Solubility Equilibrium, Ksp and more Name

Transcription

1 MC Practice F.2 (Ch 15 & ) This is practice - Do NOT cheat yourself of finding out what you are capable of doing. Be sure you follow the testing conditions outlined below. DO NOT USE A CALCULATOR. You may use ONLY the blue periodic table. Try to finish at a rate of about 1.5 min per question. Time yourself. It is important that you practice working for speed. Then when time is up, continue working and finish as necessary. Please mail me if you think you find any issues or typos. Solubility Equilibrium, Ksp and more Name 1. Temperature is often given with Ksp values because a. the solubility of solids always increases with increasing temperature. b. the solubility of solids varies with temperature changes. c. solubility changes with temperature but Ksp values do not. d. Ksp varies with temperature even though concentrations do not. e. the number of ions varies with the kind of salt that is dissolving. 2. The Ksp expression for silver phosphate is a. K sp = [Ag + [PO 4 3 b. K sp = [Ag + 2 [PO 4 3 c. K sp = [Ag + 3 [PO 4 3 d. K sp = [Ag + [PO e. K sp = [Ag+ [PO 4 3 [Ag 3 PO 4 3. The solubility of nickel(ii) hydroxide, Ksp = , is about a x10 17 b x c. 1.6x10 16 d. (1.6x10 16 )(3) e. (1.6x10 16 )(4) 4. Determine the solubility of lead(ii) fluoride, Ksp = 4.0 x 10 8, in a M lead(ii) nitrate solution. a M b M c. 4.0x10 5 M d. e. 1.0x10 5 M 2 4.0x x10 3 M 5. The solubility of magnesium hydroxide (Ksp = 8.0 x ) in a buffered solution of ph = would be a M b M c M d M e M 6. The solubility of silver sulfide is 8.0 x M. Determine the Ksp of this salt. a b. ( )( ) c. ( ) 2 ( ) d. ( )( ) e. ( ) 2 ( ) 7. Equal volumes of 1.6 x 10 5 M KCl and 1.6 x 10 5 M AgNO3 are mixed. the Ksp for silver chloride is As these two solutions intermingle, a. a precipitate of AgCl forms. b. there is no precipitate formed. c. NaCl will precipitate. d. AgNO3 will precipitate. e. the [Na + will become M. 8. A saturated solution of a strong base, MOH, has a ph of 11.00, what is the Ksp for this base. a b c d e The Ksp of the metal hydroxide, MOH, is , what is the ph of a saturated solution of MOH? a b c d e

2 Practice Multiple Test Unit F General and Solubility Equilibrium Kp Ksp page 2 of In the reaction shown below, 0.50 mole of Br2 and 0.50 moles of I2 are placed in an evacuated 1.00 liter vessel and allowed to reach equilibrium. What is the value of Kc if the vessel contains 0.84 moles of IBr at equilibrium? a. 2.0 b. 8.8 c d. 110 e. 1 I2(g) + Br2(g) 2IBr(g) 11. Which is the correct set up to determine Kc for the reaction below? 4CuO(s) + CH4(g) CO2(g) + 4Cu(s) + 2H2O(g) a. [CO 2 [Cu 4 [H 2 O 2 [CuO 4 [CH 4 b. [CuO 4 [CH 4 [CO 2 [Cu 4 [H 2 O 2 c. [CO 2 [H 2 O 2 [CH 4 d. [CO 2 [H 2 O 2 [CuO 4 [CH 4 e. [CO 2 2 [Cu 4 [H 2 O 2 [CuO 4 [CH An evacuated 1.00-liter vessel is injected with moles of sulfur trioxide gas, SO3. The vessel is then heated to a high temperature where the SO3 partially decomposes to form the products SO2 and O2 in the reaction shown below: 2SO3(g) 2SO2(g) + O2(g) The temperature is kept constant, and the amount of SO3 in the vessel at equilibrium is mol. What is the value of Kc at this temperature? a b c d e What is the molar solubility in water of PbI2? (The Ksp for PbI2 is ) a M b M c. (1.6) ½ 10 8 M d. (1.6) ⅓ 10 8 M e. 2 x 10 3 M 14. A 3.00-liter reaction vessel is filled with carbon monoxide gas, CO, and chlorine gas, Cl2. The mixture is heated to 670 K and allowed to reach equilibrium according to the balanced equation shown below: CO(g) + Cl2(g) COCl2(g) At equilibrium, the mixture contains moles CO, moles Cl2, and 1.11 moles COCl2. What is Kc at this temperature? a b c d. 411 e The reaction below is allowed to come to equilibrium at room temperature. At equilibrium, the partial pressure of H2O is 296 mm Hg, Cl2O is 15 mm Hg, and HOCl is 20 mm Hg. What is the value of Kp at this temperature? a. 222 b. 11 c d e H2O(g) + Cl2O(g) 2HOCl(g) 16. For the reaction shown below, Kp = at 400 K. 2NH3(g) N2(g) + 3H2(g) What is the value of Kc at this temperature? a b c d e At 373 K, the reaction shown below has an equilibrium constant, Kc = COCl2(g) CO(g) + Cl2(g) After placing a mixture of gases in the reaction vessel, the concentrations were measured to be [COCl2 = M, [CO = M, and [Cl2 = M. Which statement below accurately describes the reaction? a. The reaction is at equilibrium b. The reaction is not at equilibrium, and it is proceeding to the left c. The reaction is not at equilibrium, and it is proceeding to the right d. The reaction quotient is equal to Kc e. The reaction quotient is less than Kc

3 Practice Multiple Test Unit F General and Solubility Equilibrium Kp Ksp page 3 of In the reaction 3W + X 2Y + Z, all substances are gases. The reaction is initiated by adding equal number of moles of W and of X. When equilibrium is reached, a. [Y = [Z b. [X = [Y c. [W = [X d. [X > [W e. [W + [X = [Y + [Z 19. Consider the reactions 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 3 times faster than II. b. I is 1000 times faster than II. c. II is 3 times faster than I. d. II is 1000 times faster than I. e. The size of K and the time required to reach equilibrium are not directly related. 20. The reaction 3H2(g) + N2(g) 2NH3(g) has an enthalpy of change of 92 kj. Increasing the temperature of this equilibrium system causes a. an increase in [NH3 b. an increase in [N2 c. a decrease in [H2 d. an increase in K e. a decrease in pressure at constant volume 21. Consider N2(g) + O2(g) 2NO(g). The reaction was initiated by adding 15.0 moles of NO to a 1.0-L flask. At equilibrium, 3.0 moles of oxygen are present in the 1.0-L flask. The value of K must be a b. 3.0 c. 5.0 d. 9.0 e At a certain temperature, the synthesis of ammonia gas from nitrogen and hydrogen gases, shown below, has a value for K of If [H2 = [N2 = 0.10 M and [NH3 = 0.20 M N2 + 3H2 2NH3 a. the reaction would shift toward the ammonia b. the reaction would shift toward the N2 and the H2 c. the system is at equilibrium, therefore no shifting will occur d. the reaction will shift toward a new equilibrium position, but the direction cannot be determined from these data e. the equilibrium may shift but it is not possible to calculate Q without knowing the temperature 23. The equilibrium constant, K, may be used to determine the K of other reactions. I. When the equation is reversed the reciprocal of the original K becomes the value of K for the new equation. II. When the equation is doubled the square root of the original K becomes the value for the new equation. III. When the equation for a reaction is tripled the equilibrium expression for the new equation is triple the original K. Of the above three statements, those which are always true are a. I only b. II only c. III only d. I and III e. I, II, and III 24. The equilibrium P4(g) + 6Cl2(g) 4PCl3(L) is established at 10 o C. The equilibrium constant expression is a. K = [PCl 3 [P 4 [Cl 2 b. K = [PCl 3 4 [P 4 [Cl 2 6 c. K = [P 4 [Cl 2 6 [PCl 3 4 d. 1 K = [PCl 3 4 e. 1 K = [P 4 [Cl 2 6

4 Practice Multiple Test Unit F General and Solubility Equilibrium Kp Ksp page 4 of Ammonium hydrogen sulfide will decompose into ammonia gas and hydrogen sulfide gas when heated. Consider the equilibrium system NH4HS(s) NH3(g) + H2S(g) which is developed from mole of NH4HS in a 100- L cylinder. At equilibrium the total pressure is found to be atm. Kp will be equal to a b c d e. a value impossible to calculate from only these data 26. At a certain temperature, it has been determined that K = 8.0 for the reaction below. If we also determine that the equilibrium mixture contains 0.80 mole of H2O(g), mole of CO2(g), and mole of CO(g), in an 8.0-L flask, what must be the number of moles of H2(g) at equilibrium? a mole b mole c. 6.4 moles d. 8.0 moles e. 64 moles H2O(g) + CO(g) CO2(g) + H2(g) 27. Which of the following systems at equilibrium are not affected by a change in pressure caused by changing the volume at constant temperature? a. H2(g) + Cl2(g) 2HCl(g) b. H2(g) + I2(s) 2HI(g) c. N2(g) + 3H2(g) 2NH3(g) d. 2NH3(g) N2(g) + 3H2(g) e. 3O2(g) 2O3(g) 29. Which of the following changes alone would cause a decrease in the value of Keq for the reaction represented below? 2SO2(g) + O2(g) 2SO3(g) ΔH o = -197 kj a. Decreasing the temperature b. Increasing the temperature c. Decreasing the volume of the reaction vessel d. Increasing the volume of the reaction vessel e. Adding a catalyst 30. In which of the following systems would the number of moles of the substances present at equilibrium NOT be shifted by a change in the volume of the system at a constant temperature? a. 2SO2(g) + O2(g) 2SO3(g) b. N2(g) + 3H2(g) 2NH3(g) c. NO2(g) + SO2(g) SO3(g) + NO(g) d. N2O4(g) 2NO2(g) e. CO(g) + 3H2(g) CH4(g) + H2O(g) 31. For: N2(g) + 3H2(g) 2NH3(g) + energy Some N2 and H2 are mixed in a container at 200 o C, and the system reaches equilibrium according to the equation above. Which of the following causes an increase in the number of moles of NH3 present at equilibrium? I. Decreasing the volume of the container II. Raising the temperature III. Adding a mole of Ar gas at constant volume a. I only b. II only c. I and III only d. II and III only e. I, II, and III 28. After the equilibrium represented below is established, some pure O2(g) is injected into the reaction vessel at constant temperature. After equilibrium is reestablished, which of the following has a lower value compared to its value at the original equilibrium? 2CO2(g) 2CO(g) + O2(g) a. Keq for the reaction b. The total pressure in the reaction vessel c. The amount of CO2(g) in the reaction vessel d. The amount of O2(g) in the reaction vessel e. The amount of CO(g) in the reaction vessel 32. Cu2S(L) + O2(g) 2Cu(L) + SO2(g) ΔH = 250 kj The reaction above is used in the mining industry to extract copper from copper ore. Once the mixture is allowed to establish equilibrium at temperature T and pressure P, the equilibrium can be shifted to favor the products by a. increasing the pressure by decreasing the volume of the reaction vessel at constant T. b. increasing the pressure by adding an inert gas such as helium. c. decreasing the temperature. d. allowing some gases to escape at constant T and P. e. adding a catalyst.

5 Practice Multiple Test Unit F General and Solubility Equilibrium Kp Ksp page 5 of HI(g) + Cl2(g) 2HCl(g) + I2(g) + energy A gaseous reaction occurs and comes to equilibrium as shown above. Which of the following changes to the system will serve to increase the number of moles of I2 present at equilibrium? a. Increasing the volume at constant temperature. b. Decreasing the volume at constant temperature. c. Adding a mole of inert gas at constant volume. d. Increasing the temperature at constant volume. e. Decreasing the temperature at constant volume. 34. A sample of solid potassium nitrate is placed in water. The solid potassium nitrate comes to equilibrium with its dissolved ions by the endothermic process shown below. KNO3(s) + energy K + (aq) + NO3 (aq) Which of the following changes to the system would increase the concentration of K + ions at equilibrium? a. The volume of the solution is increased. b. The volume of the solution is decreased. c. Additional solid KNO3 is added to the solution. d. The temperature of the solution is increased. e. The temperature of the solution is decreased

6 Practice Test Unit F (Ch 15 & 17end) (pg 6 of 8) Keq, Kp, Kc and Ksp ANSWERS 1. b All K values always change with varying temperature. In unit C, Thermodynamics, you learned that the dissolving of different salts is sometimes exothermic and sometimes endothermic. We will study this further in chapter 13, Unit K. 2. c The formula for silver phosphate is Ag3PO4 which leads to the reaction Ag3PO4 3Ag + + PO a The formula for nickel(ii) hydroxide is Ni(OH)2 which leads to the reaction Ni(OH)2 Ni OH and the resulting Ksp expression is Ksp = [Ni 2+ [OH 2. Since there are 2 OH ions for every Ni 2+ ion upon dissolving, call Ni 2+ x and OH 2x and substitute into the Ksp expression. Thus Ksp = (x)(2x) 2 Since x = concentration of the OH, the 1:1 stoichiometry the concentration of the Ni(OH)2 that actually dissolves, this is called the molar solubility, is the same value as x. 4. d The formula for lead(ii) fluoride is PbF2 which leads to the reaction PbF2 Pb F and the resulting Ksp expression is Ksp = [Pb 2+ [F 2. This is a common ion problem. You are given the concentration of the Pb 2+ ion which you can insert directly into the Ksp expression. Thus Ksp = (0.004)[F 2, set equal to the Ksp value given, and solve for [F, without a calculator, the best you can do is [F = (1 x 10 5 ). Since there are 2 F ions for every PbF2 that actually dissolves, this is called the molar solubility, you need to divide 1 x 10 5 by 2 to get the molar solubility of PbF2 5. e A buffered solution (which you will learn more about in the next unit G) is a solution in which the ph remains fairly constant. Further, ph = 11, means poh = 3 and since poh = log[oh, you need to know in your head that log(1 x 10 3 ) = 3, thus [OH = 1 x This is a common ion problem and you can just plug in the concentration of the common ion since any difference due to LeChatelier adjustments will be insignificant compared to the relatively large concentration of the common ion. The dissolving of Mg(OH)2 leads to the reaction Mg(OH)2 Mg OH and the resulting Ksp expression is Ksp = [Mg 2+ [OH 2, so 8.0 x = [Mg 2+ [1 x c When you are told the solubility of some salt, you are being told the amount of the compound that can dissolve. For silver sulfide, Ag2S, the solubility of 8.0 x 10 17, this tells you the concentration of the Ag2S that dissolves, which is equal to the sulfide ion that forms, and the silver ion is twice that due to the dissolving reaction, Ag2S 2Ag + + S 2 and Ksp = [Ag + 2 [S 2-. Thus silver is 16 x 10 17, insert these concentrations into the Ksp expression. 7. b You need to calculate Q and compare it to Ksp. For this combination, the possible precipitate is AgCl, thus the Ksp reaction of interest is AgCl Ag + + Cl and Ksp = [Ag + [Cl. It is important to remember that when solutions combine, the concentrations of the ions will be diluted because there is more volume for the same number of moles. Since equal volumes are combined, each concentration is halved. [Ag + = [Cl - = 0.8 x 10 5, or 8 x 10 6, So Q = [8 x 10 6 [8 x 10 6 which results in 64 x or 6.4 x which is < Ksp. No precipitate forms when Q < Ksp 8. a In this problem, we assume that the saturated solution is made by putting MOH in water and allowing it to dissolve to the maximum amount possible. We know this happens if stirred for a long time, and solid still remains at the bottom. When you are told the ph = 11, you know the poh = 3, and since poh = log[oh, then you know[oh = 1 x 10 3, then insert into Ksp = [M + [OH 9. c This problem is very similar to #3 & 4. The reaction MOH M + + OH and Ksp = [M + [OH, and since [OH = [M + the square root of the Ksp will result in [OH = 1 x 10 3 and poh = log[oh which will be 3, thus the ph = d This is a classic RICE box and then solve for Kc. It is true that the math of the Kc calculation is not so simple, though it should be easy enough to calculate that d or e would be the only possible choices. I think that it is not so fair to choose among those two - I would rewrite this problem with the e choice as So, noting that the container is 1.0 L, plug in the known info in the problem (in blue) then calculate the rest of the items with stoichiometry. Kc = K c = [ [0.8[0.8x10 2 [0.84M 2 [0.08M [0.08M to do the easy math revise this you can estimate the 0.84 and 0.8 will cancel, which R I2(g) + Br2(g) 2 IBr(g) I 0.5 M 0.5 M 0 C 0.42 M 0.42 M M E 0.08 M 0.08 M 0.84 M will then clean up to 1 Kc = which is ~100 ish, choose d (if I were to rewrite the problem so e was not an 1x10 2 option so close to d 11. c In this problem, it is important to remember that solids (nor liquids) should be included in the solubility expression.

7 AP Chem Practice Multiple Choice Chapter 10 page 7 of a This is a classic RICE box and then solve for Kc. It is true that the math of the Kc calculation is not so simple, but take another look and check out my suggestions below to simplify the math - not R the only way to look at it, but it helped me get through the math without a calculator. So, noting that the container is 1.0 L, plug in the known info in the problem (in blue) then calculate the rest of the items with stoichiometry. 13. e K c = [0.257M 2 [0.1285M [0.52M 2 next I assumed that 0.257/0.52 reduces to approximately ½ giving (½) 2 (0.1285) which cleans up to ¼(0.1285) which is close enough to 0.03 ish, choice a 14. e For this problem there is no reason to do a rice box since you were given the equilibrium mole values. It is important to note that the container is 3.0 L. I plugged in the moles and the volume directly into the equilibrium expression for the equation CO(g) + Cl2(g) COCl2(g) Kc = [COCl 2 2 It is true that the math of the Kc calculation is not so simple, but take [CO[Cl 2 another look and check out my suggestions below to simplify the math - not the only way to look at it, but it helped me [ 1.11 get through the math without a calculator. Kc = 3 next I cancelled out two of the threes and divided the [ [ other into the to give Kc = and change to 0.74 (¾) so I revised to Kc = which is Kc = 1000[ 4, so the e choice seemed best. 3 2SO3(g) 2SO2(g) + O2(g) I M 0 0 C M M M E 0.52 M M M [1.11, then I wanted to cancel out the 1.11/0.012 since they are close enough [0.012[0.075 [1.11 so now I have 1 Kc = which is Kc = 1000 [1.2[0.75x10 3 [0.751x10 3 [ d Again, for this problem there is no reason to do a rice box since you were given the equilibrium pressure values. I plugged in the moles and the volume directly into the equilibrium expression for the equation H2O(g) + Cl2O(g) 2HOCl(g) K c = [HOCl 2 This time the math isn t so bad. [20 2 Kc = [H 2 O[ClO 2 [15[296 zeros and call it Kc = which approximates to Kc = [400 [15[~ 300 [4 which is Kc = [4 which is something just smaller than 1/10th or choice d [15[~ 3 [45 drop some 16. b Remember that Kc = Kp(RT) n where n = (reactant coefficients product coefficients). So plug-in and then estimate your way to an answer. Kc = 2.8 x 10 2 (0.0821*400) 2 so move the to the denominator [2.8x10 2 K p = and think of it [ as ~8*4 (move decimal on two to the right which is canceled out by moving the decimal on the 400, 2 places to the right K p = [2.8x10 2 [32 2 so then I am thinking 32 is ~30 which squares to ~900, but since it s 32, maybe it s closer to 1000 thus Kc = [2.8x10 2 [1000 which is Kc = [2.8x10 2 [10 3 which is K c = [2.8x10 2 [10 3 which is pretty close to choice b 17. b For this problem, you must plug into the equilibrium expression and solve for [3.25x10 6 [1.11x10 5 Q = [3.5x10 3 you can do some estimating since you probably will be comparing orders of magnitude. I cancelled out the 3.25 and 3.5 to get Q = [1x10 6 [1.11x10 5 Remember that 10-3 in the denominator is 10 3 in the numerator, thus [1x10 3 Q = (1 x 10 6 )(1.11 x 10 5 )(1 x 10 3 ) which is Q = (1 x 10 8 ) > Ksp given in the problem, thus the equilibrium will shift left to reduce the value of Q.

8 AP Chem Practice Multiple Choice Chapter 10 page 8 of d Starting with only W and X, Y could never be equal to Z due to the 2:1 stoichiometry. We do not have enough information to know where the equilibrium position will land, so we cannot compare X to Z. W could not be equal to X because of starting with equal quantities of them, and the 3:1 ratio between them, thus we can tell that X will be greater than W because W reacts 3x the rate of X. Choice e is certainly not what equilibrium means. 19. e rate laws and rate k constants gives us info about speed. Keq tells us nothing about speed of reaction, only where the position ends up. 20. b Since we know from the negative G that the reaction is exothermic, think of energy on the right side. Raising the temp of the system, and the reaction will shift to use some of this incoming heat, that s left. 21. d It is a 1.0 L flask, so moles are equal to molarity. This is easy math compared to those others. Kc = [92 [3[3 22. b Plug in values to solve for Q = [0.2 2 [0.1[0.1 3 which is Q = [0.04 [0.1[.001 which is Q = [4x10 2 [1x10 4 which is 400 which is greater than the K given in the problem and thus the shift will be left. (Note: If you have trouble multiplying or squaring decimals, turn 0.02 into 2 x10-1 which is might be easier to cope with.) 23. a Only I is true, refer to the summary on page 585 in your text or the Intro to Equilibrium powerpoint posted online 24. e Note that PCl3 in this problem is a liquid, so it should not be included in the equilibrium expression. 25. c In this problem, the product gases come only from the solid reactant, thus the two products must be present in equal quantities, this allows the total equilibrium pressure to be cut in half to get each gases partial pressure. Then you can square the partial pressures to solve for Kp. 26. c Plug into the Kc expression and solve for [H2 [0.01[H 8 = 2, thus [H2 = 0.8 and multiply by 8.0 L to get 6.4 moles. [0.01[ a You are looking for a reaction in which the number of moles of gases are the same on both sides. 28. e If oxygen is injected at equilibrium, stress has been applied and the system must shift to restore equilibrium. Thus a shift toward reactants will occur, which will reduce the quantity of carbon monoxide. 29. b Since we know from the negative H that the reaction is exothermic, think of energy on the right side. Raising the temperature of the system, and the reaction will shift to use some of this incoming heat, that s left, thus reducing the value of K, since K is constant only for one particular temperature. 30. c You are looking for a reaction in which 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 choice a 31. a Decreasing the volume of the container will increase the partial pressure of all the gases and cause the equilibrium to shift toward the products where there are less gas molecules. Raising the temp would cause a shift left, and adding argon would have no effect on partial pressures (even though the total pressure is increased), and thus would create no shift at all. 32. c Choice a causes no shift since there are equal number of moles of gases on both sides. Inert gases cause no shift. Decreasing temp of an exothermic reaction, as indicated by the negative H, will cause a shift toward products making c the correct choice. Adding a catalyst has no effect on the equilibrium position. Choice d is a bit odd, What do they mean by allowing some gases to escape at constant T and P? I suppose if we assume both O2 and SO2 were released equally, there would be no effect since the ratio of gases would still be constant. I suppose you might imagine that more O2 would be released since it has a higher rate of effusion, however even if the increased rate of effusion were true that caused more O2 to escape, this would cause a shift left, not right. 33. e Choices a, b, and c would all cause no shift. Choice d would cause a shift to the left. Thus e is the correct choice. 34. d Increasing the temperature of an endothermic process will cause a shift toward products, thus causing more ions to go into solution.