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Unformatted text preview: Dr. Bau Judy Chen (judyj [email protected])
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 secondorder
d. must be ﬁrstorder
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 373301 ,EZDOozglwﬂiﬁ‘ﬁpf ‘
4m “WAW V0631 ’ 9'1‘502 1 Km.
0!%= 0.wz<;’1.»+9>”
094539 = 12%” \m :7 ’2 Order for [1']: ’ 2 Order for [OIF]: 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 FITj [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 ratedetermining for each suggested mechanism
above, which is the mostlikely 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 ' ammo6) 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? Sffm? \' " c) Which reagent is in excess for these initial concentrations? Write the pseudoﬁrst
order diﬁ‘erential rate law and calculate the pseudo—ﬁrst order rate constant k’ for these conditions.
(Egg/3m is M QKCQSS‘ (fll5 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———>[email protected] 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"sailékﬂﬂ 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 PV 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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 Chemistry

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