5Kinetics and equilibrium test w solutions

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GENERAL CHEMISTRY TOPICAL: Kinetics and Equilibrium Test Time: 23 Minutes* Number of Questions: 18 * The timing restrictions for the science topical tests are optional If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit MCAT DIRECTIONS: Most of the questions in the following test are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also select the best answer to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions PERIODIC TABLE OF THE ELEMENTS H 1.0 He 4.0 Li 6.9 Be 9.0 B 10.8 C 12.0 N 14.0 O 16.0 F 19.0 10 Ne 20.2 11 Na 23.0 12 Mg 24.3 13 Al 27.0 14 Si 28.1 15 P 31.0 16 S 32.1 17 Cl 35.5 18 Ar 39.9 19 K 39.1 20 Ca 40.1 21 Sc 45.0 22 Ti 47.9 23 V 50.9 24 Cr 52.0 25 Mn 54.9 26 Fe 55.8 27 Co 58.9 28 Ni 58.7 29 Cu 63.5 30 Zn 65.4 31 Ga 69.7 32 Ge 72.6 33 As 74.9 34 Se 79.0 35 Br 79.9 36 Kr 83.8 37 Rb 85.5 38 Sr 87.6 39 Y 88.9 40 Zr 91.2 41 Nb 92.9 42 Mo 95.9 43 Tc (98) 44 Ru 101.1 45 Rh 102.9 46 Pd 106.4 47 Ag 107.9 48 Cd 112.4 49 In 114.8 50 Sn 118.7 51 Sb 121.8 52 Te 127.6 53 I 126.9 54 Xe 131.3 55 Cs 132.9 56 Ba 137.3 57 La * 138.9 72 Hf 178.5 73 Ta 180.9 74 W 183.9 75 Re 186.2 76 Os 190.2 77 Ir 192.2 78 Pt 195.1 79 Au 197.0 80 Hg 200.6 81 Tl 204.4 82 Pb 207.2 83 Bi 209.0 84 Po (209) 85 At (210) 86 Rn (222) 87 Fr (223) 88 Ra 226.0 89 Ac † 227.0 104 Unq (261) 105 Unp (262) 106 Unh (263) 107 Uns (262) 108 Uno (265) 109 Une (267) * 58 Ce 140.1 59 Pr 140.9 60 Nd 144.2 61 Pm (145) 62 Sm 150.4 63 Eu 152.0 64 Gd 157.3 65 Tb 158.9 66 Dy 162.5 67 Ho 164.9 68 Er 167.3 69 Tm 168.9 70 Yb 173.0 71 Lu 175.0 † 90 Th 232.0 91 Pa (231) 92 U 238.0 93 Np (237) 94 Pu (244) 95 Am (243) 96 Cm (247) 97 Bk (247) 98 Cf (251) 99 Es (252) 100 Fm (257) 101 Md (258) 102 No (259) 103 Lr (260) GO ON TO THE NEXT PAGE as developed by Kinetics and Equilibrium Test Passage I (Questions 1–7) H As Figure illustrates, the reaction of tert-butyl bromide with aqueous sodium hydroxide results in a substitution product (Reaction I) and an elimination product (Reaction II) CH I K=3 × 10 CH CH C CH CH H+/∆ II K=2 × 10 CH C CH C CH CH3 (1) CH H + CH CH3 C CH3 Br H H+ O III K=1 × 10 + O CH3 C CH3 H CH + Br – C OH OH– H O CH C CH + H2O (2) + CH CH3 CH + H2O + Br – Figure A student investigated the reaction kinetics for the conversion of tert-butyl bromide to tert-butyl alcohol (Reaction I) by varying the concentrations of the reactants and recording the reaction rate The results are shown in Table CH3 CH CH3 C CH2 + CH C CH + H+ (3) H Figure 2 Initial rate of reaction (mmol/[L•sec]) The energy profile for Reaction III is as follows: 6.0 6.0 12.0 Potential Energy Exp # Table Initial Initial concentration concentration of OH– of tert-Butyl bromide (mmol/L) (mmol/L) 0.02 0.05 0.02 0.10 0.04 0.05 In reaction III, tert-butyl alcohol is converted to 2methylpropene by the addition of aqueous acid followed by heating This reaction involves the formation of a protonated intermediate that loses water to form a carbonation The alkene is then generated by the loss of a proton This mechanism is illustrated in Figure Reaction Coordinate Figure l Which reaction has proceeded the farthest toward completion once equilibrium has been established? A B C D Reaction I Reaction II Reaction III It cannot be determined without more information GO ON TO THE NEXT PAGE KAPLAN MCAT Which of the following assumptions can be made about Reaction III? A The rate of formation of the carbocation is greater than the rate of protonation of the alcohol B The rate of formation of the carbonation is greater than the deprotonation of the carbonation to form the alkene C The rate of deprotonation to form the alkene is greater than the protonation of the alcohol D The rate of formation of the carbonation is less than the rate of protonation of the alcohol Which of the following is the rate equation for Reaction I? A B C D Rate = k [tert-butyl bromide] [OH–] Rate = k [OH–] Rate = k [tert-butyl bromide] Rate = k [tert-butyl bromide] [OH–]2 For Reaction II, what could be done to shift the equilibrium to the right, favoring the formation of the alkene? A Add excess water to the reaction mixture B Use tert-butyl chloride as the starting reagent instead of tert-butyl bromide C Use a lower concentration of hydroxide ions D Add AgNO3 to precipitate out the bromide ions If Reactions I, II, and III were run with an appropriate catalyst, which of the equilibrium constants would change? A The equilibrium constant for Reaction I only B The equilibrium constant for Reaction II only C The equilibrium constants for Reactions I, II, and III D None of the equilibrium constants would change What is the rate constant for Reaction I? A 3000 s–l B 300 s–l C 60 s–l D s–l If Reaction I was carried out at a higher temperature, which of the following would be observed? A Both the equilibrium constant and the constant would change B The equilibrium constant would change, but rate constant would remain the same C The rate constant would change, but equilibrium constant would remain the same D Neither the equilibrium constant nor the constant would change rate the the rate GO ON TO THE NEXT PAGE as developed by Kinetics and Equilibrium Test Passage II (Questions 8–14) A chemist interested in the reactivity of iodine concentrated his study on two reactions: the decomposition of gaseous hydrogen iodide (Reaction 1) and the reaction between iodide ions and persulfate ions (Reaction 2) slope = -9.71 x 103K log k (log k) -2 2HI(g) H2(g) + I2(g) -4 Reaction -6 3I–(aq) + S2O82–(aq) (1/T) I3–(aq) + 2SO42–(aq) 1.4 x 10 -3 1.6 x 10 -3 Reaction 1/T The value of the rate constant for Reaction was studied as a function of temperature The results are shown below T(K) 555 575 645 700 781 Table 1/T (K–1) k (l•mol–1sec–1) 1.80 × 3.52 × 10–7 –3 10 1.22 × 10–6 1.74 × 8.59 × 10–5 10–3 1.16 × 10–3 1.55 × 3.95 × 10–2 –3 10 1.43 × 10–3 1.28 × 10–3 18 x 10 -3 log k –6.453 –5.913 –4.066 –2.936 –1.403 For any reaction, the activation energy (Ea) is related to the rate constant (k) by the Arrhenius equation (Equation 1): k = A × l0(–Ea / 2.303RT) In order to determine the initial rate of Reaction 2, the following data were collected: Experiment [I–] (M) 0.21 0.21 0.42 Table [S2O8–2] (M) 0.15 0.30 0.15 Initial rate of reaction (M / sec) 1.14 2.28 2.28 What is the rate law for Reaction 2? A B C D Rate = k[I–]2[S2O82–] Rate = k[S2O82–] Rate = k[I–][S2O82–] Rate = k[I–][S2O82–]2 Equation where R = 8.314 J mol–1K–1, T is the temperature in Kelvin, and A is a constant, called the frequency factor Figure shows a graph of log k vs 1/T for Reaction What is the numerical value of the rate constant for Reaction 2? A 7.6 mol/L•sec B 36 mol/L•sec C 172 mol/L•sec D 241 mol/L•sec GO ON TO THE NEXT PAGE KAPLAN MCAT A when the temperature is held constant, the lower the activation energy, the slower the reaction B when the activation energy is held constant, the lower the value of A, the faster the reaction C when the activation energy is held constant, the lower the temperature, the faster the reaction D when the temperature is held constant, the lower the activation energy, the faster the reaction The reaction profile shown below is for an uncatalyzed reaction 120 kJ mole -1 Potential Energy According to Equation 1: 1 What is the activation energy for Reaction 1? A 9.71 kJ/mole B 22.4 kJ/mole C 80.7 kJ/mole D 186 kJ/mole In Figure 1, what does the intercept with the y-axis represent? Reaction Coordinate Which of the following is the reaction profile for the same reaction after the addition of a catalyst? A C 100 kJ mole -1 95 kJ mole -1 B D A log A B log k C –Ea D − 150 kJ mole -1 120 kJ mole -1 RT If the rate of disappearance of I– in Reaction is 2.5 × 10–3 mol/(L•s), what is the rate of formation of SO42–? A B C D 1.7 × 10–3 3.8 × 10–3 5.0 × 10–3 8.3 × 10–4 mol/(L•s) mol/(L•s) mol/(L•s) mol/(L•s) GO ON TO THE NEXT PAGE as developed by Kinetics and Equilibrium Test Questions 15 through 18 are NOT based on a descriptive passage How will the equilibrium of the following reaction be affected if more chlorine is added? PCl5(g) A B C D PCl3(g) + Cl2(g) It will be shifted to the right It will be shifted to the left It will be unaffected The effect on the equilibrium cannot be determined without more information How will the equilibrium of the following reaction be affected if the temperature is increased? N2(g) + 3H2(g) A B C D 2NH3 (g) Consider the following gas phase reaction: H2 (g) + Br2 (g) HBr(g) The concentrations of H2, Br2, and HBr are 0.05M, 0.03M, and 500.0M, respectively The equilibrium constant for this reaction at 400°C is 2.5 × 103 Is this system at equilibrium? A Yes, the system is at equilibrium B No, the reaction must shift to the right in order to reach equilibrium C No, the reaction must shift to the left in order to reach equilibrium D It cannot be determined if this system is at equilibrium without more information ∆H = –30 kJ/mole It will be shifted to the right It will be shifted to the left It will be unaffected The effect on the equilibrium cannot be determined without more information What is the value of the equilibrium constant for the following reaction if the equilibrium concentrations of nitrogen, hydrogen, and ammonia are 1M, 2M, and 15M, respectively? N2(g) + 3H2(g) 2NH3 (g) A 0.035 B 7.5 C 28 D 380 END OF TEST KAPLAN MCAT ANSWER KEY: A D C B A 10 D D C B D 11 12 13 14 15 D A A A B 16 17 18 B C C as developed by Kinetics and Equilibrium Test KINETICS AND EQUILIBRIUM TEST TRANSCRIPT Passage I (Questions 1–7) The reaction that goes to completion to the greatest degree is the one with the larger equilibrium constant You should recall that for any reaction, the equilibrium constant is equal to the product of the products each raised to their stoichiometric coefficients divided by the product of the reactants each raised to their stoichiometric coefficients So, if the equilibrium constant is large, it means that a large amount of products have been formed Thus, choice A, Reaction I with a K = x 104, would be the correct choice The answer to question is D For this question you need to be able to understand reaction profiles A reaction profile, such as that shown in Figure 3, is a plot of the potential energy of a reaction versus the reaction coordinate, which represents the progress of the reaction You should know that as reactant molecules with sufficient energy collide, they form what is known as an activated complex This activated complex can either go on to form intermediates, products, or go back to reactants Looking at Figure 3, you can see that there are three peaks representing three activated complexes and that the middle peak is the highest You should know that the higher the peak, the higher the potential energy of the activated complex, and that the higher the potential energy the slower the rate of formation of the activated complex In other words, the slowest step in a mechanism has the highest peak in the reaction profile Comparing the peaks to the steps of the mechanism (Figure 2), you can see that step two, the formation of the carbocation, has the highest peak All right, let’s look at the answer choices Choice A states that the rate of formation of the carbocation, step 2, is greater than the rate of protonation of the alcohol, step In other words, this choice says that the peak for step is lower than the peak for step This is not true Choice B states that the rate of formation of the carbocation is greater than that of the deprotonation of the carbocation to form the alkene, or the peak for step is smaller than that of step Not true Choice C, the rate of deprotonation to form the alkene is greater than the protonation of the alcohol, or the peak for step is smaller than that of step 1, is also not true That leaves choice D as the correct response: the rate of formation of the carbocation is less than the rate of protonation (The peak for step is taller than that of step 1.) The correct answer to question is choice C For this question you need to be able to construct the rate expression and be able to determine the order of the reaction from experimental data For a homogeneous reaction, the rate is proportional to the concentrations of the reactants raised to some power So, for the reaction A plus B goes to products, the rate law is equal to the concentration of A raised to some power, times the concentration of B raised to some power, times a proportionality constant The proportionality constant is called the rate constant and the powers that the concentrations of the reactants are raised to can only be determined by experimentation Okay, for the reaction in question, the rate law is equal to the concentration of tert-butyl bromide raised to some power, times the concentration of hydroxide raised to some power, times a rate constant Now, in order to determine what the exponents in the rate law are, you need to be able to interpret the data presented in Table A couple of things you should remember: if the concentration of a reactant is doubled and the rate doubles, if it is tripled and the rate triples, the exponent is equal to one If the concentration of a reactant is doubled and the rate increases by a factor of 4, if the concentration is tripled and the rate increases by a factor of 9, the exponent is equal to If the concentration of the reactant does not affect the rate, the exponent is equal to zero Armed with this information, let’s look at the data: Looking at experiments number and 2, the concentration of tert-butyl bromide is held constant at 0.02 mmol/liter and the concentration of hydroxide has been doubled We can see that the reaction rate remains unchanged: the exponent for the hydroxide concentration is zero Something raised to the zero power is equal to one, so hydroxide will NOT appear in the rate law Looking at experiments and 3, the concentration of hydroxide is held constant it doesn’t really matter since we just determined it’s not in the rate law and the concentration of tert-butyl bromide has been doubled What happens to the rate? It goes from 6.0 mmol per liter second to 12 mmol per liter second it doubles So the exponent for tert-butyl bromide is equal to one The rate law is, therefore, rate is equal to the rate constant, times the concentration of tert-butyl bromide: Choice C For question number the correct answer is B For this question, what you need to is solve the rate law for the rate constant Rearranging the rate law that we determined in question 3, you get that the rate constant is equal to the rate divided by the concentration of tert-butyl bromide Taking Experiment 1, the rate constant is equal to 6.0 mmol/liter sec divided by 0.02 mmol /liter divided by 0.02 equals 300: choice B As far as the units are concerned, mmols and liters cancel leaving inverse seconds as the units Whenever you see a rate constant of inverse seconds you can be sure that it is a first-order reaction Second-order rate constants have the units liter per mole second Again the correct answer is B KAPLAN MCAT For question the correct answer is A Rate constants and equilibrium constants are both affected by temperature The temperature dependence of the rate constant is explained by the Arrhenius equation, which is, in logarithmic form, the base-ten log of the rate constant is equal to the log of the frequency constant (A), minus the activation energy (Ea) divided by the product 2.303, times the gas constant, times the temperature If you’re not familiar with this, don’t worry we’ll talk about it in just a while You used this equation to answer some questions in the next passage Basically, what you need to know is that as the temperature increases, so does the rate constant The temperature dependence of the equilibrium constant is explained by the equation DG = –RTlnK, where DG is the standard free energy change of the reaction (remember that standard conditions are observed when each substance is pure and at atmosphere of pressure) This equation is valid for any kind of equilibrium: acid dissociation constants, base dissociation constants, equilibria involving gases, etc You should have this equation memorized Don’t worry about memorizing the Arrhenius equation; just know that the rate constant increases with increasing temperature If you need to use the Arrhenius equation it will be given to you Again, the correct answer is A The correct answer to question is choice D This question is testing your knowledge of Le Chatelier’s principle, which states if a system is subjected to a stress, it will adjust itself in order to alleviate that stress Let’s talk about of few of these stresses Adding or removing a product or reactant: the reaction will shift in order to consume the reactant that has been added and shifts in order to replace the product or reactant that has been removed In other words, if you add something to the reactant side, the equilibrium will be shifted to the right, in order to consume the excess reactant If a product is added, the equilibrium will shift to the left, in order to consume the excess product If a reactant is removed, the equilibrium will shift to the left, in order to replace the reactant If a product is removed, the equilibrium will shift to the right, in order to replace the product Another type of stress occurs when the volume of a gaseous reaction is reduced When the volume is decreased, the reaction will shift is such a way so as to reduce the total number of gaseous molecules So, if a reaction has moles of gas on the product side and moles of gas on the reactant side, it will shift to the left, favoring the side with fewer molecules Changes in temperature also stress the system If the temperature of an exothermic reaction is increased, the reaction will shift to the left In other words, the equilibrium constant becomes smaller If the temperature of an endothermic reaction is increased, it will be shifted to the right, the equilibrium constant will become larger There are also a couple of additions to a system that not affect the equilibrium: the addition of an inert gas, and the addition of a catalyst You should know that if a catalyst is added, it affects the forward and reverse reaction equally So it cancels out of the equilibrium expression If an inert gas is added, it also appears on both the product and reactant side, leaving the equilibrium unaffected All right, for this question you need to look at reaction II and decide what you can in order to increase the amount of alkene Well, the alkene is a product, and as I just said, if you need to favor the products, you need to remove a product or add a reactant Choice A, adding water to the reaction will not increase the amount of alkene formed Water is a product, and adding a product will shift the equilibrium to the left, in the opposite direction of what we need Choice B, using a different reactant, will not increase the amount of product formed Choice C, decreasing the concentration of hydroxide ions, will also not increase the amount of product: since this reaction proceeds by an E1 pathway, the concentration of hydroxide won’t affect the reaction at all Choice D, adding a material to complex out the product bromide ion will increase the amount of product formed Bromide ion is a product; if it is removed, the equilibrium will shift to favor the products side Choice D is the correct answer The correct response to question is D As I discussed in question 6, adding an inert gas or a catalyst does not alter the value of the equilibrium constant Again, the correct answer is choice D Passage II (Questions 8–14) The correct answer to question is C Determining the rate law from experimental data was discussed in question If you had a problem with this question, rewind the tape and listen to the discussion of question All right, let’s recall a couple of helpful hints: if the concentration of a reactant is doubled and the rate law doubles, the reaction is first order with respect to that reactant (In addition, if it triples and the rate law triples, it is first order as well.) If the concentration of a reactant is doubled or tripled and the rate quadruples or increases by a factor of nine, respectively, the reaction is second order with respect to that reactant You should know that the concentrations of one or more of the reactants appear in the rate law Are there any choices that can be eliminated? Are there any that contain product concentrations? Unfortunately they are all possible rate laws, so none of the choices can be eliminated Looking at the data for experiments and you can see that the concentration of iodide has been held constant and that the concentration of persulfate has been doubled So, if the concentration of persulfate is doubled and the rate doubles, it means that the reaction is first order with respect to persulfate Experiments and have the same persulfate concentrations and the concentration of iodide is doubled in experiment What happens to the rate? It doubles as well The reaction is first order with respect to iodide So, the overall rate law is the rate is equal to the rate constant times the concentration of iodide, times the concentration of persulfate: choice C 10 as developed by Kinetics and Equilibrium Test The correct answer to question is choice B In order to get the correct answer to this question, you, unfortunately, have had to answer question correctly (This does not happen on the MCAT, but keep in mind that this test is designed to test your knowledge of a particular topic, and determining rate constants is an important part of this topic.) Anyway, you can take the data from any experiment, plug it into the rate law, and solve for the rate constant Taking the data from experiment and solving for the rate constant, you get that the rate constant is equal to 2.28 divided by the product of 0.21 and 0.30, or 2.28 divided by 0.06 If you have trouble with decimals, you can multiply the numerator and the denominator by 100, which gives 228 divided by Armed with this, you should be able to tell that choice B is the correct answer: certainly goes into 228 more than 7.6 and less than 172 and 241 Again, choice B is the correct response 10 The correct answer to question 10 is D This question is simply testing your understanding of Equation If you remembered the simple relationship between the activation energy and the speed of a reaction, you could also have answered this question without even looking at Equation Anyway, looking at Equation 1, you can see that the rate constant, k, is directly proportional to 10 to the negative something So, the more negative that exponent, the smaller the rate constant, and the slower the reaction What will make this exponent more negative? Well, the larger the activation energy, Ea, the more negative the exponent Also, the lower the temperature, the more negative the exponent Let’s take a look at the answer choices Choice A states the when the temperature is held constant, the lower the activation energy, the slower the reaction Equation tells us that the HIGHER the activation energy, the slower the reaction Choice B states that when the activation energy is held constant, the lower the value of A, the faster the reaction This is clearly wrong, the HIGHER the value of A, the faster the reaction Choice C states that when the activation energy is held constant, the lower the temperature, the faster the reaction Well, if the temperature is decreased, the exponent in Equation becomes more negative, the rate constant becomes smaller, and the reaction is slower (You could have also used your intuition on this one.) Choice D states that when the temperature is held constant, the lower the activation energy the faster the reaction This is a true statement; choice D is the correct response 11 The correct answer to question 11 is D If you didn’t know how to read Figure 1, this question would be rather difficult to answer Keep in mind that the MCAT incorporates graphs and data tables into their passages, and you can count on questions like this one Anyway, Figure is a plot of the log of the rate constant versus the inverse of the temperature You are also given the slope of the line Where you see a relationship between the rate constant and the temperature? You see it in Equation 1, the Arrhenius equation What you need to realize is that Figure is a plot of Equation If you take the log of both sides of Equation 1, you get that the log of k is equal to the log of A – Ea divided by the product of 2.303, the ideal gas constant, and the temperature (As you can see, you must know how to manipulate logarithms.) This is in the form of the general equation for a straight line: y is equal to mx + b, where b is the y intercept and m is the slope Well, the slope of this line is equal to –Ea divided by the product of 2.303 and the ideal gas constant, and you are given the numerical value of the slope at the top of Figure So, all you need to is solve for the activation energy, Ea, and approximate the answer, since the choices are pretty spread out All right, what you get? You should have gotten Ea is equal to 9.71 times ten to the third Kelvin, times 2.303, times 8.314 joules per Kelvin mole (the negatives canceled each other out) The Kelvins cancel out, so you know that the answer has to be in kilojoules per mole Are there any answer choices that are not in kilojoules per mole? No there are not, so none can be eliminated In order to simplify the calculation, multiply 9.7 by 2.3 to get about 22, don’t forget to carry the ten to the third Then multiply 22 by 8.3 to get 176 So, the answer should be about 176 times ten to the third joules per mole or 176 kilojoules per mole Choice D is the closest and is , therefore, the correct choice 12 For question 12 the correct answer is A As I just stated in the explanation to question 11, in a plot of the logarithmic form of the Arrhenius equation, the intercept with the y-axis is equal to the log of A: Choice A 13 The correct answer to question 13 is A This question tests your knowledge of basic stoichiometry The first thing that you need to is make sure that Reaction is balanced Is it? Yes it is Now, the question asks you to relate the rate of disappearance of iodide to the rate of formation of sulfate How you this? Well, you need to set up the proper conversion factor and multiply it by the rate of disappearance of iodide The conversion factor needed is over 3, being the stoichiometric coefficient for sulfate and being the stoichiometric coefficient for iodide You don’t need to actually carry out this calculation; you should be able to see that the answer has to be slightly less than 2.5 times ten to the minus 3: answer choice A 14 The correct answer to question 14 is answer choice A This question is not passage related and is what we call an outside knowledge question The first thing that you should know, although you really don’t need it to answer this question, is that a catalyst speeds up a reaction without itself getting consumed The other thing that you need to know about a catalyst, and you need to know this to answer this question, is that a catalyst speeds up a chemical reaction by lower the activation energy of the reaction So, you are given a reaction profile for an uncatalyzed reaction and you need to pick out the reaction KAPLAN 11 MCAT profile that could be the catalyzed reaction Well, you need to know what reaction profiles are Reaction profiles are plots of the potential energy of a reaction as a function of the reaction coordinate, or the progress of the reaction You should know that the potential energy associated with the peak is the activation energy for the reaction The activation energy is the minimum amount of energy needed for reactants to go to products As you can see, the activation energy for the uncatalyzed reaction is 120 kilojoules per mole So, you need to look for a reaction with exactly the same profile but with a lower activation energy Looking at the answer choices you can see that choice A looks exactly the same as the reaction profile for the uncatalyzed reaction, but it has a lower activation energy Choice A is the correct answer Let’s look at the other answer choices Choice B looks the same as the uncatalyzed reaction, but it has a higher activation energy Catalysts don’t this Choice C has an activation energy of 95 kilojoules per mole, but this reaction is an endothermic reaction, not exothermic like the reaction in question Endothermic reactions have a final potential energy greater than the starting potential energy Choice D is wrong because it is the exact same reaction profile as the reaction in question Again, the correct answer is choice A Discrete Questions 15 The correct answer to question 15 is choice B This question test your knowledge of Le Chatelier’s principle Le Chatelier’s principle was discussed extensively in the explanation for question If you have forgotten that discussion, or haven’t listened to it, please rewind the tape and listen to it now All right, what will happen to this reaction if more chlorine is added? Adding more chlorine puts a stress on the system, and the system alleviates that stress by using up that added chlorine: the equilibrium shifts to the left Answer choice B is the correct response Answer choice A, it will be shifted to the right, is wrong because that would just further stress the system The equilibrium would shift to the right if we had, say, added some PCl5, the reactant Choice C, it will be unaffected, is wrong as well Le Chatelier’s principle tells us that if a stress, such as adding more product, is put on a system, it will react in such a way so as to alleviate that stress Choice D is wrong because we had all the information we needed to answer this question Again, the correct answer is choice B 16 The correct answer to question 16 is choice B As was stated earlier, if the temperature of an exothermic reaction is increased, the reaction shifts to the left If the temperature of an endothermic reaction is increased, it shifts to the right The reaction in question has a negative enthalpy of reaction, meaning that is an exothermic reaction If the temperature of this reaction is increased it will be shifted to the left: answer choice B (You can also think of heat as a product in an exothermic reaction and as a reactant in an endothermic reaction Le Chatelier’s principle can then be used as per usual.) 17 The correct answer to question 17 is choice C The equilibrium constant expression (also called the mass action equation) shows the product of the product concentrations raised to their stoichiometric coefficients over the product of the reactant concentrations raised to their stoichiometric coefficients For heterogeneous equilibria, pure solids and liquids are not included in the mass action equation; gases and aqueous components are included So, for this question you need to construct the correct equilibrium constant expression and calculate the equilibrium constant So, what is the correct equilibrium constant expression? It is the concentration of ammonia squared over the product of the nitrogen concentration and the cube of the hydrogen concentration Plugging in the appropriate values, 15 squared over the product of one times cubed This gives an equilibrium constant of 28, choice C 18 The correct answer to question 18 is choice C The first thing that you need to to answer this question correctly is to construct the equilibrium constant expression The equilibrium constant expression for this reaction is the concentration of hydrogen bromide squared over the product of the hydrogen concentration and the bromine concentration Plugging in the appropriate concentration values, You get 500 squared over the product of 0.05 times 0.03 You should be able to tell that this figures out to be in the 10 to the fifth region, much larger than the equilibrium constant In order for this reaction to reach equilibrium, it must use up all that excess hydrogen bromide Consequently, the equilibrium will shift to the left Again the correct answer is choice C 12 as developed by ... Kinetics and Equilibrium Test KINETICS AND EQUILIBRIUM TEST TRANSCRIPT Passage I (Questions 1–7) The reaction that goes to completion to the greatest degree is the one with the larger equilibrium. .. H2, Br2, and HBr are 0.05M, 0.03M, and 500.0M, respectively The equilibrium constant for this reaction at 400°C is 2.5 × 103 Is this system at equilibrium? A Yes, the system is at equilibrium. .. Both the equilibrium constant and the constant would change B The equilibrium constant would change, but rate constant would remain the same C The rate constant would change, but equilibrium
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