MC Pratie F Solubility Equilibrium, Ksp Name This is pratie - Do NOT heat yourself of finding out what you are apable of doing. Be sure you follow the testing onditions outlined below. DO NOT USE A CALCULATOR. You may use ONLY the periodi table and formula sheets. Try to finish at a rate of about 1.5 min per question. Time yourself. It is important that you pratie working for speed. Please mail me if you think you find any issues or typos. 1. Temperature is often given with Ksp values beause a. the solubility of solids always inreases with inreasing temperature. b. the solubility of solids varies with temperature hanges.. solubility hanges with temperature but Ksp values do not. d. Ksp varies with temperature even though onentrations do not. e. the number of ions varies with the kind of salt that is dissolving. 5. Determine the solubility of lead(ii) fluoride, Ksp 4.0 10 8, in a 0.0040 M lead(ii) nitrate solution. a..0 10 M b..0 10 M.! d. e.! 4.0 10 5 M 1.0 10 5 M 4.0 10 8 4.0 10 M. The Ksp expression for silver phosphate is a.! K sp [Ag + [PO 4 b.! K sp [Ag + [PO 4. d.! K sp [Ag + [PO 4 e.!. The solubility of nikel(ii) hydroxide, Ksp 1.6 10 16, is about a. b.!.! d.! e.! K sp [Ag + [PO 4 K sp [Ag+ [PO 4 [Ag PO 4 4.0 10 17 1.6 10 16 1.6 10 16 (1.6 10 16 )() (1.6 10 16 )(4) 6. The solubility of silver sulfide is 8.0 10 17 M. Determine the Ksp of this salt. a. 64 10 51 b. [16.0 10 17 [8.0 10 17. [16.0 10-17 ) [8.0 10 17 d. [8.0 10 17 [4.0 10-17 e. [4.0 10 17 ) [8.0 10 17 7. Equal volumes of 1.6 10 5 M KCl and 1.6 10 5 M AgNO are mixed. the Ksp for silver hloride is 1.6 10 10. As these two solutions are ombined, a. a preipitate of AgCl forms. b. there is no preipitate formed.. NaCl will preipitate. d. AgNO will preipitate. e. the [Na + will beome 0.00 M 4. What is the molar solubility in water of PbI? (The Ksp for PbI is. 10 8 ) a.. 10 8 M b. 8.0 10 8 M. (1.6) ½ 10 8 M d. (1.6) ⅓ 10 8 M e. 10 M 8. The solubility of metal(ii) sulfide, Ksp 1.6 10 11, is about a. b.!.! 4.0 10 6 4 10 4 1.6 10 11 d.! 8.0 10 6 e.!.0 10 6
Pratie MC Unit F Solubility Equilibrium Ksp (page of ) 9. Calulate the ph of a saturated metal hydroxide, XOH, whose Ksp 1 10 8. a. 4.0 b. 6.0. 7.0 d. 8.0 e. 10.0 1. Magnesium hydroxide, Mg(OH), has a Ksp 6. 10 10. The solubility of Mg(OH) will be the lowest in 1.0 L of whih of the following? a. 0.10 M HCl b. 0.10 M NaOH. 0.10 M MgCl d. pure HO 10. Calulate the Ksp of a saturated metal hydroxide, X(OH), solution whose ph 9.00. a..0 10 8 b. 5.0 10 16..0 10 15 d..0 10 10 e. 5.0 10 10 11. On the basis of the Ksp values listed below, what is the order of least soluble to most soluble for these ompounds? AgBr, Ksp 5.4 10 1 AgCO, Ksp 8.0 10 1 AgCl, Ksp 1.8 10 10 a. AgBr < AgCO <AgCl b. AgBr <AgCl < AgCO. AgCl < AgCO <AgBr d. AgCO < AgBr <AgCl Use the following information to answer the next questions 150 ml of saturated SrF solution is present in a 50 ml beaker at room temperature. The molar solubility of SrF at 5ºC is 1 10 M 14. What are the onentrations of Sr + and F in the beaker? a. [Sr + 1.0 10 M, [F 1.0 10 M b. [Sr + 1.0 10 M, [F.0 10 M. [Sr +.0 10 M, [F 1.0 10 M d. [Sr +.0 10 M, [F.0 10 M 15. If some of the solution evaporates overnight, whih of the following will our? a. The mass of the solid and onentration of the ions will stay the same. b. The mass of the solid and onentration of the ions will stay the inrease.. The mass of the solid will derease, and onentration of the ions will stay the same. d. The mass of the solid will inrease, and onentration of the ions will stay the same. 1. On the basis of the solubilities (in mol/l) listed below, what is the orret order of dereasing Ksp values (largest first). AgBr 7. 10 7 AgCN 7.7 10 9 AgSCN 1.0 10 6 a. AgBr > AgCN > AgSCN b. AgCN > AgBr > AgSCN. AgCN > AgSCN > AgBr d. AgSCN > AgBr > AgCN 16. How ould the onentration of Sr + ions in solution be dereased? a. Adding some NaF(s) to the beaker. b. Adding some Sr(NO)(s) to the beaker.. By heating the solution in the beaker. d. By adding a small amount of water to the beaker, but not dissolving all of the solid.
Pratie MC Unit F more pratie General Equilbrium Kp and K (page of ) 17. HI(g) + Cl(g) HCl(g) + I(g) + energy A gaseous reation ours and omes to equilibrium as shown above. Whih of the following hanges to the system will serve to inrease the number of moles of I present at equilibrium? a. Inreasing the volume at onstant temperature. b. Dereasing the volume at onstant temperature.. Adding a mole of inert gas at onstant volume. d. Inreasing the temperature at onstant volume. e. Dereasing the temperature at onstant volume. 0. An evauated 1.00-liter vessel is injeted with 1. moles of sulfur trioxide gas, SO. The vessel is then heated to a high temperature where the SO partially deomposes to form the produts SO and O in the reation shown below: SO(g) SO(g) + O(g) The temperature is kept onstant, and the amount of SO in the vessel at equilibrium is 0.40 mol. What is the value of K at this temperature? a. 0.04 b. 0.16. 0.8 d. 1.0 e. 1.6 18. In the reation shown below, 0.0 mole of Br and 0.0 moles of I are plaed in an evauated 1.00 liter vessel and allowed to reah equilibrium. What is the value of K if the vessel ontains 0.40 moles of IBr at equilibrium? a. 1.0 b..0. 4.0 d. 16 e. 40 I(g) + Br(g) IBr(g) 19. Whih is the orret set up to determine K for the reation below? 4CuO(s) + CH4(g) CO(g) + 4Cu(s) + HO(g) a.! [CO [Cu 4 [H O [CuO 4 [CH 4! b. [CuO 4 [CH 4 [CO [Cu 4 [H O. [CO [H O [CH 4 d.! [CO [H O [CuO 4 [CH 4 e.! [CO [Cu 4 [H O [CuO 4 [CH 4 1. A sample of solid potassium nitrate is plaed in water. The solid potassium nitrate omes to equilibrium with its dissolved ions by the endothermi proess shown below. KNO(s) + energy K + (aq) + NO (aq) Whih of the following hanges to the system would inrease the onentration of K + ions at equilibrium? a. The volume of the solution is inreased. b. The volume of the solution is dereased.. Additional solid KNO is added to the solution. d. The temperature of the solution is inreased. e. The temperature of the solution is dereased. A.00-liter reation vessel is filled with arbon monoxide gas, CO, and hlorine gas, Cl. The mixture is heated to 670 K and allowed to reah equilibrium aording to the balaned equation shown below: CO(g) + Cl(g) COCl(g) At equilibrium, the mixture ontains 0. moles CO, 0.5 moles Cl, and 1.5 moles COCl. What is K at this temperature? a. 0.0 b. 1. d. 10 e. 0
Pratie MC Unit F Ksp and Kp and K (page 4 of 4). The reation below is allowed to ome to equilibrium at room temperature. At equilibrium, the partial pressure of HO is 96 mm Hg, ClO is 15 mm Hg, and HOCl is 0 mm Hg. What is the value of Kp at this temperature? a. b. 11. 0.017 d. 0.090 e. 0.0045 HO(g) + ClO(g) HOCl(g) 4. At 7 K, the reation shown below has an equilibrium onstant, K.19 10 10. COCl(g) CO(g) + Cl(g) After plaing a mixture of gases in the reation vessel, the onentrations were measured to be [COCl.50 10 - M, [CO 1.11 10-5 M, and [Cl.5 10-6 M. Whih statement below aurately desribes the reation? a. The reation is at equilibrium b. The reation is not at equilibrium, and it is proeeding to the left. The reation is not at equilibrium, and it is proeeding to the right d. The reation quotient is equal to K e. The reation quotient is less than K 5. In the reation W + X Y + Z, all substanes are gases. The reation is initiated by adding equal number of moles of W and of X. When equilibrium is reahed, a. [Y [Z b. [X [Y. [W [X d. [X > [W e. [W + [X [Y + [Z 6. Consider the reations below, both at the same temperature: Rx I X Y K 1 x 10 8 Rx II Z Y K 1 x 10 5 a. I is times faster than II. b. I is 1000 times faster than II.. II is times faster than I. d. II is 1000 times faster than I. e. The size of K and the time required to reah equilibrium are not diretly related. 7. The reation H(g) + N(g) NH(g) is exothermi. Inreasing the temperature of this equilibrium system auses a. an inrease in [NH b. an inrease in [N. a derease in [H d. an inrease in K e. a derease in pressure at onstant volume 8. Consider N(g) + O(g) NO(g). The reation was initiated by adding 15.0 moles of NO to a 1.0-L flask. At equilibrium,.0 moles of oxygen are present in the 1.0-L flask. The value of K must be a. 0. b..0. 5.0 d. 9.0 e. 81 9. At a ertain temperature, the synthesis of ammonia gas from nitrogen and hydrogen gases, shown below, has a value for K of.0 10. If [H [N 0.10 M and [NH 0.0 M N + H NH a. the reation would shift toward the ammonia b. the reation would shift toward the N and the H. the system is at equilibrium, therefore no shifting will our d. the reation will shift toward a new equilibrium position, but the diretion annot be determined from these data e. the equilibrium may shift but it is not possible to alulate Q without knowing the temperature 0. The equilibrium onstant, K, may be used to determine the K of other reations. I. When the equation is reversed the reiproal of the original K beomes the value of K for the new equation. II. When the equation is doubled the square root of the original K beomes the value for the new equation. III. When reations are added, the sum of the equilibrium onstants of eah reation added will the equilibrium onstant of the new equation. Of the above three statements, those whih are always true are a. I only b. II only. III only d. I and III e. I, II, and III
Pratie MC Unit F Ksp and Kp and K (page 5 of 5) 1. The equilibrium P4(g) + 6Cl(g) 4PCl(L) is established at 10 o C. The equilibrium onstant expression is a.! K [PCl [P 4 [Cl b.! K [PCl 4 [P 4 [Cl 6.! K [P [Cl 4 6 [PCl 4 1 d.! K [PCl 4 1 e.! K [P 4 [Cl 6. Ammonium hydrogen sulfide will deompose into ammonia gas and hydrogen sulfide gas when heated. Consider the equilibrium system NH4HS(s) NH(g) + HS(g) whih is developed from 1.000 mole of NH4HS in a 100- L ylinder. At equilibrium the total pressure is found to be 0.400 atm. Kp will be equal to a..00 10 1 b. 1.00 10. 4.00 10 d. 4.00 e. a value impossible to alulate from only these data. At a ertain temperature, it has been determined that K 8.0 for the reation below. If we also determine that the equilibrium mixture ontains 0.80 mole of HO(g), 0.080 mole of CO(g), and 0.080 mole of CO(g), in an 8.0-L flask, what must be the number of moles of H(g) at equilibrium? a. 0.010 mole b. 0.80 mole. 6.4 moles d. 8.0 moles e. 64 moles HO(g) + CO(g) CO(g) + H(g) 4. Whih of the following systems at equilibrium are not affeted by a hange in pressure aused by hanging the volume at onstant temperature? a. H(g) + Cl(g) HCl(g) b. H(g) + I(s) HI(g). N(g) + H(g) NH(g) 5. After the equilibrium represented below is established, some pure O(g) is injeted into the reation vessel at onstant temperature. After equilibrium is reestablished, whih of the following has a lower value ompared to its value at the original equilibrium? CO(g) CO(g) + O(g) a. Keq for the reation b. The total pressure in the reation vessel. The amount of CO(g) in the reation vessel d. The amount of O(g) in the reation vessel e. The amount of CO(g) in the reation vessel 6. Whih of the following hanges alone would ause a derease in the value of Keq for the reation represented below? SO(g) + O(g) SO(g) Exothermi a. Dereasing the temperature b. Inreasing the temperature. Dereasing the volume of the reation vessel d. Inreasing the volume of the reation vessel e. Adding a atalyst 7. In whih of the following systems would the number of moles of the substanes present at equilibrium NOT be shifted by a hange in the volume of the system at a onstant temperature? a. SO(g) + O(g) SO(g) b. N(g) + H(g) NH(g). NO(g) + SO(g) SO(g) + NO(g) d. NO4(g) NO(g) e. CO(g) + H(g) CH4(g) + HO(g) 8. For: N(g) + H(g) NH(g) + energy Some N and H are mixed in a ontainer at 00 o C, and the system reahes equilibrium aording to the equation above. Whih of the following auses an inrease in the number of moles of NH present at equilibrium? I. Dereasing the volume of the ontainer II. Raising the temperature III. Adding a mole of Ar gas at onstant volume a. I only b. II only. I and III only d. II and III only e. I, II, and III d. NH(g) N(g) + H(g) e. O(g) O(g)
Pratie MC Unit F Ksp and Kp and K ANSWERS (page 6 of 6) 1. b All K values always hange with varying temperature. Whether or not the equilibrium onstant is larger or smaller is a funtion of whether the reation is exothermi or endothermi. If the reation is exothermi K will be smaller value at higher temperatures, and a larger value for lower temperatures. The onverse is also true.. The formula for silver phosphate is AgPO4 whih leads to the reation AgPO4 Ag + + PO4. a The formula for nikel(ii) hydroxide is Ni(OH) whih leads to the reation Ni(OH) Ni + + OH and the resulting solubility expression is Ksp [Ni + [OH. Sine there are OH ions for every Ni + ion upon dissolving, we an identify Ni + as s and OH as s and substitute into the Ksp expression. Thus Ksp (s)(s) and Ksp 4s solve for s s 4.0 10 17 without a alulator, we an t 4 4 really lean this up any further. Sine s onentration of the Ni +, the 1:1 stoihiometry the onentration of the Ni(OH) that atually dissolves, s would be the molar solubility. 4. e This problem is nearly the same as #, however, this time the numbers are simple enough to finish the solve. The formula for lead(ii) iodide is PbI whih leads to the reation PbI Pb + + I and the resulting solubility expression is Ksp [Pb + [I. Sine there are I ions for every Pb + ion upon dissolving, we an identify Pb + as s and I as s and substitute into the Ksp expression. 5. d This problem has a twist ompared to # s and 4 beause instead of dissolving the insoluble salt, lead (II) fluoride in water, it is being dissolved in a solution that already ontains some lead ions, lead(ii) nitrate. We all this type of problem a ommon ion problem, meaning the solution you are dissolving into already has an ion in it that is ommon to the salt that you are trying to dissolve. More information on this in your text on page 681. The formula for lead(ii) fluoride is PbF whih leads to the reation PbF Pb + + F and the resulting Ksp expression is Ksp [Pb + [F. Up till now we have substituted s and s into this equation, and solved for s whih will be the solubility. Sine you are given a onentration of the Pb + ion that is already dissolved in the water (and is muh larger, when onsidering signifiant figures, than any lead ions that would be added to it) whih you an insert diretly into the Ksp expression. Thus Ksp (0.004)[s 4.0 10 8 Solve: 4s Sine s onentration of the Pb +, the 1:1 stoihiometry the onentration of the PbF that atually dissolves, this s would be the molar solubility. (If you needed a gram solubility, you would have to multiply by the molar mass of PbF.) 6. When you are told the solubility of some salt, you are being told the maximum amount of the ompound that dissolves to ahieve a saturated solution. For silver sulfide, AgS, the solubility of 8.0 10 17, this tells you the onentration of the AgS that dissolves, whih is equal to the sulfide ion that forms, and the silver ion is twie that due to the dissolving reation, AgS Ag + + S thus Ksp [Ag + [S. Substitute 1.6 10 16 16 10 17 Thus Ksp (s)(s) and Ksp 4s solve for s. 10 8 10 9 s 8 10 9 10 4 4 Sine s onentration of the Pb +, the 1:1 stoihiometry the onentration of the PbI that atually dissolves, this s would be the molar solubility. (If you needed a gram solubility, you would have to multiply by the molar mass of PbI.) 4.0 10 8 1 10 5 1 10 5 1 10 5 4 10 4 K sp 16 10 17 8 10 17 7. b In order to answer this question, you need to alulate Q and ompare it to Ksp. For this ombination, the possible preipitate is AgCl, thus the Ksp reation of interest is AgCl Ag + + Cl and Ksp [Ag + [Cl. The trik in this problem is to remember that when solutions ombine, the onentrations of the ions will be diluted beause there is more volume for the same number of moles. Sine equal volumes are ombined, eah onentration is halved. [Ag + [Cl 0.8 10 5, or 8 10 6, So Q [8 10 6 [8 10 6 whih results in Q 64 10 1 or 6.4 10 11 In effet, reation shifts to the right and no preipitate forms when Ksp > Q.
! Pratie MC Unit F Ksp and Kp and K ANSWERS (page 7 of 7) 8. a The formula for metal(ii) sulfide is MS whih leads to the reation MS M + + S and the resulting Ksp expression is Ksp [M + [S and thus Ksp s Change Ksp to 16 10 1 to make it easier without a alulator, and then solve for s whih will be the solubility. 9. e The formula for metal hydroxide, XOH leads to the reation XOH X + + OH and the resulting Ksp expression is Ksp [X + [OH and thus Ksp s Solve for s whih will be the solubility, s 1 10 4, and then take the poh log[oh 4, then 14-it to solve for ph 10. 10. b The formula for metal hydroxide is X(OH) whih leads to the reation M(OH) M + + OH and the resulting Ksp expression is Ksp [M + [OH. ph of 9 should be onverted to a poh 5, whih an be unlogged to [OH 1 10 5. Due to the stoihiometry of the reation above, the [M + 0.5 10 5. Now insert the values bak into the Ksp formula. Ksp [0.5 10 5 [1 10 5 11. b It is tempting to just look at the Ksp and assume that the smallest Ksp means the lowest solubility, however, you must onsider the amount of ions that the salt dissoiates into. Sine the AgCO salt has ions, Ksp 4s, whereas for the other two salts, the Ksp s. Thus for AgBr s 54 10 14 ~ 7 10 7, for AgCl s 1.8 10 10 ~ 1 10 5, but for 4s 8 10 1, thus 10 1 ~ 1 10 4 1. d There is no need to make this problem any harder than neessary. Sine all three salts are ion salts, alulating the Ksp will be a simple squaring of the solubility. Thus the dereasing order of the Ksp values will be in the same order as the solubility values. 1. b HCl will reat with the OH ion, lowering the OH, ausing a shift to the right resulting in greater solubility, thus not the answer. This problem requires you to pik out the substane with a ommon ion, and both NaOH and MgCl will satisfy that riteria. So there is a twist. Remember the solubility an be determined by solving for s from the equilibrium expression. K sp Mg + OH. For the introdution of 0.1 M MgCl, you an simply substitute the onentration of [Mg + into the equilibrium expression, and the equation to solve for s would look like this! 6. 10 10. Divide by 0.1 for, 0.1 s 6. 10 9, onvert to 6 10 10, then divide by 4, to get ~16 10 10 and square root to solve for s ~4 10 5. For the introdution of 0.1 M Mg(OH), you an simply substitute the onentration of [OH into the equilibrium expression, and the equation to solve for s would look like this 6. 10 10. Square 0.1, then divide Ksp by s 0.1 0.01, and solve for s 6. 10 8. It should now be obvious that the squared effet of the onentration of the ommon ion [OH auses the NaOH to have the greater effet on solubility. 14. b Remember that solubility desribes the amount of the hemial that does dissolve. Sine SrF dissolves and dissoiates aording to the following equation, SrF Sr + + F the onentration of Sr + will be the same as the solubilty, and the onentration of F will be the solubility. 15. d When evaporation ours, only water will evaporate, the salt will stay in the solution. Sine there an NOT be a greater onentration in solution, that the saturation quantity, preipitation will have to our to maintain the saturation onentrations. 16. a Addition of NaF(s) provides the ommon ion [F,whih will serve to ause the equilibrium to shift left, thus reduing the [Sr +. Addition of Sr(NO)(s) would also ause a shift to the left, but the addition of the [Sr + ions will inrease [Sr +, and while the [F will shift below the original [F, the [Sr + will never be able to redue below the original [Sr + onentration before the addition of Sr(NO)(s). 17. e Changing the volume will NOT ause any shift in the equilibrium position sine Q will not hange due to the fat that the stoihiometry of the gas moleules is the same on both sides of the equation. Just as adding an inert gas will not Choies would all ause no shift. Choie d would ause a shift to the left. Thus e is the orret hoie.
! Pratie MC Unit F Ksp and Kp and K ANSWERS (page 8 of 8) 18. d This is a lassi RICE box and then solve for K. It is true that the math of the K alulation is not so simple, yet the answers are quite spread, so approximations make it possible. So, noting that the ontainer is 1.0 L, plug in the known info in the problem (in blue) then alulate the rest of the items with stoihiometry.! [0.4M K to do the easy math revise this! you an [0.4 [0.1M [0.1M [0.1[0.1 whih will then lean up to! 0.16 whih is 16, hoose d 1 10 R I(g) + Br(g) IBr(g) I 0. M 0. M 0 C 0. M 0. M +0. M E 0.1 M 0.1 M 0.4 M 19. In this problem, it is important to remember that solids (nor liquids) should be inluded in the solubility expression. 0. e This is a lassi RICE box and then solve for K. So, noting that the ontainer is 1.0 L, plug in the known info in the problem (in bold) then alulate the rest of the items with stoihiometry. [0.8M [0.4M [0.8M [0.4M [0.4M [0.8M [0.8M [M [0.8M 1.6 [0.4M R SO(g) SO(g) + O(g) I 1. M 0 0 C 0.8 M +0.8 M +0.4 M E 0.4 M 0.8 M 0.4 M 1. d Inreasing the temperature of an endothermi proess will ause a shift toward produts, thus ausing more ions to go into solution.. e For this problem there is no reason to do a rie box sine you were given the equilibrium mole values. It is important to note that the ontainer is.0 L. I plugged in the moles and the volume diretly into the equilibrium expression for the equation CO(g) + Cl(g) COCl(g) K Yikes, the math looks to be a mess, and you don t have a [COCl [ 1.5 K [CO[Cl [ 0. [0.5 alulator, so look for fators that are easy to anel and end up with 0. d Again, for this problem there is no reason to do a rie box sine you were given the equilibrium pressure values. I plugged in the moles and the volume diretly into the equilibrium expression for the equation HO(g) + ClO(g) HOCl(g)! [HOCl This time the math isn t so bad.! [0 K whih approximates to! [400 drop some zeros [H O[ClO [15[96 [15[~ 00 and all it! [4 K whih is! whih is something just smaller than 1/10th or hoie d [4 [15[~ [45 4. b For this problem, you must plug into the equilibrium expression and solve for! Q [.5 10 6 [1.11 10 5 you an do some [.5 10 estimating sine you probably will be omparing orders of magnitude. I anelled out the.5 and.5 to get Q [1 10 6 [1.11 10 5 Remember that 10 in the denominator is 10 in the numerator, thus [1 10 Q (1 10 6 )(1.11 10 5 )(1 10 ) whih is Ksp < Q (1 10 8 ) given in the problem, thus the equilibrium will shift left to redue the value of Q. 5. d Starting with only W and X, Y ould never be equal to Z due to the :1 stoihiometry. We do not have enough information to know where the equilibrium position will land, so we annot ompare X to Z. W ould not be equal to X beause of starting with equal quantities of them, and the :1 ratio between them, thus we an tell that X will be greater than W beause W reats x the rate of X. Choie e is ertainly not what equilibrium means. 6. e rate laws and rate k onstants gives us info about speed. Keq tells us nothing about speed of reation, only where the position ends up. 7. b Sine we know from the reation is exothermi, think of energy on the right side. Raising the temp of the system, and the reation will shift to use some of this inoming heat, that s left.
Pratie MC Unit F Ksp and Kp and K ANSWERS (page 9 of 9) 8. d It is a 1.0 L flask, so moles are equal to molarity. This is easy math ompared to those others.! 9. b Plug in values to solve for! [0. whih is! [0.04 Q Q whih is! Q [4 10 whih is 400 whih is greater than [0.1[0.1 [0.1[.001 [1 10 4 the K given in the problem and thus the shift will be left. (Note: If you have trouble multiplying or squaring deimals, turn 0.0 into 10 1 whih is might be easier to ope with.) 0. a Only I is true. For II when the stoihiometry of a reation is hanged, the equilibrium onstant of the original equation should be raised to the power of the stoihiometry hange. For III, when reations are added, the produt of the equilibrium onstants of eah reation added will the equilibrium onstant of the new equation. 1. e Note that PCl in this problem is a liquid, so it should not be inluded in the equilibrium expression.. In this problem, take note that the reatant is a solid. The produt gases ome only from the solid reatant, thus the two produts must be present in equal quantities, this allows the total equilibrium pressure to be ut in half to get eah gases partial pressure, 0. atm for eah gas. Then you an square the partial pressures to solve for Kp. (0.)(0.). Plug into the K expression and solve for [H 8 [0.01[H, thus [H 0.8 and multiply by 8.0 L to get 6.4 moles. [0.01[0.1 4. a You are looking for a reation in whih the number of moles of gases are the same on both sides. [9 [[ 5. e If oxygen is injeted at equilibrium, stress has been applied and the system must shift to restore equilibrium. Thus a shift toward reatants will our, whih will redue the quantity of arbon monoxide. 6. b Sine we know that the reation is exothermi, think of energy on the right side. Raising the temperature of the system, and the reation will shift to use some of this inoming heat, that s left, thus reduing the value of K, sine K is onstant only for one partiular temperature. 7. You are looking for a reation in whih the number of moles of gases are the same on both sides. (Note there was a typo in the original version of the test in hoie a 8. a Dereasing the volume of the ontainer will inrease the partial pressure of all the gases and ause the equilibrium to shift toward the produts where there are less gas moleules. Raising the temp would ause a shift left, and adding argon would have no effet on partial pressures (even though the total pressure is inreased), and thus would reate no shift at all.