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- Title: 2006 Exam 2
- Type: Notes
- School: Rutgers
- Course: CHEM 162 162
- Term: Spring
162 Chemistry Exam II March 8, 2006 Student Name (Print): ____TAVSS________________ Recitation Section Number: Recitation Instructor: ____TAVSS______________ The exam booklet has 25 questions for credit and one additional question to check the color of your exam booklet. Please answer all 26 questions on the OpScan sheet. Use the best answer of the choices available. There is no penalty for guessing. At the end of the 80-minute exam period, please hand in only this top sheet and your OpScan form. If you finish early, please do not disturb your fellow students. A proctor will check your picture ID, OpScan form, signature and calculator during the exam. The use of calculators with permanent memories (graphing calculators), cell phones, pagers, PDAs or other electronic devices other than a basic scientific calculator is expressly forbidden. The last page of the booklet contains a periodic table along with other useful data. The use of any other notes or information on this test will be considered a violation of the Academic Honesty provisions of the student code. Exam scores will be posted as soon as possible. ON THE OpScan FORM (Use a #2 pencil or darker) Sec Per. Instr Sec Per. 1. SIGN your name across the top of the form. 2. Code the following information (blacken circles) Your Name (LAST NAME FIRST) Your SOCIAL SECURITY NUMBER [Start under Box A and continue to Box I] Your RECITATION SECTION NUMBER in K & L [Sections 01-08, code a 0 under box K] Your EXAM FORM NUMBER under box P LH = L. Huebner, RP = R. Porcja, RA = R. Agarwal HS = H. Sangari, NM = N. Marky, ET = E. Tavss Periods: 1 = 9:30-10:25, 3 = 12:15-1:10 or 12:50-1:45 5 = 3:35-4:30, 6 = 5:15-6:10, 8 = 8:25p-9:20p 01 03 04 06 07 08 11 17 19 20 23 25 26 M3 M5 M5 W3 W3 W5 F3 M1 M3 M3 W1 W3 W3 RA RA LH LH HS LH NM HS NM HS ET ET NM 30 31 33 34 36 37 39 Th3 Th3 M6 M6 W6 M8 W8 Instr HS ET RP RA NM RP NM Your EXAM FORM is C162s06e2v1 1 1 Chem 162-2006 Exam II Acids and Bases - Chapter 14 Solutions of acids and bases, Ka,Kb,pH,pOH,% dissoc. ET: Molar % dissociation A 0.020 M acid HX is10.0% ionized. What is the Ka for this acid? A. 1.8 x 10-5 B. 2.2 x 10-4 C. 3.6 x 10-4 D. 7.2 x 10-5 E. 4.0 x 10-3 HX + H2O H3O+ + X+ HO 2 H3O+ 0 0.002 HX Initial Change Equilibrium 0.020 -X 0.020-X HX Initial Change Equilibrium 0.020 -0.002 0.020-0.002 = 0.018 + H2O + HO 2 H3O+ 0 +X 0.002 H3O+ 0 +0.002 0.002 + + + X0 0.002 X0 +X 0.002 X0 +0.002 0.002 0.10 X 0.020 = 0.0020 HX Initial Change Equilibrium 0.020 Ka = ([H3O+][X-])/[HX] Ka = ([0.002][0.002])/[0.018] Ka = 2.22 x 10-4 C162s06e2v1 2 2 Chem 162-2006 Exam II Chemical Kinetics - Chapter 12 Reaction mechanisms ET: Molecularity Which one of the following is least likely to be a step in a reaction mechanism? A. H2(g) + 2I(g) 2HI(g) B. NO2(g) + CO(g) CO2(g) + NO(g) C. I2(g) 2I(g) D. N2(g) + 3H2(g) E. NO2(g) + F2(g) 2NH3(g) NO2F(g) + F(g) The elementary step that is least likely is one which has the most number of molecules colliding at the same time in the transition state. (a) has 3 molecules coming together. (b) has 2 molecules coming together. (c) has 1 molecule that will decompose. (d) has 4 molecules coming together (e) has 2 molecules coming together Therefore, (d) is the least likely. C162s06e2v1 3 3 Chem 162-2006 Exam II Acids and Bases - Chapter 14 Solutions of acids and bases, Ka,Kb,pH,pOH,% dissoc. ET: Given initial concentration of strong base and change, find final concentration. A solution of NaOH has pH 13.40. If 1.00 mL of this solution is diluted with water to a total of 100 mL, what is the pH of the diluted solution? A. 10.40 B. 10.90 C. 11.40 D. 11.90 E. 12.40 pH = 13.40 pOH = 14.00 pH = 14.00 13.40 = 0.60 [OH-] = 10-pOH = 10-0.60 = 0.251 M Dilution of [OH-] 100 fold makes the [OH-] = 0.00251M pOH = -log[OH-] = -log(0.00251) = 2.60 pH = 14 pOH = 14 2.60 = 11.40 If we try to solve this problem by converting the pH into [H+], and then diluting the [H+], the results would make no sense. The [H+] is very low. Diluting it would make the [H+] lower, and correspondingly the [OH-] higher. This would result in a high [OH-] increasing further as the solution becomes more dilute, which makes no sense. The calculations must be done in such a manner that dilution of an acid or base solution would result in the pH approaching neutrality, which is the case when the calculations are done by way of the high [OH -] getting diluted. C162s06e2v1 4 4 Chem 162-2006 Exam II Acids and Bases - Chapter 14 Nature of acids and bases, conjugate acids and bases ET: Nature of strong acids Assume that acid HX is a strong acid. Which one of the following statements is false? A. 0.10 M HX solution will have pH = 1.00. B. Almost all of the HX will be in the form of H+ and X-. C. HX almost completely transfers its protons to H2O. D. X will have significant basic properties in H O. 2 E. A solution of HX will be a good electrical conductor. (a) HX + H2O H3O+ + XHX Initial 0.10 Change Equilibrium HX 0.10 -0.10 0 + H2O H3O+ 0 + X0 + H2O Initial Change Equilibrium H3O+ 0 +0.10 +0.10 + X0 +0.10 +0.10 The entire HX dissociates into H3O+ and X-, so that at equilibrium there is 0 M HX and 0.10 M H3O+. [H+] = 0.10 M pH = -log(0.10) = 1.0 Therefore, (a) is true. (b) HX + H2O H3O+ + XThe HX changes virtually 100% into H3O+ and X-. Therefore, (b) is true. (c) HX + H2O H3O+ + XThe HX transfers virtually all of its protons to H2O. Therefore, (c) is true. (d) HX + H2O H3O+ + XIf X- has significant basic properties in H2O, the above reaction would be reversible; i.e., X- would react with H3O+ to form HX + H2O. However, the reaction is not reversible. The X- doesn't react with H3O+ to form HX + H2O. If X- doesn't react with H3O+, then it certainly wouldn't react with H2O, a weaker acid than H3O+. Therefore, X- doesn't have significant basic properties in water. Therefore, (d) is false. (e) HX + H2O H3O+ + XA good electrical conductor is one which provides a large concentration of ions. Since HX completely dissociates in water it will provide a large concentration of ions, and is therefore a good electrical conductor. Therefore, (e) is true. C162s06e2v1 5 5 Chem 162-2006 Exam II Chemical Equilibrium - Chapter 13 Le Chatelier ET: Typical Le Chatelier Four substances, HCl(g), I2(s), HI(g), and Cl2(g), are mixed in a reaction vessel and allowed to reach equilibrium in the reaction: 2HCl(g) + I2(s) 2HI(g) + Cl2(g) Which of the following changes will result in an increased amount of HI(g) when equilibrium is reached again? X. Add HCl(g) to the reaction vessel. Y. Add I2(s) to the reaction vessel. Z. Decrease the volume of the reaction vessel. A. X only B. C. D. E. X and Y only X and Z only Z only X, Y, and Z X. If HCl is increased, it will impose a stress on the system, and the reaction will shift to the right, resulting in an increased concentration of HI, and therefore an increased amount of HI. Y. Assuming there is an excess of I2 to begin with, addition of I2 to the reaction vessel will not affect the concentration of HI, because the concentration of I 2 won't increase since it's a solid. Z. If the volume of the reaction vessel is decreased, then the pressure of the system is increased. Since the pressure can be reduced by converting three molecules (2HI + 1Cl2) to 2 molecules (2HCl), the equilibrium will shift to the left to have less moles of substance. This will result in a decrease amount of HI. C162s06e2v1 6 6 Chem 162-2006 Exam II Acids and Bases - Chapter 14 Solutions of acids and bases, Ka,Kb,pH,pOH,% dissoc. ET: Given initial and equilibrium concentrations of base, find Kb A 0.25 M solution of methylamine, CH3NH2, has a pH of 12.01. Calculate Kb of methylamine. A. 1.8 x 10-5 B. 3.6 x 10-4 C. 3.6 x 10-5 D. 4.4 x 10-4 E. 2.2 x 10-4 CH3NH2 + H2O CH3NH3+ + OH- pH = 12.01 [H+] = 10-pH = 10-12.01 = 9.77 x 10-13 [OH-] = (1 x 10-14)/(9.77 x 10-13) = 1.82 x 10-2 CH3NH2 Initial Change Equilibrium 0.25 + H2O CH3NH3+ 0 + OH0 1.82 x 10-2 CH3NH3+ 0 +X +X CH3NH3+ 0 +1.02 x 10-2 +1.02 x 10-2 + OH0 +X 1.82 x 10-2 + OH0 1.02 x 10-2 1.02 x 10-2 CH3NH2 Initial Change Equilibrium 0.25 -X 0.25-X + H2O CH3NH2 Initial Change Equilibrium 0.25 -1.02 x 10-2 0.2398 + H2O Kb = ([CH3NH3+][OH-])/[CH3NH2] Kb = ([1.02 x 10-2][1.02 x 10-2])/[0.2398] = 4.34 x 10-4 C162s06e2v1 7 7 Chem 162-2006 Exam II Acids and Bases - Chapter 14 Autoionization of water and the pH scale ET: Relationship between endothermic, exothermic and Kw for water Consider the autoionization of water: H2O() H+(aq) + OH-(aq) Kw = 1.0 x 10-14 at 25 C Kw = 3.8 x 10-14 at 40 C Which of the following statements must be true based on this information? X. The autoionization reaction is endothermic. Y. The pH of pure water at 40 oC is less than 7.0 Z. Pure water at 40 oC contains more ions than pure water at 25 C. A. B. C. D. X only Y and Z only Y only X and Y only X, Y and Z E. X. If the reaction is endothermic: + 2H2O() H3O+(aq) + OH-(aq) An increase in heat would drive the reaction to the right, increasing the K w. Since Kw does increase at the higher temperature the reaction is endothermic. Y. 2H2O() Initial Change Equilibrium Kw = [H3O+] x [OH-] 3.8 x 10-14 = [X] x [X] X = 1.95 x 10-7 = [H3O+] pH = -log[H3O+] pH = -log(1.95 x 10-7) = 6.71 The pH of pure water at 40oC is less than 7.0. Z. At 40oC: [H3O+] = 1.95 x 10-7 At 40oC: [OH-] = 1.95 x 10-7 At 25oC: [H3O+] = 1.00 x 10-7 At 25oC: [OH-] = 1.00 x 10-7 Therefore, pure water at 40oC contains more ions than pure water at 25oC. H3O+ + OH- X X C162s06e2v1 8 8 Chem 162-2006 Exam II Acids and Bases - Chapter 14 Relationship between structure and strengths of acids and bases ET: Relationship between structure and strengths of acids Consider the following pairs of acids. HF or HCl HClO3 or HClO2 HClO or HIO Which choice lists the stronger of the two acids in each pair? A. HF, HClO3, HIO B. HCl, HClO3, HClO C. HF, HClO2, HClO D. HCl, HClO2, HIO E. HF, HClO3, HClO The vertical groups of acids increase in strength in descending order because the H-X bond becomes weaker, thereby allowing the H+ ion to separate from the H-X molecule. Hence, HCl is a stronger acid than HF. The strengths of all of the other acids (in this case, two oxy-acids) increase in order of polarity (electronegativity) of the groups attached to the H. This is because the H-O bond strength of the two acids are similar, leaving electronegativity as the next most important factor. Hence, since HClO3 has 4 electronegative groups attached, and HClO2 has 3 electronegative groups attached, then HClO3 is the stronger acid. The strengths of all of the other acids (in this case, two oxy-acids) increase in order of polarity (electronegativity) of the groups attached to the H. This is because the H-O bond strength of the two acids are similar, leaving electronegativity as the next most important factor. Hence, since HClO (really H-O-Cl) has an oxygen and a chlorine group attached to the hydrogen, while HIO (really H-O-I) has an oxygen and an iodine group attached to the hydrogen, and Cl is more electronegative than I, then HClO is the stronger acid. (b): HCl, HClO3, HClO C162s06e2v1 9 9 Chem 162-2006 Exam II Chemical Equilibrium - Chapter 13 Equilibrium calc'n/concepts, Kp,Kc,Q,K, Heterogeneous equilibria ET: Given pH, find moles What amount of barium hydroxide, Ba(OH)2, must be dissolved in a total of 5.00 L of solution in order to have a pH of 12.20? A. 0.025 mol B. 0.040 mol C. 0.060 mol D. 0.075 mol E. 0.095 mol Let X = moles of OH(moles of X)/(5.00L) = concentration of OHpH = 12.20 [H+] = 10-pH = 10-12.20 = 6.31 x 10-13 Kw = [H+][OH-] [OH-] = (1 x 10-14)/(6.31 x 10-13) = 0.0158 M X/5.00L = 0.0158M X = 0.0792mol OHBa(OH)2 Ba2+ + 2OH0.0792 mol OH- x (1 mol Ba(OH)2/2 mol OH-) = 0.0396 mol Ba(OH)2 C162s06e2v1 10 10 Chem 162-2006 Exam II Chemical Kinetics - Chapter 12 Temp depend (Arrhenius, Ea) ET: Recognizing graphical form of Arrhenius equation In order to calculate the activation energy for a reaction, which of the following plots would be made? A. k vs T B. ln k vs T C. ln k vs 1 T 1 T D. k vs ln E. k vs 1 T ln(k) = -Ea/RT + ln(A) ln(k) = ((-Ea/R) x (1/T)) + ln(A) Y = ( m x X) + b A plot of ln(k) vs 1/T fits a straight line. The slope is Ea/R, which will provide the energy of activation. C162s06e2v1 11 11 Chem 162-2006 Exam II Acids and Bases - Chapter 14 Solutions of acids and bases, Ka,Kb,pH,pOH,% dissoc. ET: Given equilibrium concentration and Ka for an acid, find initial conc. A saturated solution of benzoic acid, C6H5COOH, has pH = 2.93. What amount of benzoic acid is dissolved in a total of 1.00 L of this solution? Ka = 6.3 x 10-5 for C6H5COOH A. 0.16 mol B. 0.30 mol C. 0.023 mol D. 0.080 mol E. 0.016 mol HB + H2O B- + H3O+ pH = 2.93 [H+] = 10-pH = 10-2.93 = 1.17 x 10-3 HB Initial Change Equilibrium Y -X Y-X HB Initial Change Equilibrium + H2O + H2O B0 +X +X B0 1.17 x 10-3 1.17 x 10-3 + + H3O+ 0 +X 1.17 x 10-3 H3O+ 0 1.17 x 10-3 1.17 x 10-3 Y -1.17 x 10-3 Y-1.17 x 10-3 = Ka [B-][H3O+]/[HB] 6.3 x 10-5 = [1.17 x 10-3]x[1.17 x 10-3]/[Y-1.17 x 10-3] Y = 2.29 x 10-2 moles C162s06e2v1 12 12 Chem 162-2006 Exam II Chemical Kinetics - Chapter 12 Temp depend (Arrhenius, Ea) ET: Understanding terms in Arrhenius equation, e.g., temperature A rough rule says that a reaction rate will double for a 10 C increase in temperature. This rule works, because when the temperature is increased by 10 C: A. B. C. D. the activation energy doubles the activation energy halves the collision frequency doubles the average kinetic energy doubles the number of molecules having sufficient activation energy doubles E. (a) The activation energy depends on the difference in enthalpy between the reactants and transition state. It is temperature independent. (b) The activation energy depends on the difference in enthalpy between the reactants and transition state. It is temperature independent. (c) Although the collision frequency does increase with an increase in temperature, it's increase is quite small. (It is due to this minimal increase in collisional frequency that ln(A) drops out when we subtract the Arrhenius equation graphical form from itself at two different temperatures.) (d) KE = (3/2)RT KE = (3/2) x 8.314 x 273 = 3405 J KE = (3/2) x 8.314 x 283 = 3529 J Therefore, the KE doesn't double for a 10o increase in temperature. (e) As the temperature is increased, the fraction of collisions with the required energy to get over the Ea increases dramatically. C162s06e2v1 13 13 Chem 162-2006 Exam II Chemical Equilibrium - Chapter 13 Equilibrium calc'n/concepts, Kp,Kc,Q,K, Heterogeneous equilibria ET: Given an initial concentration and K, find equilibrium concentration. (Ignore solid.) Bringing reaction to completion and then back to equilibrium. Consider the reaction: Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s) K = 2.0 x 1015 Starting with an initial [Ag+] = 1.0 M and excess Cu(s), when equilibrium is reached, the concentration of Cu2+ would be: A. 2.0 x 10-15 M B. 2.0 M C. 1.5 M D. 0.50 M E. 2.0 x 1015 M Cu(s) XS + 2Ag+(aq) 1.0 -2X 1.0 2X Cu2+(aq) 0 +X +X + 2Ag(s) 0 Initial Change Equilibrium K = [Cu2+]/[Ag+]2 2.0 x 1015 = [X]/[1.0 2X]2 K is too large to simplify the equation by dropping the 2X. Instead, since K is soooo large, bring the reaction to completion and then bring it back to equilibrium: 2Ag+(aq) Cu2+(aq) + 2Ag(s) Initial 1.0 0 0 Change -1.0 +0.5 Equilibrium 0 +0.5 We can stop here, since the equilibrium concentration is sooo huge that the reaction goes virtually to completion to Cu2+; i.e., there is virtually no [Ag+] present at equilibrium. If we want to be exact then we can bring the reaction back to equilibrium: Cu(s) XS + Bring to equilibrium: Cu(s) New initial Change Equilibrium Cu2+(aq) 0.5 -X 0.5-X + 2Ag+(aq) 0 +2X +2X + 2Ag(s) K = [Cu2+]/[Ag+]2 2.0 x 1015 = [0.5-X]/[+2X]2 "-X" is very small considering the huge K. Therefore, simplify the equation by dropping "-X". 2.0 x 1015 = [0.5]/[+2X]2 X = 7.9 x 10-9 [Cu2+] = 0.5 7.9x10-9 = 0.5M C162s06e2v1 14 14 Chem 162-2006 Exam II Acids and Bases - Chapter 14 Acid-base properties of salts ET: Qualitative pH of salts Place the following 0.10 M solutions in order of increasing pH (from lowest pH to highest pH). X. AlCl3 Y. KNO3 Z. Na2CO3 A. B. C. D. E. X, Y, Z Y, Z, X Z, Y, X Y, X, Z Z, X, Y Al3+ + 3Cl- X. AlCl3 Since Cl- is the conjugate base of a strong acid, it is not basic. Al3+ is a Lewis acid, so it is acidic in water. Proof: Al3+ + 6H2O Al(H2O)63+ B. KNO3 Al(H2O)63+ Al(H2O)5OH2+ + H+ K+ + NO3- K+ is the conjugate acid of a strong base, so it is not an acid. NO3- is the conjugate base of a strong acid, so it is not a base. K+ and NO3- have no effect on the pH of water. C. Na2CO3 2Na+ + CO32- Na+ is the conjugate acid of a strong base, so it is not an acid. CO32- is the conjugate base of a weak acid; therefore, it is basic. CO32- + H2O HCO3- + OH- C162s06e2v1 15 15 Chem 162-2006 Exam II Chemical Kinetics - Chapter 12 Temp depend (Arrhenius, Ea) ET: Understanding terms in Arrhenius equation, e.g., temperature What would the proper labels for the x- and y-axes in the diagram below: T1 T2 A. x-axis: reaction progress y-axis: potential energy y-axis: y-axis: y-axis: y-axis: number of molecules time concentration activation energy B. x-axis: kinetic energy C. x-axis: concentration D. x_axis: time E. x-axis: rate constant (b) The y-axis is the fraction of molecules/collisions having a certain KE, whereas the xaxis is the kinetic energy. (Figure 12.12, Zumdahl & Zumdahl 6th edition) C162s06e2v1 16 16 Chem 162-2006 Exam II Chemical Equilibrium - Chapter 13 Equilibrium calc'n/concepts, Kp,Kc,Q,K, Heterogeneous equilibria ET: Given initial concentrations and K, find equilibrium concentrations. Consider the following reaction: 2NO(g) + I2(g) 2NOI(g) Kp = 1.6 x 10-5 Calculate the pressure of NOI(g) at equilibrium from initial pressures of 0.50 atm of NO(g) and 0.30 atm of I2(g) A. B. C. D. E. 1.1 10-3 atm 7.2 10-4 atm 3.6 10-4 atm 1.8 10-3 atm 5.9 10-3 atm 2NO(g) Initial Change Equilibrium 0.50 -2X 0.50 2X + I2(g) 0.30 -X 0.30 - X 2NOI(g) 0 +2X +2X K = [NOI]2/([NO]2 x [I2]) = 1.6 x 10-5 K = [2X]2/([0.50-2X]2 x [0.30-X]) 1.6 x 10-5 = [2X]2/([0.50-2X]2 x [0.30-X]) Simplify: 1.6 x 10-5 = [2X]2/([0.50]2 x [0.30]) X = 5.48 x 10-4 atm 2X = 1.096 x 10-3 atm C162s06e2v1 17 17 Chem 162-2006 Exam II Chemical Equilibrium - Chapter 13 Equilibrium calc'n/concepts, Kp,Kc,Q,K, Heterogeneous equilibria ET: Given initial pressure and final pressure, find K. 2N2O5 4NO2(g) + O2(g) A 1.00 atm sample of N2O5 is placed in a container, and reacts to reach equilibrium in the above reaction. The partial pressure of NO2 at equilibrium is found to be 1.60 atm. Calculate Kp for this reaction. A. B. C. D. 4.10 7.29 13.1 35.8 65.5 2N2O5(g) Initial Change Equilibrium 1.00 1.60 4NO2(g) + O2(g) E. 2N2O5(g) Initial Change Equilibrium 1.00 -2X 4NO2(g) 0 +4X 1.60 + O2(g) 0 +X 2N2O5 Initial Change Equilibrium 1.00 -0.80 0.20 4NO2(g) 0 +4 x 0.40 1.60 + O2(g) 0 +0.40 +0.40 Kp = ([NO2]4[O2])/[N2O5]2 Kp = ([1.60]4[0.40])/([0.20]2) = 65.5 C162s06e2v1 18 18 Chem 162-2006 Exam II Chemical Kinetics - Chapter 12 Temp depend (Arrhenius, Ea) ET: Quantitative problems with long form of Arrhenius equation What is the activation energy of a reaction whose rate triples when the temperature is increased from 20 C to 32 C? A. B. C. D. E. kJ mol kJ 42 mol kJ 53 mol kJ 82 mol kJ 102 mol 68 ln(k2/k1) = -(Ea/R) x [(1/T2) (1/T1)] ln(3X/1X) = -(Ea/8.314) x [(1/305) (1/293)] Ea = 6.8 x 104 J = 68 kJ/mol C162s06e2v1 19 19 Chem 162-2006 Exam II Chemical Equilibrium - Chapter 13 Equilibrium calc'n/concepts, Kp,Kc,Q,K, Heterogeneous equilibria ET: Given initial pressures and K, find equilibrium pressures. Consider the reaction: 2NH3(g) N2(g) + 3H2(g) Kp = equilibrium constant. Initial partial pressures of 0.500 atm NH3 and 0.800 atm H2 are placed in a container of fixed volume at fixed temperature. Let x = the equilibrium partial pressure of N 2. Which one of the following equations would yield the correct value for x when solved? A. B. C. D. x 0.80 0.50 3 2 = Kp 3 2 x 0.80 x 0.50 x = Kp = Kp 3 2 0.80x 4 0.50 2 x 0.50 2 x 2 x 0.80 3x = Kp 3 E. x 0.80 3x 0.50 2 x 2 = Kp 2NH3 N2 + 3H2 0.800 X Initial Change Equilibrium 0.500 2NH3 Initial Change Equilibrium 0.500 -2X 0.500 2X N2 0 +X X + 3H2 0.800 +3X 0.800 + 3X Kp = ([N2][H2]3)/[NH3]2 Kp = ([X][(0.800+3X)]3)/[(0.500-2X)]2 C162s06e2v1 20 20 Chem 162-2006 Exam II Chemical Equilibrium - Chapter 13 Equilibrium calc'n/concepts, Kp,Kc,Q,K, Heterogeneous equilibria ET: Solve for Q Consider the reaction: N2O4(g) 2NO2(g) Kp = 80 In which of the following systems will the reaction proceed in a direction to use up some of the NO 2 (from right to left in the above equation). Partial Pressure of N2O4 0.0020 atm 0.0040 atm 0.0040 atm Partial Pressure of NO2 0.400 atm 0.800 atm 0.300 atm X. Y. Z. A. X and Y only B. Y only C. Z only D. Y and Z only E. X, Y, and Z N2O4 Initial Change Equilibrium 0.0020 2NO2 0.400 Q = ([NO2]2)/[N2O4] Compare Q to K ([0.400]2)/[0.0020] = 80 Q = K; therefore, the system is at equilibrium. N2O4 Initial Change Equilibrium 0.0040 2NO2 0.800 Q = ([NO2]2)/[N2O4] Compare Q to K ([0.800]2)/[0.0040] = 160 Q > K; therefore, the numerator (products) is too large, so the reaction will go from right to left, and some of the NO 2 will be used up. N2O4 Initial Change Equilibrium 0.0040 2NO2 0.300 Q = ([NO2]2)/[N2O4] Compare Q to K ([0.300]2)/[0.0040] = 22.5 Q < K; therefore, the numerator (products) is too small, so the reaction will go from left to right, and additional NO 2 will be formed. C162s06e2v1 21 21 Chem 162-2006 Exam II Chemical Kinetics - Chapter 12 Reaction mechanisms ET: Defining terms, e.g., catalyst Which is a catalyst in the following mechanism? 1. H2O2 + I- H2O + IO2. H2O2 + IO- H2O + O2 + IA. H2O2 B. IC. H2O D. IOE. O2 1. H2O2 + I- H2O + IO2. H2O2 + IO- H2O + O2 + I2H2O2 + I- + IO- 2H2O + IO- + O2 + I2H2O2 2H2O + O2 A catalyst is a substance that is present at the beginning and end of a reaction. This is I - in this reaction. H2O2 is a reactant; H2O and O2 are products. IO-, which forms and disappears during the reaction, and isn't present at the beginning or end of the reaction, is a reactive intermediate. C162s06e2v1 22 22 Chem 162-2006 Exam II Acids and Bases - Chapter 14 Solutions of acids and bases, Ka,Kb,pH,pOH,% dissoc. ET: Polyprotic acids; K conversions Phosphoric acid, H3PO4 is a polyprotic acid with Ka values: Ka1 = 7.5 x 10-3 Ka2 = 6.2 x 10-8 Ka3 = 4.8 x 10-13 What is the value of the equilibrium constant for the reaction: HPO42-+ H2O H2PO4- + OHA. 6.2 x 10-8 B. 4.8 x 10-13 C. 1.6 x 10-7 D. 1.3 x 10-12 E. 2.1 x 10-2 H3PO4 + H2O H2PO4- + H2O HPO42- + H2O H3O+ + H2PO4H3O+ + HPO42H3O+ + PO43Ka1 = 7.5 x 10-3 Ka2 = 6.2 x 10-8 Ka3 = 4.8 x 10-13 Note that HPO42- is acting as a base in this reaction. (It is accepting a proton and forming hydroxide ion.). It is the conjugate base of H2PO4Kw = Ka x Kcb Kcb = Kw/Ka = (1 x 10-14)/(6.2 x 10-8) = 1.6 x 10-7 C162s06e2v1 23 23 Chem 162-2006 Exam II Chemical Kinetics - Chapter 12 Reaction mechamisms ET: Rapid reversible step, then slow step The reaction: Cl2(g) + CHCl3(g) HCl(g) + CCl4(g) has the following mechanism: Fast equilibrium Slow Fast 1. Cl2(g) 2Cl(g) 2. Cl(g) + CHCl3(g) HCl(g) + CCl3(g) 3. CCl3(g) + Cl(g) CCl4(g) What rate law is consistent with this mechanism? A. B. C. D. Rate = k[Cl2] Rate = k[Cl2][CHCl3] Rate = k[Cl2]2 Rate = k[Cl2]2[CHCl3] Rate = k[Cl2]1/2[CHCl3] E. The slow step is the only important step in determining the rate of the reaction. Rate = k[Cl][CHCl3] Get rid of Cl as it isn't in the equilibrium equation. Fast equilibrium step: Ratef = k1[Cl2] Rater = k-1[Cl]2 Ratef = Rater k1[Cl2] = k-1[Cl]2 [Cl]2 = (k1/k-1)[Cl2] [Cl] = ((k1/k-1)1/2 [Cl2]1/2) Rate = k((k1/k-1)1/2 [Cl2]1/2) [CHCl3] Rate = k'[Cl2]1/2[CHCl3] C162s06e2v1 24 24 Chem 162-2006 Exam II Chemical Equilibrium - Chapter 13 Equilibrium calc'n/concepts, Kp,Kc,Q,K, Heterogeneous equilibria ET: Given equilibrium concentration and no initial concentration, find K 2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g) Solid sodium bicarbonate, NaHCO3, decomposes on heating according to the equation above. When equilibrium is reached, some of the original solid remains, and the total pressure in the container is 7.76 atm. Calculate the value of Kp for this reaction. A. 3.88 B. 15.1 C. 60.2 D. 30.1 E. 7.76 2NaHCO3(s) Na2CO3(s) + H2O(g) + 0 +X +X CO2(g) 0 +X +X Initial 2Y 0 Change -2X +X Equilibrium X = the equilibrium concentration of H2O(g). X = the equilibrium concentration of CO2(g) X + X = 7.76 atm, X = 3.88 atm = PH2O = PCO2 KP = PH2O x PCO2 KP = 3.88 x 3.88 = 15.1 C162s06e2v1 25 25 Chem 162-2006 Exam II Acids and Bases - Chapter 14 Acid-base properties of salts ET: Given initial conc of salt and Ka, find equilibrium concentration. Calculate the pH of 0.20 M NaCH3COO (sodium acetate) Ka = 1.8 x 10-5 for CH3COOH (acetic acid) A. B. C. D. 11.48 9.47 7.49 10.60 9.02 E. NaA in water = Na+ + ANa+ is the conjugate acid of a weak base; therefore it has no effect on pH. A- + H2O HA + OHAInitial Change Equilibrium 0.20 -X 0.20 - X + H2O HA 0 +X +X + OH0 +X +X Kw = Ka x Kcb Kcb = Kw/Ka Kcb = (1 x 10-14)/(1.8 x 10-5) = 5.56 x 10-10 Kcb = ([HA] x [OH-])/[A-] 5.56 x 10-10 = ([X][X])/[0.20 - X] Simplify: 5.56 x 10-10 = ([X][X])/[0.20] X = 1.055 x 10-5 [OH-] = 1.055 x 10-5 [H+] = (1 x 10-14)/(1.055 x 10-5) = 9.48 x 10-10 pH = -log(9.48 x 10-10) = 9.02 C162s06e2v1 26 26 What color is your exam? A. white B. yellow C. blue D. pink C162s06e2v1 27 NA = 6.022 10 mol 23 -1 Constants and Formulae R = 0.08206 L atm mol-1 K-1 = 8.3145 J mol-1 K-1 PV = nRT KP = Kc(RT)n Kw = Ka Kb Kw = 1 10-14 at 25 C x b b2 2a 4ac C162s06e2v1 28
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