Create assignment, 48975, Quiz 5, Apr 19 at 12:52 pm 1This print-out should have 8 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore making your selection. The due time isCentral time.PLEASE remember to bubble in yourname, student ID number, and version num-ber on the scantron!Msci 16 010116:01, general, multiple choice, > 1 min, fixed.001The rate of the reaction2 O3→ 3 O2is equal to1. −∆[O3]∆t.2. +12∆[O3]∆t.3. −∆[O2]∆t.4. −12∆[O3]∆t. correct5. −13∆[O2]∆t.Explanation:The rate of a chemical reaction can eitherbe expressed as the rate of disappearance ofa reactant (as in this case) or as the rateof appearance of a product. However, thisrate is multiplied by the inverse of the coef-ficient of the species in question. The nega-tive sign here reminds us that as the reactionprogresses, this species is consumed and de-creases in concentration.Msci 16 031116:03, general, multiple choice, > 1 min, fixed.002What are the units of the rate constant forthe reaction2 Cl + O2→ Cl2+ O2(M = molarity and s = seconds) rate = k[Cl][O2], where k = rate constant.1. M−2·s−12. M−3·s3. M2·s−14. s−15. M−1·s−1correctExplanation:The unit of rate will always take the formAmount (in mol or concentration)unit timeAs the concentrations are expressed in mo-larity, the rate constant will have units ofM−1·s−1in order for the rate to have units ofM/s.Msci 16 031916:03, general, multiple choice, > 1 min, fixed.003The following data were collected for the re-actionA + B → Cat a particular temperature.Three experiments giveInitial Initial Initial rateTrial [A] [B] ∆[C]/∆tM M M/min1 0.1 0.1 4.0 × 10−42 0.2 0.2 3.2 × 10−33 0.1 0.2 1.6 × 10−3What is the rate-law expression for thisreaction?1. 0.4 M−2·min−1[A][B]2correct2. 0.4 M−2·min−1[B][A]23. 10−4M−2·min−1[A][B]2Create assignment, 48975, Quiz 5, Apr 19 at 12:52 pm 24. 10−4M−2·min−1[B][A]2Explanation:To determine the order of reaction withrespect to B, compare trials 1 and 3:rate3rate1=µinitial [B]3initial [B]1¶x1.6 × 10−3M/min4.0 × 10−4M/min=µ0.2 M0.1 M¶x4 = 2xx = 2Thus the reaction is second order with respectto B.Likewise, to determine the order of reactionwith respect to A, compare trials 2 and 3:rate3rate2=µinitial [A]3initial [A]2¶y3.2 × 10−3M/min1.6 × 10−3M/min=µ0.2 M0.1 M¶y2 = 2yy = 1Thus the reaction is first order with respectto A. The rate law will take the formRate = k [A][B]2.To determine the rate law expression, in-cluding the value of k, the data from any ofthe trials can be used. Using trial 1 as anexample:Rate = k [A][B]24.0 × 10−4M/min = k [0.1 M][0.1 M]2k = 0.4 M−2· min−1,so the rate-law expression here isRate = 0.4 M−2· min−1[A][B]2Msci 16 041316:04, general, multiple choice, > 1 min, fixed.004The reaction2 N2O5→ 2 N2O4+ O2is first order. At a certain temperature it hasa rate constant of 0.0840 sec−1.At this temperature, how long would it takefor 95% of a given sample to decompose?1. 35.7 sec.2. 0.305 sec.3. 17.8 sec. correct4. 7.74 sec.5. 0.595 sec.6. −17.8 sec.7. −7.74 sec.8. −35.7 sec.Explanation:k = 0.0840 % decomposed = 95%The integrated 1storder rate equationneeds to be used:ln·A0A¸= a k tSince 95% of the sample has been decom-posed, A0= 100, A = 5. (These are arbitrarynumbers; the units are not important becausethey cancel out).Substitute these values, along with thosefrom the problem, and solve for time:ln·1005¸= 2(0.0840 sec−1)t2.9960.168 sec−1= tt = 17.8 secMlib 50 452116:09, basic, multiple choice, > 1 min, fixed.005In what way does a catalyst speed up reac-tions?Create assignment, 48975, Quiz 5, Apr 19 at 12:52 pm 31. They lower the energy of the products.2. They lower the energy of the reactants.3. They lower the activation energy of thereaction. correct4. They add energy to the system.Explanation:Catalysts speed up a reaction by providingan alternative mechanism with a lower energyof activation.Msci 16 101316:08, general, multiple choice, > 1 min, fixed.006For a reaction the rate constant is 6 × 105sec−1at 25◦C.If Ea= 40 kJ/mol, what is A, the Arrheniusfactor?1. 6.2 × 1012s−1correct2. 6.2 × 109s−13. 155000 × s−14. 3.1 × 107s−15. 1.24 × 1014s−1Explanation:k = 6.0 × 105s−1Ea= 40000 J/molT = (25 + 273) K= 298 Kk = A e−Ea/(R T )A = k eEa/(R T )=¡6 × 105s−1¢× exp£(40000J/mol)(1.987 J/mol·K)(298 K)¤= 6.1627 × 1012s−1Msci 16 090816:07, general, multiple choice, > 1 min, fixed.007Consider the multistep reaction that has theoverall reaction2 A + B → 2 C + D.What would be the observed rate expres-sion?Mechanism:2 A → C + I (slow)I + B → C + D (fast)1. Rate = k [A]2[B]2. Rate = k [I] [B]3. Rate = k [A]2correct4. Rate = k [A] [I] [B]5. Rate = k [A]Explanation:The first step is the slowest step (the ratedetermining step), so the rate law can bewritten based on the stoichiometry of the re-actant.Poten E diag16:06, general, multiple choice, < 1 min, fixed.008ABReaction coordinateEnergy (kJ)100200300What is the activation energy for A → B ?1. 100 kJ/mol rxn2. 200 kJ/mol rxn correct3. 300 kJ/mol rxn4. 150 kJ/mol rxn5. 250 kJ/mol rxnCreate assignment, 48975, Quiz 5, Apr 19 at 12:52 pm 46. More information is needed.Explanation:The activation energy is the energy differ-ence the reactants (shown at A above) mustovercome (the maximum of the peak) beforethey can proceed to form products (at B). Inthis case this difference is300 kJ − 100 kJ = 200
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