Ghapter 13 Number 4 (13.9 - 13.13)4. Consider: 2 NOr(g) ) 2 NO(g) + Oz(9).

The rate constant increases with temperature assummarized here. Use the Arrhenius equation tocalculate the energy of activation for this reaction

Temperature k (L/mol s)330& K 0.773s4Æ R 1.8

3839 K 4.7

1. Consider these four reaction profiles:

A. B.

c.pro|'asr

D.Pre¡nt3

p.ol¡ll¡3 pmtlL3

O..O.a

¡+a

(Unit 1) 29 January 20182f . For this particular

reaction, draw avertical line torepresent E"6tod andE".1tuu on the energypopulation graphbelow. Note that"speed" is related tokinetic energy and collision energy.

3. The following mechanism has been proposed forthe reaction between nitrogen monoxide andchlorine gases.

Step 1. NO + Cl2) y}eb Slow stepstep 2. r'ro + ryøef,à 2 Noct

3a. \lúhat is the overall reaction?

L No r Cr¿ ^> 2 NoCl3b. Circle any intermediates in the mechanism.3c. Given the first step is slow, what is the rate law?

rq+e * þ &t03 €b33d. Sketch the

reactants àtransition state) products forStep 1.

3e. \Â/hich of the reaction profiles given in Question1 best describes this mechanisrp, given that theoverall reaction is exothermicZ@)A C O

3f. Suppose instead that the second step is the slowstep. \rl/hat is the rate law in this case?

Do J?^r I eur- a'Ê*ø¡ -ti.P€trp urr lr [¡v'tvt n ^ chæp*e r''

Questions in finalexam format:5. What is the minimum energy barrier that must be

overcome for a chemical reaction to occur?

C. potentialenergy D. rate limiting energy

6. lf a plot of [A] vs time gives a linear.plot...A. reaction is first order and slope is the rate.B. reaction is O-order and slope = rate constant.C. reaction is 1"t order and the slope is the halflife.

@tne reaction is zero order and the slope is thenegative rate constant.

Now try these problems from the book:Problems 23-32,48,92,94,96,98, 100, 106, 108 and

112

#{wn,

ì4v,.

W,&Hot

ln(k2lk1) = (-Ea/RX1iT t - 1 fi z)

Se¿ PrPvtoot¡' | ú¡rp-/,reù*

€lla. Which ones are exothermic?6@ C61b. Which reaction is the slowest? A B @ D1c. For which of the two^steplpactions is the first

step the slow step? @ e (q O

1d. For which reaction is there the best chance ofdetecting an intermediate? A e OO

le. Forwhich reaction is there no intermediate?A B ct

2. Let A = red circlesandB=bluecircles.

2a. ls this reaction F,,'õñtherm¡c or CLËñlothermic? t¡¡

2b. Label E""ttud, E""1'"u,and ÂH''n.

2c. Write a formula relating AHon in terms of E"6tudand E""¡t"u

2d. Write a balanced chemical reaction for thisreaction.

*_ tå -;er A l. ,+-&

Reaction progress à Sctivatio n energy B. net energy^

O+aO

rl tlo-MÌ,, î.,

o-P0*'cl

4ûA Flnn È

2e. Would increasing t$#qnperature increase therate of the reaction?(þJor No

Chapter 14 Number 1 (14.1 - 14.41 (Unit 2)1. Considerthe reaction given below. At425 oC, the

initial concentration of Hl, [Hl], is 0.4000 M and theother concentrations are 0.0000 M. Afterequilibrium is reached, [Hl]E = 0.0254 M. Whatare the equilibrium concentrations of H2(g) + lz(9)and what is the value for K"?

M 2qt(s) à Hz{g) + tz@)

t Ot4vo ü üc _-?{ "J1._x ^E 0,*ü-2x X = 0, ¡8?J

2 February 2018and 0.00 M CH3OH. At equilibrium, [CO] = 0.096M. What are the equilibrium concentrations of allthree gases and what is the value,þlKc?M CO(s) + 2Hz@) ) CH3OH(g)t Ø¡ zàþâ a,vll oc -x :Zx TTE O,A6-X O,ml -2x ^ = O,lS

Questions in finalexam format:3. Which one of the following statements does not

describe the equilibrium state?A. Equilibrium is dynamic and there is no net

conversion to reactants and products.

6ån" concentration of the reactants is equat tothe concentration of the products.

C. The concentration of the reactants and productsreach a constant level.

D. The rate of the forward reaction is equal to therate of the reverse reaction.

4. lf Kc= 7 .04 x 10-2 for the reaction:

2 HB(g) + Hz(g) + Br2(g)

What is the value of K. for this reaction?

%Hzß)+% Brz(9) + HB(g)

A. 3.52 x 1O-2 B. 0.265

Ç)ztt D 2s.4

5. Which of the following is the correct equilibriumconstant expression for this reaction? (Hint:Balance the reaction first)

*re(s) Sortgl +fe2o3(s)

A. K. = [Fe2O3]/([Fe][O2]) B. K. = 1/[O2]

C. K. = [Fe2O3]/([Fe] + [O2]) D. K" = [Or]-3Now try these problems from the book:Section 14.2. (Equilibrium constant, K") Problems: I -6,

40, and 42Section 14.3 (Equilibrium constant, Ko) Problems:7 - 10Section 14.4 (Heterogeneous) Problems: 11 - 12Overall Sections 1 - 4. Problems 50 - 72(even)

Ê

6t *OO-Z)< s ü' Az{¿fX : Ot lg?"

+

1b. Plot these concentrations on the y-axis as afunction of time on this graph. The kinetics regionis not zero order. Assume equilibrium is reachedabout haltway along your x-axis.

C,¡

ot3

0tz

0tl

time )1c. What would be the equilibrium constant if the

initial concentrations of Hl, [Hl], is 0.600 M and theother concentrations are 0.00 M?

l+'l I s+¡ll be 5 t, 51d. What is the value of K. for the reverse reaction?

- Hz('s) + tzl.g) + z{t(g)

2. Consider the reaction given below. At 250 oC, theinitial concentrations are 0.246 M CO, 0.491 M H2,

*.tr

rl"f a

O, Z¿t6 -)4 = A , 016 .E-Ð l= Ot

*ol(c; E*aa# 3?* 0' l5

Ker . .&sç#*#$7E- þ

Ghapter 14 Number 2 (14,5 -'14.71 (Unit 2)1. Consider the reaction given below. At425 oC, the

equilibrium constant, Kc= 54.4. Starting with, [Hl], =0.300 M Hl, what are the equilibrium concentrations ofallthree gases, [Hl]E, [H2lE, and !2]E?M 2Hl(g¡) ) Hz(ù + l2@)

I 0,7tt O e

^ -2X L åt--

,01¿tO( E=Ot

5 February 20184. Another LeChatelier Principle predicts the results of a

volume change. lmagine the reaction and equilibriumconcentrations: + ,:::, ,&M, iE 5.0

l\30'l tly!". What is the value of K"?

0,3A4b. Now suppose that the volume of the contained was

decreased to just half its original value. \Mtat are thenew initial concentrations of A and B? Calculate Q". lnwhich direction must a shift occur?M A(s) + 2B(g)

E 0' 'iii{;))Ot

yz-&t tÑoq -¿xf1b. Suppose that [Hl]' = 0.0700 M, [H2J¡ = [lz]l = 0.400 M

Which direction must the equilibrium shift in order toattain equilibrium? \lVhat are the equilibriumconcentrations of all threë gases, [Hl]E, [H2]¡, and

ltzle?

3. \Â/ithout doing any Q. calculations, predict whichdirection the following equilibrium would shifr if...sH2(9) + N2(9) + zNH3(9)

For each of these, write "Shiff Right", "Shift Left" or "Noshift"

3a. Some nitrogen gas was added. ft3b. Some hydrogen gas was removed. L3c. Some ammonia gas was removed. R,3d. Some ammonia gas was added. l?3e. Some argon gas was added. MO

E 5.0 2.O

t /A' t 4rêc +N -uE lO fY 4,0 -U

¿2-Qc5fr: 4shl*"I LM 2{l(g\ ) H2{g) + 12(ù ,,(OL

t ,a7 ,tl7 ,10 Qc= fr' E

c -þ *x* l- f'=327 c

Ê o¡o?-Lt¿ Ão'*r ã''f, f^ft*'= Q¡ 0157 Q T:rori ¡sot,?atg'z= æ

2. considerthe reaction atzsol KP' çtf¿Tfry^

Repeat the problem for this equilibrium. Again, thevolume is decreased by half.M c(s) + D(s)

E 4.0 3.0

t h0 lûcã= lrtl öû

co(s) + 2H2@) ) cH3oH(sf) Kc= 42.8 6. Which direction willeach reaction shift if the volume isdoubled. 'Shift R'"Sh¡ft L'or "No shift"4a. 2CO(s) + o2(9) = 2 Co2(9) Lab. 3H2(9) + N2(S) =2NH3(S) l-

2a. \Mrat is the relationship between K" and ?

2b. Solve for Ko Given: R = 0.0821 L arm mot-l K-\ *}t5ø'C 4c.3H2(s) + wo3(s) =W(s)+3Hzo(9) NO

K"= * â 0,7fQc= êt7f

ltEËT

Kp = tLtE J+(,øtzt *'52Å)-¿' -- OtgllZQuestions in finalexam format:7. What best describes what would happen to the system

given in Question 1, at equilibrium, if some Hl wereadded?

A. The system would no longer be at equilibrium andwould shift to the left in order to-reestãblishequilibrium.

(îÀn"requiibrium is reestablished, the concentration ofv all three gases would be-greater than they were priorto the addition of more Hl(g).

C. The system would remain at equilibrium, however theamount-of H2(S) and l2(9) would increase and the amountof Hl(g) would decrease.

D. The concentration of Hl(g) would end up lower thanbefore the addition of Hl(g).

Fp = K"i7-r] Anl ¡¡A L

8. What best describes what would happen to the systemgiven in Question 2, at equilibrium, the volume of thevessel were decreased?A. The system would no longer be at equilibrium and

would shift to the lefr in order to reestablishequilibrium.

B. After equiibrium is reestablished, the value of K" .

would be larger.

C. After equllbrlum is reestablished, the value of K"would be smaller.A

Qnne system would no longer be at equilibrium and e,. Kc.

9. The equilibrium constant is equal to 5.00 at 1300 K forthe reaction:

2 SO2(9) + O2@) ¡== 2 SO3(e)lf initial concentrations are [SO2] = 1.20 M, [O2] = 0.4S M,

and [SO3] = 1.80 M, the system is^t\þå,teOuilibrium.B. not at equilibrium and will remain in an unequilibrated

state.C. not at equilibrium and will shift to the left to achieve

an equilibrium state.D. not at equilibrium and will shift to the right to achieve

an equilibrium state.Now try these problems from the book:Section 14.5. (Using the equilibrium constant) problems

13 -22, and74 - 84(even)Section 16.6 - 16.8. (Shifring equitibria) 23 -26,98 - 108

(Wait on parts 104(d) and 1 06(a) that refer totemperature until Day 3)

Available at Chm 205 website:Quadratic Equation Solver Excel

qurdraüc fqu¡tþn tolve¡

Êntê?., b¡ âñd c lor tha equttion arr + bt + c = t

Enter a

Ente¡ c

x is elther or 5.87Ê-02

f{otr: ÊomF¡rG dra tro aaotr. fof ot¡f Furpo¡ai....-if boú ¡oc¡ se gqltlug chooæ tlre rmrllsr one..,.f onc þot lr Fodihae rnd onê ir nÊ$üve, Ghoara $e pûf¡tlve rôot^

4

6

Iv

tl

Qr=#îlr¿

= f,00¿0,'lJa'l þ

4.4

u Go*Jz'/\s" - ffiã¡øo7t+0 = ø:-aoÊ

¿aâ]t * ùoat I[coJ = o,ôtzz &¡\,

. 2a. Suppose [N2Oo], = 0.30 M, [Nz]r = [Oz]l = 0 M. At-- some temperature suppose ¡trl2OalE = 0.10 MFurther suppose that it takes about 30 s to reachequilibrium. Plot [N2Oo], [Nz]and [Ozlthru 50 s.

Chapter 14 Number 3 (14.9 - 14.1011. An equilibrium mixture of CO, 02 and CO2 at a

certain temperature contains 0.0010 M CO2 and0.0015 M 02. At this temperature, Ç equals 140for the reaction:M 2CO(g) + oz(9) + 2Cor(g)E l'oôls o'oôl0What is the equilibrium concentration of CO?

1b. To the reaction mixture at equilibrium, supposesome additional CO2(g) was added so that [CO2]= 0.0020 M. Solve for Q. in order to predict thedirection in which the shift will occur as equilibriumin established. Without actually solving for x, setup the MICE table to reflect how the predictedchanges.M 2co(g) + oz(9) + 2Co2@)

I tooLz 0,0olî Otool?c Qc'¿on:^

0,oO2*r 0,oolC: tSOE

Qc 2 Ke l, shrê* L1c. To the reaction mixture at equilibrium as

described in Question 1a, suppose the volume ofthe vessel could be adjusted so that in was nowhalf the size it was in Question la. What are thenew initial concentrations? Solve for Q. in orderto predict the direction in which the shift will occuras equilibrium in established. Without actuallysolving for x, set up the MICE table to reflect howthe predicted changes.M 2CO(g) + Oz(g) + 2COr(g)L 0o4rf 0,00?0 2t\azi)

(Unit 2) 7 February 20182.Ihe third LeChalelier Principle involves changing

the temperature. The "rule" is that "lncreasing thetemperature always favors the endothermicdirection." Changing the temperature is the onlyway to change the value of K". For endothermicreactions, increasing the temperature will cause ashift is towards the right and K" increases. For

' exothermic reactionrl tn" opporite is true.

NzO¿(g) + Nz(g) + 2 Oz(9) LH= -22.2kJ

0'f0,3

0,L'

0tlÁ/¿lLoa

01020304050time (s) à

2b. What is the numericalvalue of Kc?

2c. At t = 50 s, suppose the temperature was raised.With little arrows, show which direction eachequilibrium concentration will shift. Also will it takemore than 30 s or less than 30 s to reachequilibrium again? Or about 30 s?

l1.- l\rrll rhrf* À 'A 1,.. t, ttot, l-g

60

c

E

l* hoqtd Þes

l,v,J1n

oa l#le+.9

t lthflur*..,

û"< Ko

'm#> tCIûota-- 6a les¡ 7ï^n 3CI sl'?,-¿..r|¡f lrt og'r

ii

Éa L

Yc. --

I'' rh' ${'

,t * R

3. Suppose for a certain reaction, the value of K"decreases as the temperature is increased. ls thereaction exothermic or endothermic? Sketch areaction profile (using one hump because we donot know the mechanism) to describe the reaction

4. ln general, what effect will a change intemperature have on the value of K.?

A. lt will have no effect on the value of Ç.B. The value of K" always decreases with an

increase in temperature.C. The value of K. always increases with an

increase in temperature.ß.hn"value of K^ will decrease or increase with\-/ u

an increase in temperature, depending onwhether the reaction is exothermic orendothermic.

5. The reaction of nitrogen monoxide with oxygen isexothermic. What happens when the system isheated?

2No(g)+Oz(9)+2NOr(g)e amount of NO2(g) decreases.

B. The amount of NO2(g) increases.

C. The amount of NO2(g) remains unchanged.

D. The NO(g) is completely converted into NO2(g).

6. A reaction reaches dynamic equilibrium at a giventemperature whenA. the amount of products exceeds the amount of

reactants.B. k¡"0 equals kr"u.

C. opposing reactions cease and the system isstatic.

{prlhe relative amounts of reactants and productsare constant and rate¡r¿ = rater"u.

Now try theee problems from the book:Section 14.9. (Temperature) Problems 27 - 30, 44, 46,

104(d), and 106(a)Section 14.10. (Kinetics and equilibrium) 31 -32,48, and

110-118(even)

ru .1,

@n

E ,*

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