SI_Midterm_1_ANSWERS Bau

SI_Midterm_1_ANSWERS Bau - Dr. Bau Judy Chen (judyj...

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Unformatted text preview: Dr. Bau Judy Chen (judyj che@usc.edu) CHEM 105b— Spring ’08 http://www.usc.edu/si SI Midterm 1 Review 02/03/2008 H . Multiple Choice 1) A given reaction is found to be exothermic (gives off a significant amount of heat). What can be said about this reaction? a. Since so much energy is given off the reaction will be very fast b. The activation energy will be small c. The reaction will slow because energy is given off (1. The potential energy of the products is great than that of the reaction. @ None of the above. 2) What are the units for the rate constant of a third order reaction? a. mol L'1 s‘1 b. mol'1 L s'1 3) A reaction that has a very low activation energy a. gives a curved Arrhenius plot b. has a rate that is very sensitive to temperature c. must be second-order d. must be first-order Ce: has a rate that does not change much with temperature 4) An elementary reaction has an activation energy of 88 kJ mol’1 and AH= —60 kJ mol'l. The activation energy of the reverse reaction is a. -88 kJ mol’1 b. 28 kJ mol'1 (A\ 1 148 kJ mol' \ . 60 kJ mol'1 6. -60 kJ mol" II. a) The following data was obtained for the redox reaction between I" and OCl‘ in the presence of base. From the data given below, determine the rate law: m — 5 0.009 00! 0,037 4,,grloz [175mman ' E! 2 [0w]: ewerimgmg 3 ill “; [OCT]: experimw‘fi Bag} 2 _ I PM A 0597b 1’ 7"“‘1 mass 373-301 ,EZD-Oozglwflifi‘fipf ‘- 4m “WAW V0631 ’ 9'1‘502 1 Km. 0-!%= 0.w-z<;’1.»+9>” 094539 = 12%” \m :7 ’2 Order for [1']: ’ 2 Order for [OI-F]: I Order for [OCl’]: O b) What is the overall rate law and the value of k (with units)? Me = k W" m ‘ [00: 7’1 32492”? k {0' l) 0(O. ©0?§'M)ff$.OZSM)’Z MW...____MW 2 - Answerzk= M 5 Overall Rate law: W30? 5 FIT-j [00"] 2W III. Consider the reaction 3 A + 4 B —) 2 C Is the rate of consumption of A (-A[A]/At): greater then (>), smaller than (<) or equal to (=) the rate of consumption of B (—A[B]/At)? _|_—A ___J__—A[3 3 At" H at :éi_ 33495. 4M L: At Thereaction N0 (g) + 03 (g) 9 N02 (g) + 02 (g) Obeys the following rate law: — AfNO] / At = (0.8 mol'1 L s") [03]2 a) Four possible mechanisms for the decomposition are: I 11 N0 + 03 a No2 + 02 03 —> 9’+ 02 NO +6 —> N02 N0 + 03 -> “01 a O; 034» NO ——> N02 .4. 02 III A IV NO + 03 '9 N02 2/ 03 '9 ,0; O3+N0l1 9 N02+202+0 N0+9€ '9 N02+p3/+02 N0+203 we Magi/303% Q 034M) was M32 4 0; b) What mechanisms are consistent with the overall balanced equation? tot/WA 1/131, 11 c) If the first step is assumed to be rate-determining for each suggested mechanism above, which is the most-likely mechanism? Why? WW9 EDIE 1V Haws? Route [3. UIQVIBFMMM/ 5an [6:35, NWCA {m AIMS rag? is first Raw, am ONE? m EM. {/53 Wed V. VI. At elevated temperatures, nitrous oxide decomposes according to the equation 2 N20 (5%) 9 2 N2 (g) + 02 (g) a) Given the following data, determine the rate law for the reaction dealt“? MW? b) What is the rate constant for the reaction? (Make sure to include units) glo e z 2%}, "1.303133 M , P Q'f “I ‘80 “.0 :”2.3"O j K 2: 0.0023 3‘?“ 4 lg Oldrfl' c) What is the half life? ‘ Gite??? - ~ t7: *’ «Jim 9,] d) How much time would it take for 88% of the N20 to be depleted? A [N20] : ~kt + M [Nzojor at. twelve” 0/207, wk 1;. a? fix. (J1) ;—(o,00235"}€f [FIEFEE mnw—wmnn—vammwwhm I In Houston (near sea level), water boils at 100.0°C. In Cripple Creek, Colorado (near 9500 ft), it boils at 90.0°C. If it takes 4.8 minutes to cook an egg in Cripple Creek and 4.5 minutes in Houston, what is Ea for this process? . \ (< (5 olwagg 4mg \: CyEPFEP/Crfek ‘52 ~— r Ea. _L_ l 2»; wagaorx r /@L( k‘) " “fl «@(fli.§)\vea I me‘ Or : — Egan—mm; a? ’9‘ J 353%} if. J p 2.7g3‘ \D ‘ th‘é ' ammo-6) Ea: 72Hb J/Mol = 7 .WWZETW‘" .«W 74...”...W— 7v-~.-....._r VII. Hemoglobin molecules in blood bind oxygen and carry it to cells, where it takes part in metabolism. The binding of oxygen via Hemoglobin (aq) + 02 (g) -) Hemoglobin - 02 (aq) Is 1st order in dissolved oxygen and lst order in hemoglobin with a rate constant of 4 x 107 L mol"1 s'l. a) Write the differential rate law for the reaction. KM ’* [4 [HE] [027 b) Calculate the initial rate at which oxygen will be bound to hemoglobin, if the concentration of hemoglobin is 2 x 10‘9 mol L'1 and that of oxygen is 5x10’5 mol L". «I «l . '9 , HI]. » 3“» Fate 2 (Mo1 ism! s )5?" 45‘ ‘m? S-ffm? \' " c) Which reagent is in excess for these initial concentrations? Write the pseudo-first order difi‘erential rate law and calculate the pseudo—first order rate constant k’ for these conditions. (Egg/3m is M QKCQSS‘ (fl-l5 1' l5 )[Hzoj bub—(3N kl :7 l< [027 k": (Li'lxfi'l‘ornoi "\S'llth'lO’SMOl L”) £- : 20 407‘s” : )ZOOOT‘ ,, M d) How long would it take for half the hemoglobin to become oxygenated for these initial concentrations? nae ‘2 , 0.90% “"11"” {'V2' " m~ : 3.540937 VIII. The reactions catalyzed by enzymes involve the following mechanism: H B/ + s <:=\> (fast equilibrium) 55> [E] + “A/ (slow) 9/ + F 13?» P + B (fast) Where E is the enzyme, S is the substrate, and P is the product molecule. a) Determine the overall reaction. S+F———>P+@ b) What are the intermediates involved in the reaction? 93% c What is the catal st involved inthe reaction? Howcan ou tell? y y _ ‘ E -~> H; is hol- imolw‘d ankiini, Gwyn“ it"s-ailékflfl flirt?! 54“ I ,,.K __ l g. f- {a «if! ‘15,! J} ,2 {1’4} 6:55;) in f}?! ammo us ,i a, . fr . i i * r, i . "a “wild? m FWifiSS; d) What is the rate determining step? ?“d 34630» e) What is the molecularity of this reaction? W ’1 ‘ {7 33m: r D What is the rate law? \\\ k r : k, - \ : [E_S kzfifliéi’!‘ ] x 3" J E j k {213ij USEFUL PHYSICAL CONSTANTS AND EQUATIONS Temperature conversion Avogadro’s number Atmospheric pressure Energy units Universal gas constant Faraday’s constant Ideal gas law Work P-V work Heat transfer 1St law of thermodynamics Enthalpy Arrhenius equation Arrhenius equation Quadratic formula Order Integrated rate law Half—life Tp = (Tc x 9/5) + 32 TX = TC+273.15 NA = 6.022 x 10+23 1 atm = 760.00 torr = 101.325 kPa 1 cal = 4.184 J 1 L—atm= 101.325 J R = 0.08206 L atm 1cI mol‘1 : 8.314 J K“ mor1 F = 96,485 C mor1 PV = nRT w = —FAL w = —PextAV q = mCSAT AE=q+ H=E+PV W k = A exp( ——Ea/RT) m k_2 =Ea .1": k1 R T1 T2 x_—-bi\/b2—4ac 2a zero first [AL— [Mo = —k1 t1/2 = [A]o /2k 10 111M]: = 111 [Mo —kt t1/2 = 0.693/k second 1/[A]t = 1/[A]0 + kt t1/2 = 1/kIAlo ...
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SI_Midterm_1_ANSWERS Bau - Dr. Bau Judy Chen (judyj...

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