moles for all the species? [HI]2 [H2] = nH2÷V= 0.050÷1.00= 0.050 M Also [I2] = 0.050 M H2 + I2 2 HI Initial 0.050 0.050 0 Change - x -x +2x Equil. 0.050-x 0.050-x 2x Equilibrium Calculations Question (a) [HI] = 2x = 2*0.0394 = 0.080M Final answer: At equilibrium, the 1 L vessel contains: 0.011 0.011 mole of each H2 and I2 0.080 0.080 mole of HI Equilibrium Calculations Question (b) Kc = [ HI ] 2 [ I2 ] [ H2 ] H2(g) + I2(g) ⇔ 2 HI(g) initially 0.040M 0.025M 0 during -x -x +2x at equil. 0.040-x 0.025-x 2x x = 0.0228 or 0.0472 X = 0.0228 M Final answer: [H2] = 0.040 – x = 0.017 M 0.017 [I2] = 0.025 – x = 0.002 M 0.002 [HI] = 2x = 0.046 M 0.046 Fall 2009 UM-SJTU JI 55.5= [0.050[0.050-x]2 Equilibrium Calculations Question (b) [H2] = 0.040 M H2 + Initial 0.040 Change - x Equil. 0.040-x K= [I2] = 0.025 M I2 2 HI 0.025 0 -x +2x 0.025-x 2x [2x]2 = 55.5 [0.040-x][0.025-x] [ I2 ] [ H2 ] √55.5= 55.5 = [0.0638+2x]2 √55.5= solve for x: 7.45 * (0.050-x) = 2x (0.050x = 0.0394 M Solve for x (using quadratic equation) then find M of each gas. [ 0.0638+2x ]2 [ 0.00212-x ]2 7.45 = x= - 5.08 x 10-3 [0.050-x]2 7.45 = [2x] [0.050-x] H2(g) + I2(g) ⇔ 2 HI(g) initially 0.040M 0.025M 0 change -x -x +2x at equil. 0.040-x 0.025-x 2x [ HI ] 2 [0.0638+2x] [0.00212-x] [ I2 ] [ H2 ] (4) * x2 (0.001-0.065x+x2) 0.0721x2 = 0.001 – 0.065x + x2 0.928x2 – 0.065x +0.001 = 0 x= -b±√b2-4ac 2a [0.00212-x]2 [2x]2 Equilibrium Calculations Question (b) 55.5 = Equilibrium Calculations Question (c) Kc = [0.050-x]2 Kc = (c) [H2] = [I2] = 2.12×10-3 M and [HI] = 1.58×10-2M. If 0.0480 mol of HI are added, calculate new equilibrium concentrations. H2(g) + I2(g) ⇔ 2 HI(g) Initial 0.00212M 0.00212M 0.0158+0.048 Change -x -x +2x Equil. 0.00212-x 0.00212-x 0.0638+2x [ HI ] 2 [2x]2 55.5= [2x]2 K = 55.5 for: H2(g) + I2(g) ⇔ 2 HI(g) at 425°C (b) If 0.040 mol of H2 and 0.025 mol of I2 are placed in a 1.00 L vessel and heated to 425°C and allowed to reach equilibrium, what are the equilibrium values for all the species? I2 2 HI 0.050 0 -x +2x 0.050-x 2x x = 0.0394 M [H2] = [I2] = 0.050 - x = 0.050- 0.0394 0.050= 0.011 M =55.5 Solve for x (take the square-root on both sides) (take squarethen find moles of each gas! moles K = 55.5 for: H2(g) + I2(g) ⇔ 2 HI(g) at 425°C H2 + Initial 0.050 Change - x Equil. 0.050-x [H2][I2] Equilibrium Calculations Question (a) K = 55.5 for: H2(g) + I2(g) ⇔ 2 HI(g) at 425°C (a) If 0.050 mol of H2 and 0.050 mol of I2 are placed in a 1.00 L vessel and heated to 425°C and allowed to reach equilibrium, what are the equilibrium values in moles for all the species? H2(g) + I2(g) ⇔ 2 HI(g) x = 0.0228 or 0.0472 Equilibrium Calculations Question (c) initially during at equil. Kc = H2(g) + I2(g) ⇔ 2 HI(g) 0.00212M 0.00212M 0.0158+0.048 -x -x +2x 0.00212-x 0.00212-x 0.0638+2x [ HI ] 2 [ I2 ] [ H2 ] 55.5 = [ 0.0638+2x ]2 [ 0.00212-x ]2 x= - 5.08 x 10-3 [Negative value of x means the reaction mixture reached equilibrium with more reactants and proportionally less more and less product than what was present in the “initial” condition.] condition.] [H2] = 0.00212-x = 7.20 x10-3M [I2] = 0.00212-x = 7.20 x10-3M [HI] = 0.0638+2x = 5.36 x10-2 M page 1 Chem 210 / Wang / Chapter 15B Equilibrium Calculations & LeChatelier's Principle Question: Consider the reaction A + B AB. AB The vessel on the right contains an equilibrium mixture of A (red), B (blue), and AB molecules. (red), (blue), AB If the stopcock is opened and the contents of the two vessels are allowed to mix, predict how predict many AB molecules will be at the new equilibrium? at at equilibrium K = [AB]/[A][B] = [4/V]÷([2/V][2/V]) = V-1 assume assume V = 1 L K=1 A +B AB AB Initial 2/2 2/2 12/2 Change +x +x -x Equil. 1+x 1+x 6-x K = 1 = (6-x)/(1+x)(1+x) (61+2x+x2 = 6-x 62 +3x -5 = 0 x x = (-3± 29)/2 (29)/2 Le Chatelier’s principle Add more NH3 to this equilibrium N2 + 3 H2 2 NH3 + 92 kJ Too much NH3 Will react to use it up … speed up the reverse reaction: “system shifts left” As excess NH3 is being consumed: Heat will be absorbed. N2 will increase; H2 also increases. Stoichiometry Stoichiometry: 1 N2 to 3 H2 to 2 NH3 to 92 kJ H2 increases at 3/2 the rate of decrease of NH3. 3/2 N2 increases at 1/2 the rate of decrease of NH3. 1/2 2 NH3 + 92 kJ (a) Heating the system: • = raise temperature of system • system shifts to left to use up extra heat. • As a result: [N2] will …increase [H2] will…increase 3X as fast as [N2] [NH3] will…decrease 2X as fast as [N2] Fall 2009 UM-SJTU JI x= -b±√b2-4ac If a system at equilibrium is disturbed, that equilibrium system will respond by attempting to minimize minimize the effect of the disruption. 2a 2a You can disturb an equilibrium by adding / removing chemicals or heat / chill the system. x = 1.2 6-x = 4.8 AB = 4.8 × 2L = 9.6 or round to 10 9.6 N2 + 3 H2 2 NH3 + 92 kJ Too much N2 Will react to use it up … the → reaction will speed up more (system shifts right) As N2 is being consumed: Heat will be released NH3 will increase, at 2 times the rate of N2 decrease. H2 will also decrease, 3 times faster than N2! What effect will there be if you added more NH3? How about adding more N2 or heat? Le Chatelier’s principle Temperature can also have an effect. For exothermic reactions reactants products + heat Raising the temperature shifts system to the left. For endothermic reactions heat + reactants 2 NH3 + 92 kJ (b) Add Helium gas – inert gas Not a disturbance to the system. No change. Presence of He does not change the pressure of other gases… each gas behaves as if it is alone in the container. (c) Add a catalyst – reaction rates increase Rate increases in both directions at the same time. System stays in equilibrium. No change. products Raising the temperature shifts system to the right. Le Chatelier’s principle N2 + 3 H2 2 NH3 + 92 kJ N2 + 3 H2 Le Chatelier’s principle Add more N2 to this equilibrium Le Chatelier’s principle questions N2 + 3 H2 +1.193 +1.193 - 4.193 Le Chatelier’s principle Le Chatelier’s principle (d) Increasing pressure at the same temperature Pressure affects an equilibrium involving gaseous reactants or products. 3 H2(g) + N2(g) 2 NH3(g) + 92 kJ 92 4 mol of gases 2 mol of gases P↑ causes the system to shift to the side with the least moles of gas (to reduce total pressure) The The above system shifts…to the right! page 2 Chem 210 / Wang / Chapter 15B Equilibrium Calculations & LeChatelier's Principle Le Chatelier’s principle N2 + 3 H2 2 NH3 +92 kJ (e) Increase volume – Partial pressure of each gas decreases. Each gas will try to “retain” original pressure; System attempts to “restore” total pressure Shifts left to make more gas molecules (2 4 molecules). NOTE: NOTE: Expansion of gas requires heat. If If volume increases suddenly, the system may shift to right to produce more heat. (If temperature of the system is kept constant, the system will shift left.) Haber process: N2 (g) + 3H2 (g) ↔ 2 NH3 (g) +92 kJ How does each step maximize production? 1) Compress gases. 2) Preheat feed gases. 3) Cool gases. 4) Recycle unreacted gases. Ref. Page 651 Figure 15.12 This picture represents two equilibrium mixtures of solid CaCO3, CaCO solid CaO, and gaseous CaO CO2, as a result of the CO endothermic endothermic decomposition of CaCO3: Predict and explain how temperature may affect this equilibrium. Endothermic when T increases, forward when reaction speeds up per LeChatelier’s principle More products (CO2 & CaO) formed at equilibrium CaO) Higher Temp larger larger K for endothermic reactions!! Fall 2009 UM-SJTU JI N2 + 3 H2 2 NH3 +92 kJ N2 is abundant in the atmosphere, but useless to plants and people. The The above reaction turns N2 to an useful form: NH3. Liquid Liquid ammonia (NH3) can then be sprayed directly into soil as fertilizer. fertilizer. How to maximize the yield of NH3? Apply LeChatelier's principle!! N2 + 3 H2 2 NH3 +92 kJ Key stages in the Haber process for synthesizing ammonia. Haber for N2 + 3 H2 2 NH3 +92 kJ % yield of ammonia vs. temperature (0C) at five temperature at pressures different operating pressures. 2CrO4-2 + 2 H+ Cr2O7-2 + H2O Cr Description: Yellow chromate ion is turned orange orange by addition of acid. The orange dichromate is orange returned to yellow by the addition of base. http://en.wikipedia.org/wiki/Fritz_Haber German chemist of Jewish ethnicity. Fritz Haber (1868 – 1934) Nobel Received 1918 Nobel Prize for the Haber process. Credited as the "father of chemical warfare" for developing / deploying Cl2(g) and other poisonous gases during World War I. Personally oversaw the first successful use of Cl2(g) at the 2nd Battle of Ypres on April 22, 1915. First wife, Clara Immerwahr (also a PhD chemist), opposed his work on poison gas and committed suicide with his service weapon in their garden. He is also credited with proposing the Born-Haber cycle as a Bornmethod for evaluating the lattice energy of an ionic solid. Sample Problem 1 For the reaction: X(g) + Y2(g) + heat XY(g) + Y(g) the following molecular scenes depict different reaction mixtures. (X = green, Y = purple) Yellow chromate ion + acid orange orange dichromate 2CrO4-2 + 2 H+ Cr2O7-2 + H2O Cr Sample Problem 2 Effect of Temperature on the Equilibrium Position How does an increase in temperature affect Kc for the following reactions? (a) CaO(s) + H2O(l) (b) CaCO3(s) + 178kJ (a) If Kc = 2 at the temperature of the reaction, which scene represents the mixture at equilibrium? (b) Will the reaction mixtures in the other two scenes proceed toward reactant or toward products to reach equilibrium? (c) For the mixture at equilibrium, how will a rise in temperature affect [Y2]? (c) SO2(g) + 297kJ Ca(OH)2(aq + 82kJ CaO(s) + CO2(g) S(s) + O2(g) page 3 Chem 210 / Wang / Chapter 15B Equilibrium Calculations & LeChatelier's Principle Question: The following pictures represent the composition of the equilibrium mixture for the AB at 300 K and at 400 K: reaction A + B Is Co(H2 O)62+ O) CoCl endoCoCl42- endo- or exothermic? Co(H2 O)62+ + heat ↔ CoCl42O) Is the reaction exothermic or endothermic? Explain using Le Châtelier’s principle. 2NO2 (g) N2O4 (g) Room temp Both present Heating up Mostly CoCl4-2 2NO2 (g) N2O4 (g) (g) Is the above reaction endo- or exothermic? Low temp medium temp Cooling with ice Mostly Co(H2O)6+2 Question: The following pictures represent equilibrium mixtures at 325 K and 350 K for a reaction involving A atoms atoms atoms (red), B atoms (blue), and AB molecules: Can you explain how heat is involved in this reaction by considering molecular structures? 1. Write a balanced equation for the reaction. 2. Is the reaction exothermic or endothermic? Predict using Le Châtelier’s principle. 3. If the volume of the container is increased, will the number of A atoms increase, decrease, or remain the same? atoms Explain. Fall 2009 UM-SJTU JI page 4 high temp ...
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