Chem 104 Exam I Spring 2012

C O D E - 43867 - C H E M 104CD Spring 2012 Exam I - Paper Exam

1

Name:

7 pt Consider the following reaction performed at 273 K :

2 N O (g) + CL, (g) - 2 NOC1 (g)

E x p e r i m e n t [ N O ] ( M ) [C1 2 ] (M) R a t e of reac-t i o n ( M / s )

1 0.00100 0.00100 1.4

2 0.00100 0.00200 2.8

3 0.00200 0.00200 2.8

What is the rate law for this reaction?

1. A O rate=k[NO][Cl 2]

C ^ L i a i e s k J C y ? / : f c ) ^ n ^ I e ^ k f C l a ^ - 3

7 pt The reaction:

A - B + C

MS is known to be second order with respect to A and to have a rate constant of 0 .243 M / s at 2 8 9 K . A n experiment was run at this temperature where [A] = 0 . 4 1 7 M . Calculate the concentration of B after 0.151 seconds has elapsed.

2 . A Q 0.00478 / B Q 0.00628 C Q 0.00704 D Q 0.380 E Q 0.402

7 pt The rate of the reaction between hemoglobin (Hb) and carbon monoxide (CO) was studied at 229 K and the rate law was determined to be rate = k[Hb][CO]. When the [Hb] = 0.00871 M and [CO] = 0.00317 M the initial rate of the reaction was found to be 0.2697 M / s . What would the initial rate of the reaction be for an experiment at 229 K where [Hb] c = 0.0373 M and [C0] o = 0.0225 M ?

(in K/s)

3. A Q 7.26 C Q 9.26 D Q 10.5 E Q H - 8

C.ooJtfflt -00*17] | ^ - # 2697

2

7 pt Consider the following information for the reaction A —• products, where [A]D = 1.17 M .

R e a c t i o n E a ( k J / m o l ) A E ( k J / m o l ) rxn 1 75.0 50.0 rxn 2 100. | 7<T -75.0 rxn 3 100. ' • 25.0 rxn 4 ,50.0 25.0 rxn 5 75.0 -50.0

t the reactions wi l l have the fastest initial rate of reaction?

ft 4. / A O rxn 5 B O rxn 3 C O rxn 1 D O rxn 4 E Q rxn 2

Q 5 o I n

7pt Balance the following unbalanced redox reaction (assume acidic solution if necessary):

M n 2 + + CLj -> M n 0 4 " + C l "

Indicate the coefficient that wi l l be used for each species in the overall balanced equation. _

> M n 2 +

5. A O 1 B O 2 C O 3 D O 4 E O 5

> M n 0 4 " 6. A Q 1 B O 2 C O 3 D O 4 E O 5 >Cl2 7. A Q 1 B O 2 C Q 3 D Q 4 E Q 5

It

7 pt Consider the following statements based on collision theory. Indicate if the statement is true or false.

t> A catalyst works by lowering the activation energy of a reaction, allowing a greater number of collisions to overcome the actiyation barrier. 8. [jLQ True B O False

> For a given reaction, collisions of any energy can lead to the formation of a product. 9. A O True B O False

> Increasing the concentration of the reactants in a first order reaction wil l increase the number of successful collisions that take place in the reaction. f - jr\ ^ ^ % 10. A O True B O False L ^

> Collisions with the appropriate orientation wil l always lead to product formation. 11- A O True B O False

> Increasing the temperature of a reaction increases the energy in the random collisions which will lead to a larger percentage of collisions yielding products. 12. A Q True B Q False

3

7 pt For the following two problems use the given half reactions to "construct" a galvanic cell.

M G 2 + + 2 e- - » M g E° c e l l =i2.37 V

Cu 2 + + 2 e- - » C u &>„n=0M V

Predict the standard potential of the galvanic cell at 298 K .

(in Vj

13. A Q 1-45 B Q 1-69 C Q 1-98 D Q 2.32 / " E Q 2.71

7pt

Predict which species wi l l be found in the following portion of the galvanic cell.

9 > Aapde-Btectrode Vf-AQjUg^ B O Cu

C O Mg2+ D O Cu2+ E O Pt

> Cathode solution 15. A O Mg B O Cu

C O Mg2+ Cu2+ E O Pt

> Cathode electrode, 16. A O Mg / ^ ^ C ^

C O Mg2+ D O Cu2+ E O Pt

> Anode solution 17. A O Mg B O Cu

/ / S Q Mg2+ D Q Cu2+ E Q Pt


7 pt Consider a galvanic cell that is set up under standard conditions and allowed to react. Evaluate if the following statements are true or false.

> Electrons wi l l flow from the cathode half cell to the anode half cell. 18. A O True B O False

6 > The half reaction with the lower standard reduction potential will be found in the cathode portion of the cell. 19. A O True B O False

/ i z - f -> Electrons are produced spontaneously at the anode electrode. / 1 ' ^ ' 20. A O True B O False

> The mass of the anode electrode wil l increase as the galvanic cell approaches being "dead". 21. A O True B O False

t> The salt bridge acts as a second pathway for electrons to flow, thereby "completing" the electrical circuit. 22. A Q True B Q False

7 pt la your lab you are studying the kinetics of the degradation of a pain killer in the human liver. You are monitoring the concentration of the pain killer over a period of time. The initial concentration of the pain kilter in your experiment was 1.69 M . After 9.437 hours the concentration was found to be 0.8450 M . In another 18.87 hours the concentration was found to be 0.4225 M (t = 28.307 hours overall).

If another experiment were set up where the initial concentration of the pain killer was 0.281 M , how long would it take for the pain killer concentration to reach 0.0213 M ?

(in hours)


5

7 pt Consider the following data for the decomposition of nitrous oxide ( N 2 0 ) to N 2 and 0 2 gas at 315 K :

2 t/.o 0. E x p e r i m e n t [ N 2 0 ] ( M ) I n i t i a l R a t e ( M / m i n ) 1 0.337 0.00296 2 0.674 0.0119 3 2.359 0.145

Mechanism 1

Mechanism 2

Mechanism 3

Mechanism 4

> Mechanism 4 24. A O Plausible



t> Mechanism 1 27. A Q Plausible

N 2 0 ^ N 2 + O. y/" fast equihbrium N 2 0 + O ^ N 2 + 0 2

2 N 2 0 ^ N 4 0 2

2 N 2 + 0 2 N 4 0 2

2 N 2 O ^ N 4 0 2

1/2 N 4 0 2 N 2 + O a

slow

fast equilibrium slow

fast equilibrium slow

N 2 0 + N 2 0 2 N 2 + 0 + slow ^ 0 I tszo O + O ^ 0 2 fast equilibrium

B O Not valid

B O Not valid

B O Not valid

B Q Not valid

V 7 * ftJ

0

7 pt Consider the cell described below at 293 K :

• e l l Sn | Sn 2+ (0.989 M) || Pb 2 + (0.963 M) | P b -I . II3

Given E 0P B 2 + _ P B = -0.131 V , E ° S N 2 + _ ( S N = -0.143 V . Calculate the cell potential after the reaction has operated long

enough for the Sn 2+ to have changed by 0.381 mo l /L . (in V)

28. A Q 0.00101 B Q 0.00111 D Q 0.00129 E Q 0.00133

i t

0li


7 pt A n experiment you are performing in your lab requires you to generate fluorine gas through an electrolytic hydrolysis set up. The current experiment you wish to perform requires that you supply 7.57 moles of F 2 gas per hour from your set up. What is the minimum amperage current your electrolysis apparatus wi l l need?

2 H j O + 2 e - - » H 2 + 2 0 H - E° = -0.83 V 2+1,0 + Hi f 2off' I F 2 + 2 e - - » 2 F - E° = 2.87 V _ ^

(ink) h : (|j 2 9 . A Q 208 B Q 260. C Q 325 D Q 406 E Q 507

7 pt A concentration cell is built based on the following half reactions by using two pieces of zinc as electrodes, two Z n 2 + solutions, 0.143 M and 0.425 M , and all other materials needed for a galvanic cell. What wil l the potential of this cell be when the cathode concentration of Z n 2 + has changed by 0.045 M at 299 K ?

Zn 2 + + 2 e- Zn E° = -0.761 V

(in V)

30. A Q -0-752 B Q -0.00907 /€Q Q.QMOj/ D Q 0.0181 E Q 0.0202

7pt ' Consider the cell described below at 295 K :

Fe | Fe 2+(1.35 M) || Cd 2 +(2.05 M) | C d

< e > j ?

Given E ° C D 2 + + 2 e - _ C d = -0.403 V and E ° F E 2 + + 2 e - _ F e = -0.441 V . What wi l l the concentration of the C d 2 + solution be when the cell is dead? (ihH)

35. AO 0.145 BO 0.148 / p O 0.163^£>O 0.239 EO 1-89

7 pt Consider the fol lowingWfreactions at 298 K dd^^^d ^ " f f c ? ^

Fe 2+ + 2 e- - Fe E° = --0441 V ' ~ -™ ( F ' Cd 2 + -I- 2 e- — C d E° = -0.403 V £ J z+ -f- Z C * C c /

A galvanice cell based on these half reactions is set up under standard conditions where each solution is 1.00 L and each electrode weighs exactly 100.0 g. How much wil l the C d electrode weigh when the nonstandard potential of the cell is 0.02749 V ? ^ ^ . 017 W - ,03 2 - £ T / M

36. A Q 58.9 B Q 118 C Q 1 4 4 / D Q 174 E Q 190.

7 In your lab you are stbdymga certain biochemical bacterial metabolic pathway (shown below). You have initial data that suggests that if you could inhibit this pathway you would have the "next pemcillinj 1. but first you need to determine the activation energy for the rate determining step of this process. ^ — - " ,

enzyme + substrate —* product 1 + product 2

It is known that under the conditions you are performing your experiment the rate law for the reaction is: rate = k[enzyme] [substrate]2.

You set up an experiment at 311 K where [enzyme] = 0.00387 M and [substrate] = 1.51 M . A plot of ln[enzyrne] versus time (seconds) gives a straight line relationship with a slope of -0.007331. Previous studies have shown that a plot of ln(k) at a series of different temperatures gives a y-intercept of 0.48674. What is the activation energy for this reaction?

(in k J j

37. A O 5.28 B O 7.66 C O H I rl D O / l E O 23.3

too $~2-l5*-0*


9

7pt Consider the fictional reaction:

which is catalyzed by reagent C.

A + B - . F + G

I

The rate of this reaction was monitored by looking at the disappearance of B under several different experimental setups at 298 K .

E x p e r i m e n t [A ] (M) [B] (M) [C ] (M) R a t e of d i s appearance of B ( M / s )

1 0.300 0.0500 0.0500 4.75e-05

2 0.300 0.100 0.0500 4.75e-05

3 0.300 0.0500 0.100 9.51e-05

4 0.400 0.0500 0.200 0.000254

5 0.400 0.0500 0.0500 6.34e-05

Step 1: A + C - » D Step 2: D -+ E + C

Step 3: E + B ^ F + G

Given the previous information, which of the following situations would lead to a plausible reaction mechanism?

> Step 2 as the rate determining step. Step 1 is a fast equihbrium step while step 3 is a fast step. 38. A O Plausible Mechanism B O Not a valid mechanism

t> Step 2 as the rate determining step. A l l other steps fast steps. 39. A O Plausible Mechanism B O Not a valid mechanism

> Step 3 as the rate determining step. A l l other steps fast steps. 40. A O Plausible Mechanism B O Not a valid mechanism

> Step 3 as the rate determining step. Step 2 is a fast equihbrium step while step 1 is a fast step. 41. A Q Plausible Mechanism B O Not a valid mechanism

> Step 1 as the rate determining step. A l l other steps fast steps. 42. A Q Plausible Mechanism B Q Not a valid mechanism

7pt Consider a cell based on the following line notation and half reactions:

Cd|Cd2+||Fe2+|Fe U 1'

where E ° C D 2 + + 2 = -0.403 and E ° F E 2 + + 2 e - _ F e = -0.441.

Evaluate whether the following statements are true or false.

D> Electrons spontaneously flow from anode to cathode 43. A O True B O False

t> The half reaction with the higher standard reduction potential will be the anode portion of the cell 44. A O True B O False

t> Oxidation takes place at the anode 45. A Q True B Q False

7 pt In class we watched a demonstration where the color of the solution oscillated between orange and blue. This is a very comphcated reaction with multiple pathways the reaction can take. One pathway can be described as follows:

H 2 0 2 + 3 I- + 2 H+ -> I 3 - + 2 H 2 0

Step 1. Hs%+ tfy'-> H 3 0 2 + Step 2. H 3 0 2 + -t*fr - » ^ 0 + HOI Step 3. HOI O H - + I 2

Step 4. O H - + ft* -+/*yt> Step 5. Ia - » if

How many of the following statements are correct based on the information given?

> There are no catalysts in this reaction 46. A O Correct B O Incorrect

> If step 2 is the rate determining step, and step 1 is a fast equilibrium step, this reaction would be third order overall 47. A O Correct B O Incorrect

> The reaction is first order with respect to H 2 0 2

48. A O Correct B O Incorrect

> A n energy diagram of this reaction would have 5 peaks and the starting materials would be higher in energy than the products 49. A Q Correct B Q Incorrect

P r i n t e d from L O N - C A P A © M S U L i c e n s e d u n d e r G N U G e n e r a l P u b l i c L i c e n s e

Useful Information Hour Exam II

R = 8.314 J/mol K - 0.0821 L atm/mol K A G = A G 0 + RT in(Q)

A G = -nFE

AG° = -nFE°

k = AeA *

K = °C + 273

E°cell * E°cell (cathode) - E°cell (anode) F = 96485 C/mol e" RT

+ ln(A) In V _ l i s ri r

w = - P e x t ( V f - V i ) 1 L a t i n = 1 0 1 . 3 J

Order Integrated Rate Law

Differential Half-life Rate law

0

1

2

[A]t= -kt + [A]„ rate = k

ln[A]t = -kt + l n [ A ] 0 rate = k| A| 2k

In 2

1 = 1

14 +PI rate = k l A f

17.1 Standard Reduction Potentials at 26'C (298 K) for Many Cornwion Half-Reactions

Half-Reaction %° (V)

^ r t + c" - * Ag* Co3" + r - * Co 2 " H A 1 2H" + 2c' -0>*+ 4-. e" Ce !* pbOj + «H* + S O / H A ' + + J * " p- + 2H + 4- I0 4- -MnCV + * M * * 5c A H " * 3e~ —• Au ftOj + 411* * 2e~ -Oj * 2«" -•acr CrvO,* + I4H+ + fa" -+ 2Cr'* + 7HjO

• 2H,0

+ 2c" PbSO, + 2I%0 -* MrO* + 2HjO - l O f * H 2 0 -*> M r . " + 4HjO

* Pb** • 2H/J

2H20 - * M B * * + 2Hj.o * | i , + 3HjO

Oj + 4H* + 4c' - i MnO, + 4H* + 2e" 10,- + 6H + + 3»" Btj + 2e" -+ 2Br" VOj* + 2H 4 4 t -+ VO** + HjO ABQ4" + 3e" - * AM + 4CT 1*V +- 4H 4 + i»* ~» NO • 2HjO a a + e ciOj fflg5* + 2e" -~* Hgi2* Ag* + «" - * Ag Hfc5* +• 2e* - * 2Hg rV* f e~ - * Fe*+

0, + 2ir + 2e" ~* H A IfaO. + e" -+ Mn04

2~ lj + 2c > 21" C»* + e -> Cu

2.87 1.99 1.82 1.78 1.70 1.69 1.68 1.60 1.51 1.50 1.46 1.36 1.33 1.23 1.21 1.20 1.09 1.00 0.99 0.96 0.954 0.91 0.80 0.80 0.77 0.68 0.56 0.54 0.52

Half-Reaction

4QH Cu 2 * + 2e - » Cu HgjCfj 4- 2e - » 2Hg + 2CV AgCl + e"' -# Ag + i t S O / - + 4iV + 2c - 4 H3SO, + HjO Q f * + e - _» Cu* 2H' + 2e H; l V v t + 3e" - » Fe pt*2* + | r - * Pb Sn 1 ' + 2e" ->• Sn Ni** + 2e" Ni PbS04 + 2c" - » Pb + S O / " Cd2*' + 2c" «* Cd F e , + + 2e" -> Pe Q 3 * + e" —> C r " Cr 3 * +• 3e- - * Or Zn*' + 2c' / n 2H,0 + 2o" - » Hj + 20H" Kfa2* + 2e~ -* Mn A l 3 * + 3e" —* A! Hj + 2e" - * 2H +

Mfe2* + 2e" - * Mg La'* + 3e~ - * U Na + +• e" - » Na Ca 2~ + 2e' - » Ca B a J + + 2e" - * Ba K* + e K L i * + e~ - * U

0.40 0.34. 0.27 0.22 0.20 0.16 0.00

-0.036 -0.13 -0.14 -0.23 -0.35 -0.40 -0.44 -0.50 -0.73 -0.76 -0.83 -1.18 -1.66 -2.23 -2.37 -2.37 -2.71 -2.76 -2.90 -2.92 -3.05

Chem 104 Exam I Spring 2012

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Transcript of Chem 104 Exam I Spring 2012