1 The solubility of Cd(IO 3 ) 2 is g/l at 25 C. What is K sp for this compound? A B C D E. 7.

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1 version: master Exam 3 - McCord This exam should have 20 questions. The point values are given with each question. Bubble in your answer choices on the bubblehseet provided. Your score is based on what you bubble on the bubblesheet and not what is circled on the exam. Below are some constants you might want to use. Compound Ag 2 SO 4 BaCO 3 BaSO 4 K sp Compound Ca(OH) 2 CuCO 3 PbBr 2 K sp Compound PbCl 2 PbI 2 Y(OH) 3 K sp The solubility of Cd(IO 3 ) 2 is g/l at 25 C. What is K sp for this compound? A B C D E Explanation: The molecular mass of cadmium iodate is g/mol.(0.529g/l)/(462.2g/mol) = M. [Cd 2+ ] = M, [IO 3 ] = M. K sp = [Cd 2+ ][IO 3 ]2 = ( )( ) 2 = The closest choice is the Which of the following compounds has the highest molar solubility in water? A. Co(OH) 2, K sp = B. FeCO 3, K sp = C. Mg 3 (AsO 4 ) 2, K sp = D. CuI, K sp = Explanation: CuI and FeCO 3 both have x = Ksp the larger one is FeCO 3 and gives you x = M. Co(OH)2 will have x = (Ksp/4) 1/3 = But Mg 3 (AsO 4 ) 2 will have the formula x = (Ksp/108) 1/5 = which is the largest value of the four choices given.

2 version: master Exam 3 - McCord page 2 3 The Setup: An aqueous solution is allowed to saturated with the salt MX(s) and it turns out that the resulting solution has a X entration of M. After sitting for 24 hours a 100 ml sample of this saturated solution is pipetted into another container. No solid on the bottom of the flask was transferred, only the solution part. Now a small portion (between 20 and 35 ml) of a M solution of NaX is added to the pipetted solution. Question: What is the K sp for MX and does more solid precipitate when this other solution is added? A. K sp = and no precipitate forms. B. K sp = and a precipitate does form. C. K sp = and no precipitate forms. D. K sp = and a precipitate does form. E. K sp = and no precipitate forms. F. K sp = and a precipitate does form. Explanation: The Ksp is just the squared x 2 = (.001) 2 = There can be no precipitate because the X entration never changes (both solutions are M) and the M + entration drops due to dilution which means the solutions goes undersaturated and no precipitation occurs. 4 Yttrium hydroxide is added to water to make a saturated solution 25 C. What is the ph of this solution? A B C D E F G H I J Explanation: The rice table for Y(OH) 3 leads to K s p = (x)(3x) 3. Solving for x you get x = (K s p/27) 1/4, where x = (10 22 /27) 1/4 = [OH-] = 3x = M. poh = 5.38 and ph = 8.62

3 version: master Exam 3 - McCord page 3 5 A large scoop of Ag 2 SO 4 is added to a liter of water and stirred to make a saturated solution. What is the entration of silver ion in this solution? A M B M C M D M E M Explanation: K sp = [Ag + ] 2 [SO 2 4 ], [Ag+ ] = 2x, [SO 2 4 ] = x. Therefore K sp = 4x 3 = 1.2 x Solving for x you get M. [Ag + ] = 2x = M. 6 If you mix 300 ml of a 0.05 M solution of Pb(NO 3 ) 2 solution with 200 ml of a M solution of CaCl 2, will you form a precipitate? and Why? A. no, because Q sp < K sp B. yes, because Q sp < K sp C. no, because Q sp > K sp D. yes, because Q sp > K sp E. no, because Q sp = K sp F. yes, because Q sp = K sp Explanation: You need to compare Q sp to K sp. Q sp = [Pb 2+ ][Cl ] 2. You must first account for the diluting of adding 300mL to 200mL. [Pb 2+ ] = (.05M)(300mL/500mL) = 0.03 M. For the chloride ion, the entration in the original solution is double that of the formula unit (0.05M), accounting for dilution, [Cl ] = (0.05M)(200mL/500mL) = 0.02M. Q sp = [Pb 2+ ][Cl ] 2 = (.03)(0.02) 2 = which is less than When Q sp < K sp no precipitate is formed, solution is undersaturated. 7 What is the molar solubility of lead(ii) bromide in a solution of M NaBr? A B C D E Explanation: K sp = [Pb2+][Br ] 2. The bromide ion is set via the NaBr to So the amount that dissolves is the lead ion and [Pb2+] = Ksp/(0.255) 2 = M.

4 version: master Exam 3 - McCord page 4 8 A solution contains both the anions sulfate and carbonate at entrations of M each. Barium ion is introduced to this solution to selectively precipitate out the sulfate ion and leave the carbonate behind. What is the maximum percentage of the sulfate ion that can be precipitated with no interference from the carbonate ion? A % B % C % D % E % F % Explanation: [Ba2+] = 2.6e-9/(.001) = 2.6e-6 (the carbonate precip onset). [SO42-] = 1.1e-10/2.6e-6 = 4.23e-5 M sulfate still in solution /.001 x 100% = 4.23% still in solution which means that 95.77% has been precipitated. 9 The reaction: 2NO(g) + Cl 2 (g) 2NOCl is first order in chlorine and second order in nitrogen monoxide. What are the units for the rate constant? A. M 1 s 1 B. M s 1 C. M 1 D. M 2 s 1 E. s 1 Explanation: The reaction is third order overall. rate = k[no][cl 2 ] In order to get the desired M/s units for rate, k must have units of M 2 s 1 10 For the reaction: 2HI(g) H 2 (g) + I 2 (g) the rate constant at 650 K was determined to be When the temperature was increased to 700 K, the rate constant was found to be What is the activation energy for this reaction? A. 182 kj/mol B kj/mol C kj/mol D. 203 kj/mol E kj/mol Explanation: ln = E ( a ) 700 E a = 182,344 J/mol = 182 kj/mol

5 version: master Exam 3 - McCord page 5 11 The overall reaction 3A + B 2C + 5D is proceeding such that the rate of change in product D is M/min. What is the corresponding rate of change in reactant A? A M/min B M/min C M/min D M/min E M/min F M/min G M/min H M/min Explanation: The reactant A is running at 3/5 that of D and opposite in sign. 12 A decomposition reaction has the rate law k[a] 2 where the rate constant is M 1 s 1. About how long will it take for a 0.25 M sample of A to decompose down to M? A. 13 hours B. 96 days C. 48 days D. 35 minutes E. 122 days F. 21 hours Explanation: use 2nd order integrated rate law: t = (1/.001 1/.25)/ = 8.3e6 seconds. /60 /60 /24 = 96 days. 13 The decomposition of aqueous hydrogen peroxide to gaseous oxygen and water is a first-order reaction with a rate constant of k = hr 1. If it takes 6.5 hours for the entration of H 2 O 2 to decrease from 0.50M to 0.25M, how many hours are required for the entration to decrease from 0.40M to 0.10M? A. 13 hours B. 6.5 hours C hours D. 30 hours E hours F hours Explanation: There are two possible ways to solve this problem. In the general method, solve for using the first-order integrated [A] rate equation: 0 [A] = e kt which is equivalent to ln(a) = kt + ln(a 0 ). ln 0.10 = - ( hr 1 )(t) + ln 0.40, = - ( hr 1 )(t) , = - ( hr 1 )(t), t = 13 hr Alternatively, one could recognize that a decrease from 0.50 to 0.25 represents one half-life. A decrease from 0.40 to 0.10 is two half-lives. Since we are informed that one half-life equals 6.5 hrs, the to go from 0.40 to 0.10 must 13 hrs.

6 version: master Exam 3 - McCord page 6 14 A table of initial rate data are shown for the reaction 2A + B C Use the data in the table to determine the overall order for this reaction. A. zero B. first C. second D. third E. fourth F. fifth G. it is not possible to determine the order from this data [A] [B] rate (M) (M) (M/min) Explanation: As [A] doubles, the rate doubles, indicating the reaction is first order in A (2 1 = 2). As [B] doubles, the rate goes up by a factor of 8, indicating it is third order in B. (2 3 = 8). First order in A and third order in B means that the reaction is fourth order overall. 15 Which of the following factors does not ultimately affect the reaction rate? A. temperature B. pressure of a reactant C. amount of catalyst D. magnitude of G E. size of reactant particles Explanation: G does not affect reaction rates - it affects spontaneity and the equilibrium condition. 16 Balance the reaction:?h 2 S(g)+?O 2 (g)?s(s)+?h 2 O(l) Now, after further inspection, which one of the following is the proper lusion pertaining to the rate law? A. The reaction is first order with respect to H 2 S and second order with respect to O 2. B. The reaction is fourth order overall. C. The rate law is: rate = k[h 2 S] 2 [O 2 ]. D. The reaction is catalyzed by the O 2. E. The rate law cannot be determined from the information given. Explanation: A rate law can only be determined experimentally.

7 version: master Exam 3 - McCord page 7 17 Consider the simple reaction: A B At 25 C, the forward reaction has a rate constant equal to min 1, while the reverse reaction has a rate constant equal to min 1. Which of the following plots is the correct one for this reaction? A. B. C. D. E. Explanation: The ratio of the rate constants is 3 which is what the equilibrium constant, K, must equal. There is only one plot where the products finish at a 3:1 ratio more than the reactants. 18 Look at the entration vs plot to the right for a simple X Y reaction scheme. What is the rate of the reaction at 120 seconds? 0.15 [X] vs A M/min B M/min C M/min D M/min E M/min F M/min (mol/l) (min) Explanation: The absolute value of the tangential slope at 2 minutes is M/min. Notice the red line shown over the graph.

8 version: master Exam 3 - McCord page 8 19 How is it that a catalyst is able to speed up a reaction? A. By participating in the reaction mechanism and lowering the activation energy. B. By physically shifting the reaction to the right (products). C. By lowering the enthalpy of the reaction, H rxn. D. This is a false statement. Catalysts do not speed up reactions. E. By increasing the enthalpy of the reaction, H rxn. F. By raising the activation energy via reactant-catalyst interaction. G. By changing the energies of all species in the reaction scheme. Explanation: A catalyst alters the mechanism of a reaction by participating in the reaction mechanism and lowering the barrier of the rate limiting step which is the activation energy. 20 Consider the following 2-step mechanism for combining Cl atoms to make Cl 2. N 2 + Cl N 2 Cl (fast) N 2 Cl + Cl Cl 2 + N 2 (slow) Which of the following rate laws does this mechanism support? A. rate = k[n 2 Cl][Cl] B. rate = k[cl] 2 C. rate = k[n 2 ][Cl] 2 D. rate = k [Cl] [Cl 2 ] E. rate = k[n 2 ][Cl] Explanation: none Remember to bubble in ALL your answers BEFORE is called. Please SIGN your bubblesheet. Turn in your bubblesheet and KEEP your exam copy for reference.