Equilibrium Lesson Notes

Equilibrium 1977 D For the system 2 SO2(g) + O2(g) ↔ 2 SO3(g) , ∆H is negative for the production of SO3. Assume that one has an equilibrium mixture of these substances. Predict the effect of each of the following changes on the value of the equilibrium constant and on the number of moles of SO3 present in the mixture at equilibrium. Briefly account for each of your predictions. (Assume that in each case all other factors remain constant.) (a) Decreasing the volume of the system. (b) Adding oxygen to the equilibrium mixture. (c) Raising the temperature of the system. Answer: (a) As volume decreases, pressure increases and the reaction shifts in the direction of fewer molecules (less volume; more SO3) to relieve the stress. Value of Keq does not change. (b) Additional O2 disturbs the equilibrium and SO3 is formed to relieve the stress. Value of Keq does not change. (c) Increase in temperature shifts the reaction to the left to “use up” some of the added heat. Less SO3 remains. Value of Keq decreases due to the relative greater increase in the rate of the endothermic reaction (reaction to the left). 1980 D NH4Cl(s) ↔ NH3(g) + HCl(g) ∆H = +42.1 kilocalories Suppose the substances in the reaction above are at equilibrium at 600K in volume V and at pressure P. State whether the partial pressure of NH3(g) will have increased, decreased, or remained the same when equilibrium is reestablished after each of the following disturbances of the original system. Some solid NH4Cl remains in the flask at all times. Justify each answer with a one-or-two sentence explanation. (a) A small quantity of NH4Cl is added. (b) The temperature of the system is increased. (c) The volume of the system is increased. (d) A quantity of gaseous HCl is added. (e) A quantity of gaseous NH3 is added. Answer: page 1 (a) PNH does not change. Since NH4Cl(s) has constant concentration (a = 1), equilibrium does not shift. (b) PNH increases. Since the reaction is endothermic, increasing the temperature shifts the equilibrium to the right and more NH3 is present. (c) PNH does not change. As V increases, some solid NH4Cl decomposes to produce more NH3. But as the volume increases, PNH3 remains constant due to the additional decomposition. (d) PNH decreases. Some NH3 reacts with the added HCl to relieve the stress from the HCl addition. (e) PNH increases. Some of the added NH3 reacts with HCl to relieve the stress, but only a part of the added NH3 reacts, so PNH3 increases. 3 3 3 3 3 1981 A Ammonium hydrogen sulfide is a crystalline solid that decomposes as follows: NH4HS(s) ↔ NH3(g) + H2S(g) (a) Some solid NH4HS is placed in an evacuated vessel at 25ºC. After equilibrium is attained, the total pressure inside the vessel is found to be 0.659 atmosphere. Some solid NH4HS remains in the vessel at equilibrium. For this decomposition, write the expression for KP and calculate its numerical value at 25ºC. (b) Some extra NH3 gas is injected into the vessel containing the sample described in part (a). When equilibrium is reestablished at 25ºC, the partial pressure of NH3 in the vessel is twice the partial pressure of H2S. Calculate the numerical value of the partial pressure of NH3 and the partial pressure of H2S in the vessel after the NH3 has been added and the equilibrium has been reestablished. (c) In a different experiment, NH3 gas and H2S gas are introduced into an empty 1.00 liter vessel at 25ºC. The initial partial pressure of each gas is 0.500 atmospheres. Calculate the number of moles of solid NH4HS that is present when equilibrium is established. Answer: (a) KP = (PNH3)(PH2S) PNH3 = PH2S = 0.659/2 atm = 0.330 atm KP = (0.330)2 = 0.109 Equilibrium (b) PNH3 = 2 PH2S (2x)(x) = 0.109 ; x = 0.233 atm = PH2S PNH3 = 0.466 atm (c) Equilibrium pressures of NH3 and H2S are each 0.330 atm. Amounts of each NH3 and H2S that have reacted correspond to (0.500 - 0.330) = 0.170 atm. n = mol of each reactant = mol of solid product (0.170 atm )(1.00L ) PV = 6.95∞10− 3 mol = n= RT (0.08205 L_atm )( 298 K) mol _K 1983 A Sulfuryl chloride, SO2Cl2, is a highly reactive gaseous compound. When heated, it decomposes as follows: SO2Cl2(g) → SO2(g) + Cl2(g). This decomposition is endothermic. A sample of 3.509 grams of SO2Cl2 is placed in an evacuated 1.00 litre bulb and the temperature is raised to 375K. (a) What would be the pressure in atmospheres in the bulb if no dissociation of the SO2Cl2(g) occurred? (b) When the system has come to equilibrium at 375K, the total pressure in the bulb is found to be 1.43 atmospheres. Calculate the partial pressures of SO2, Cl2, and SO2Cl2 at equilibrium at 375K. (c) Give the expression for the equilibrium constant (either Kp or Kc) for the decomposition of SO2Cl2(g) at 375K. Calculate the value of the equilibrium constant you have given, and specify its units. (d) If the temperature were raised to 500K, what effect would this have on the equilibrium constant? Explain briefly. Answer: (a) ⊇ 1 mol _ 0.0821 L_atm )( 375 K) ℑ3.509 g ∞ ( mol _K 135.0g ↓ nRT ⊄ P= = 1.00L V = 0.800 atm (b) PSO2Cl2 = (0.800 - y) atm PSO2 = PCl2 = y atm PT = PSO2Cl2 + PSO2 + PCl2 1.43 atm = (0.800 - y + y + y) atm page 2 y = 0.63 atm = PSO2 = PCl2 PSO2Cl2 = (0.800 - 0.63) atm = 0.17 atm (PSO )( PCl ) ( 0.63atm )2 2 2 Kp = = = 2.3atm PSO Cl 0.17atm 2 2 (c) (d) Heat is absorbed during the dissociation and so K500 > K375. A stress is place on the system and K increases, which reduces the stress associated with the higher temperature. 1985 A At 25ºC the solubility product constant, Ksp, for strontium sulfate, SrSO4, is 7.6×10-7. The solubility product constant for strontium fluoride, SrF2, is 7.9×10-10. (a) What is the molar solubility of SrSO4 in pure water at 25ºC? (b) What is the molar solubility of SrF2 in pure water at 25ºC? (c) An aqueous solution of Sr(NO3)2 is added slowly to 1.0 litre of a well-stirred solution containing 0.020 mole F- and 0.10 mole SO42- at 25ºC. (You may assume that the added Sr(NO3)2 solution does not materially affect the total volume of the system.) 1. Which salt precipitates first? 2. What is the concentration of strontium ion, Sr2+, in the solution when the first precipitate begins to form? (d) As more Sr(NO3)2 is added to the mixture in (c) a second precipitate begins to form. At that stage, what percent of the anion of the first precipitate remains in solution? Answer: (a) SrSO4(s) ↔ Sr2+(aq) + SO42-(aq) At equilibrium: [Sr2+] = X M = [SO42-] X2 = Ksp = 7.6×10-7 X = 8.7×10-4 mol/L, solubility of SrSO4 (b) SrF2(s) ↔ Sr2+(aq) + 2 F-(aq) At equilibrium: [Sr2+] = X M = [F-] = 2X M KSP = [Sr2+][F-]2 = (X)(2X)2 = 7.9×10-10 X = 5.8×10-4 mol/L, solubility of SrF2 (c) Solve for [Sr2+] required for precipitation of each salt. Equilibrium Ksp = [Sr2+][F-]2 = 7.9×10-10 ⊇( 0.020 mol )2 _ = 7.9 ∞10− 10 ; x = 2.0 ∞10− 6 M =( x )ℑ 1 . 0 L ⊄ ↓ Ksp = [Sr ][SO4 ] = 7.6×10 ⊇0.10mol ) _ = 7.6 ∞10− 7 ; y = 7.6 ∞10− 6 M =( y )ℑ ⊄ 1.0L ↓ 2+ 2- -7 Since 2.0×10 M < 7.6×10 M, SrF2 must precipitate first. When SrF2 precipitates, [Sr2+] = 2.0×10-6 M (d) The second precipitate to form is SrSO4, which appears when [Sr2+] = 7.6×10-6 M (based on calculations in Part c.) When [Sr2+] = 7.6×10-6 M, [F-] is determined as follows: Ksp = [Sr2+][F-]2 = 7.9×10-10 = (7.6×10-6)(z)2 = 7.9×10-10 ; z = 1.0×10-2 M 1.0 ∞10 -2 %F − still in solution = ∞100 = 50.% 2.0 ∞10 −2 -6 -6 1988 A At elevated temperatures, SbCl5 gas decomposes into SbCl3 gas and Cl2 gas as shown by the following equation: SbCl5(g) ↔ SbCl3(g) + Cl2(g) (a) An 89.7 gram sample of SbCl5 (molecular weight 299.0) is placed in an evacuated 15.0 litre container at 182ºC. 1. What is the concentration in moles per litre of SbCl5 in the container before any decomposition occurs? 2. What is the pressure in atmospheres of SbCl5 in the container before any decomposition occurs? (b) If the SbCl5 is 29.2 percent decomposed when equilibrium is established at 182ºC, calculate the value for either equilibrium constant Kp or Kc, for this decomposition reaction. Indicated whether you are calculating Kp or Kc. (c) In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first placed in an empty 2.00 litre container maintained at a temperature different from 182ºC. At this temperature, Kc, equals 0.117. How many moles of Cl2 must be added to this container to reduce the number of moles of SbCl3 to 0.700 mole at equilibrium? page 3 Answer: (a) (1) 89.7g S bCl 5 ∞ [SbCl 5 ]init . = 1 mol = 0.300 mol SbCl 5 299.0g 0.300 mol 15.0L = 0.200 M (2) T = 182ºC + 273 = 455K L_atm nRT (0.300 mol )( 0.0821 mol _K )( 455 K) = P= 15.00L V = 0.747 atm OR 273 K 30.0L 15.0L at std. temp. = ∞ mol 0.300 mol 455 K 1 atm × f(22.4 L/mol,30.0 L/mol) = 0.747 atm (b) [SbCl3] = [Cl2] = (0.0200 mol/L)(0.292) = 5.84×10-3M [SbCl5] = (0.0200 mol/L)(0.708) = 1.42×10-2M [SbCl 3 ][ Cl 2 ] ( 5.84 ∞10 −3 )2 Kc = = 2.41 ∞10 −3 = − 2 [ SbCl 5 ] 1.42 ∞10 OR PSbCl3 = PCl2 = (0.747 atm)(0.292) = 0.218 atm PSbCl5= (0.747 atm)(0.708) = 0.529 atm (PSbCl )( PCl ) ( 0.218 )2 3 2 KP = = = 8.98 ∞10 −2 PSbCl 0 . 529 5 [SbCl 3 ][ Cl 2 ] K= = 0.117 [ SbCl 5 ] (c) (1.00 − 0.70)mol = 0.15M [SbCl 5 ]equil . = 2.00L 0.700 mol = 0.350 M [SbCl 3 ]equil . = 2.00L [Cl2]equil. = X M (0.350 )( X) = 0.117 ; X = 0.050 M =[ Cl 2 ] KC = 0.15 Moles Cl2(equil.) = (0.0050 mol/L)(2.00L) = 0.10 mol Cl2 Moles Cl2 needed to make 0.300 mol SbCl3 into SbCl5 = 0.30 mol Moles Cl2 that must be added = 0.40 mol 1988 D NH4HS(s) ↔ NH3(g) + H2S(g) ∆Hº = +93 kilojoules Equilibrium The equilibrium above is established by placing solid NH4HS in an evacuated container at 25ºC. At equilibrium, some solid NH4HS remains in the container. Predict and explain each of the following. (a) The effect on the equilibrium partial pressure of NH3 gas when additional solid NH4HS is introduced into the container (b) The effect on the equilibrium partial pressure of NH3 gas when additional solid H2S is introduced into the container (c) The effect on the mass of solid NH4HS present when the volume of the container is decreased (d) The effect on the mass of solid NH4HS present when the temperature is increased. Answer: (a) The equilibrium pressure of NH3 gas would be unaffected. KP = (PNH3)(PH2S). Thus the amount of solid NH4HS present does not affect the equilibrium. (b) The equilibrium pressure of NH3 gas would decrease. In order for the pressure equilibrium constant, KP, to remain constant, the equilibrium pressure of NH3 must decrease when the pressure of H2S is increased. KP = (PNH3)(PH2S). (A complete explanation based on LeChatelierÆs principle is also acceptable.) (c) The mass of NH4HS increases. A decrease in volume causes the pressure of each gas to increase. To maintain the value of the pressure equilibrium constant, Kp, the pressure of each of the gases must decrease. The decrease is realized by the formation of more solid NH4HS. Kp = (PNH3)(PH2S). (A complete explanation based on LeChatelier’s principle is also acceptable.) (d) The mass of NH4HS decreases because the endothermic reaction absorbs heat and goes nearer to completion (to the right) as the temperature increases. 1992 A 2 NaHCO3(s) ↔ Na2CO3(s) + H2O(g) + CO2(g) Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating according to the equation above. (a) A sample of 100. grams of solid NaHCO3 was placed in a previously evacuated rigid 5.00-liter container and heated to 160ºC. Some of the page 4 original solid remained and the total pressure in the container was 7.76 atmospheres when equilibrium was reached. Calculate the number of moles of H2O(g) present at equilibrium. (b) How many grams of the original solid remain in the container under the conditions described in (a)? (c) Write the equilibrium expression for the equilibrium constant, KP, and calculate its value for the reaction under the conditions in (a). (d) If 110. grams of solid NaHCO3 had been placed in the 5.00-liter container and heated to 160ºC, what would the total pressure have been at equilibrium? Explain. Answer: PV (7.76atm )( 5.00L ) n gas = = =1.09mol L_atm RT ( 433 K ) 0 . 0821 ( mol _K ) (a) mol H2O = (1/2)(1.09 mol) = 0.545 mol H2O(g) (b) 2mo l NaHCO 3 84.0gNaHCO 3 = ∞ 0.545 mo l H 2O ∞ 1 mol 1mo l H 2O = 91.9 g NaHCO3 decomposed remaining = 100.g - 91.6g = 8.4g OR 100 18.0g 44.0g _ ⊇ ℑ0.545 molH 2O ∞ + 0.545 molCO 2 ∞ ⊄ 1 mol 1 mol ↓ = 100g - 33.8g = 66g (or 66.2g) [includes Na2CO3 solid in this mass] (c) Kp = (PH2O)(PCO2) =(3.88)(3.88)atm2 = 15.1 atm2 (d) Pressure would remain at 7.76 atm. Since some solid remained when 100.g was used (and there has been no temperature change), then using 110g will not affect the equilibrium. 1994 A MgF2(s) ↔ Mg2+(aq) + 2 F-(aq) In a saturated solution of MgF2 at 18ºC, the concentration of Mg2+ is 1.21×10-3 molar. The equilibrium is represented by the equation above. (a) Write the expression for the solubility-product constant, Ksp, and calculate its value at 18ºC. Equilibrium (b) Calculate the equilibrium concentration of Mg2+ in 1.000 liter of saturated MgF2 solution at 18ºC to which 0.100 mole of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible. (c) Predict whether a precipitate of MgF2 will form when 100.0 milliliters of a 3.00×10-3-molar Mg(NO3)2 solution is mixed with 200.0 milliliters of a 2.00×l0-3-molar NaF solution at 18ºC. Calculations to support your prediction must be shown. (d) At 27ºC the concentration of Mg2+ in a saturated solution of MgF2 is 1.17×10-3 molar. Is the dissolving of MgF2 in water an endothermic or an exothermic process? Give an explanation to support your conclusion. Answer: (a) Ksp = [Mg2+][F-]2 = (1.21×10-3)(2.42×10-3)2 = 7.09×10-9 (b) X = concentration loss by Mg2+ ion 2X = concentration loss by F- ion [Mg2+] = (1.21×10-3 - X) M [F-] = (0.100 + 2.42×10-3 - 2X) M since X is a small number then (0.100 + 2.42×10-3 - 2X) ≈ 0.100 Ksp = 7.09×10-9 = (1.21×10-3 - X)(0.100)2 X = 1.2092914×10-3 [Mg2+]= 1.21×10-3-1.20929×10-3 = 7.09×10-7M (c) [Mg2+] = 3.00×10-3M × 100.0 mL/300.0 mL = 1.00×10-3M [F-] = 2.00×10-3 M × 200.0 mL/300.0 mL = 1.33×10-3 M trial Ksp = (1.00×10-3)(1.33×10-3)2 = 1.78×10-9 trial Ksp < = 7.09×10-9, ∴ no ppt. (d) @ 18ºC, 1.21×10-3 M MgF2 dissolves @ 27ºC, 1.17×10-3 M MgF2 dissolves MgF2 ↔ Mg2+ + 2 F- + heat dissolving is exothermic; if heat is increased it forces the equilibrium to shift left (according to LeChatelier’s Principle) and less MgF2 will dissolve. 1995 A page 5 CO2(g) + H2(g) ↔ H2O(g) + CO(g) When H2(g) is mixed with CO2(g) at 2,000 K, equilibrium is achieved according to the equation above. In one experiment, the following equilibrium concentrations were measured. [H2] = 0.20 mol/L [CO2] = 0.30 mol/L [H2O] = [CO] = 0.55 mol/L (a) What is the mole fraction of CO(g) in the equilibrium mixture? (b) Using the equilibrium concentrations given above, calculate the value of Kc, the equilibrium constant for the reaction. (c) Determine Kp in terms of Kc for this system. (d) When the system is cooled from 2,000 K to a lower temperature, 30.0 percent of the CO(g) is converted back to CO2(g). Calculate the value of Kc at this lower temperature. (e) In a different experiment, 0.50 mole of H2(g) is mixed with 0.50 mole of CO2(g) in a 3.0-liter reaction vessel at 2,000 K. Calculate the equilibrium concentration, in moles per liter, of CO(g) at this temperature. Answer: (a) CO = f(0.55 mol, 1.6 mol) = 0.34 (b) Kc = ([H2O][CO])/([H2][CO2]) = (0.55×0.55)/(0.20×0.30) = 5.04 (c) since ∆n = 0, Kc = Kp (d) [CO] = 0.55 - 30.0% = 0.55 - 0.165 = 0.385 M [H2O] = 0.55 - 0.165 = 0.385 M [H2] = 0.20 + 0.165 = 0.365 M [CO2] = 0.30 + 0.165 = 0.465 M K = (0.385)2/(0.365×0.465) = 0.87 (e) let X = ∆[H2] to reach equilibrium [H2] = 0.50 mol/3.0L - X = 0.167 - X [CO2] = 0.50 mol/3.0L - X = 0.167 - X [CO] = +X ; [H2O] = +X K = X2/(0.167 - X)2 = 5.04 ; X = [CO] = 0.12 M 1998 D C(s) + H2O(g) ↔ CO(g) + H2(g) ∆Hº = +131kJ Equilibrium A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(g) at equilibrium. State whether the number of moles of CO(g) in the container will increase, decrease, or remain the same after each of the following disturbances is applied to the original mixture. For each case, assume that all other variables remain constant except for the given disturbance. Explain each answer with a short statement. (a) Additional H2(g) is added to the equilibrium mixture at constant volume. (b) The temperature of the equilibrium mixture is increased at constant volume. (c) The volume of the container is decreased at constant temperature. (d) The graphite pellets are pulverized. Answer (a) CO will decrease. An increase of hydrogen gas molecule will increase the rate of the reverse reaction which consumes CO. A LeChatelier Principle shift to the left. (b) CO will increase. Since the forward reaction is endothermic (a ∆H > 0) an increase in temperature will cause the forward reaction to increase its rate and produce more CO. A LeChatelier Principle shift to the right. (c) CO will decrease. A decrease in volume will result in an increase in pressure, the equilibrium will shift to the side with fewer gas molecules to decrease the pressure, ∴, a shift to the left. (d) CO will remain the same. Once at equilibrium, the size of the solid will affect neither the reaction rates nor the equilibrium nor the concentrations of reactants or products. 2000 A Required 1. 2 H2S(g) → 2 H2(g) + S2(g) When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of H2S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72×10–2 mol of S2(g) is present at equilibrium. page 6 (a) Write the expression for the equilibrium constant, Kc, for the decomposition reaction represented above. (b) Calculate the equilibrium concentration, in mol•L-1, of the following gases in the container at 483 K. (i) H2(g) (ii) H2S(g) (c) Calculate the value of the equilibrium constant, Kc, for the decomposition reaction at 483 K. (d) Calculate the partial pressure of S2(g) in the container at equilibrium at 483 K. (e) For the reaction H2(g) + Ошибка! S2(g) → H2S(g) at 483 K, calculate the value of the equilibrium constant, Kc. Answer: (a) Kc = Ошибка! (b) (i) Ошибка! × Ошибка! = 5.95×10–2 M H2 (ii)  1 mol  2 mol H2S 3.40 g H2S ×  −  3.72 ×10−2 mol S2 ×   34.0 g  1 mol S2  = 2.05×10–2 M H2S 1.25 L  (c) Kc = [5.952 ×10−2 ]2 3.721.×2510 [0.02048]2 (d) PV=nRT = 1.18 (e) K’c = 1 Kc = 2.00 −2    = 0.251