Final

Version 056 – Final – Mccord – (52450) 1

This print-out should have 52 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

McCord CH302 1pm

This exam is only forMcCord’s MWF 1pm CH302 class.

R = 8.314 J/mol·K

R = 0.08206 L atm/mol·K

R = 62.36 L·torr/mol·K

1 L·atm = 101.325 J

∆G = ∆H − T∆S

∆Tf = kf · m ∆Tb = kb · m

PA = xA · PA,pure Π = cRT

ln

(

P2

P1

)

=∆Hvap

R

(

1

T1− 1

T2

)

ln

(

K2

K1

)

=∆Hrxn

R

(

1

T1− 1

T2

)

ln

(

k2

k1

)

=Ea

R

(

1

T1− 1

T2

)

pH = -log[H+] Kw =[H+][OH−]

Kw = 1.0 × 10−14 at 25◦C

pH = pKa + log[A−]

[HA]

∆G = −nFE◦ ∆G◦ = −RT lnK

anode | solution || solution | cathode

E = E◦ − RT

nFlnQ E = E◦ − 0.0257

nlnQ

E = E◦ − 0.05916

nlog Q

I · tn · F = moles

ln

(

[A]0[A]

)

= kt t1/2 =ln 2

ak

1

[A]− 1

[A]0= kt t1/2 =

1

ak[A]0

[A]0 − [A] = kt t1/2 =[A]02ak

001 10.0 pointsConsider five generic acids (HA, HB, HC, HD,and HE) that have the following ionizationconstants.

IonizationAcid Constant

Ka value

HA 3.6 × 10−3

HB 8.3 × 10−4

HC 2.6 × 10−6

HD 9.3 × 10−6

HE 7.3 × 10−7

Which of the following anions will be theWEAKEST base?

1. C−

2. B−

3. E−

4. A− correct

5. D−

Explanation:

The larger the Ka, the greater the dissoci-ation and the stronger the acid. In general,a conjugate base is opposite in strength fromits parent acid strength. HA is the strongest


acid listed (largest Ka), therefore A− is theweakest base.

002 10.0 pointsWe know that the rate expression for the re-action

2 NO + O2 → 2 NO2

at a certain temperature is Rate =k [NO]2 [O2]. We carry out two experimentsinvolving this reaction at the same tempera-ture, but in the second experiment the initialconcentration of NO is doubled while the ini-tial concentration of O2 is halved. The initialrate in the second experiment will be howmany times that of the first?

1. 2 correct

2. 4

3. 8

4. 1

Explanation:

[NO]2 = 2 [NO]1 ; [O2]2 =1

2[O2]1

Rate = k[NO]2[O2]

Rate2

Rate1=

k2 [NO]22 [O2]2k1 [NO]21 [O2]1

=(2 [NO]1)

2 12 [O2]1

[NO]21 [O2]1=

4

2= 2

003 10.0 pointsWhat is the concentration of OH− ions ina 0.40 M solution of KCN? The ionizationconstant of HCN is 4.0 × 10−10.

1. 4.08 × 10−12

2. 3.16 × 10−3 correct

3. 1.26 × 10−5

4. 6.25 × 10−5

5. 1.60 × 10−10

Explanation:

004 10.0 pointsConsider two liquids A and B. The vaporpressure of pure A (molecular weight = 50g/mol) is 225 torr at 25◦C and the vaporpressure of pure B (molecular weight = 75g/mol) is 90 torr at the same temperature.What is the total vapor pressure at 25◦C of asolution that is 70% A and 30% B by weight?

1. 225 torr

2. 124 torr

3. 108 torr

4. 135 torr

5. 195 torr correct

6. 203 torr

7. 76 torr

8. 115 torr

9. 335 torr

Explanation:

For A,P 0 = 255 torr MW = 50 g/mol

For B,P 0 = 90 torr MW = 75 g/mol

The mole fractions are7

9for A and

2

9for B.

(7

9

)

(225) +(2

9

)

(90) = 175 + 20 = 195 torr

005 10.0 pointsThe solubility product constant of barium sul-fite (BaSO3) is 8.0 × 10−7. Will a precipitateof BaSO3 form if 100 mL of a 4.0 × 10−4 mo-lar solution of BaCl2 is added to 100 mL of a3.0 × 10−3 molar solution of Na2SO3?

1. yes, because the solubility product con-stant is exceeded


2. yes, because the solubility product con-stant is not exceeded

3. no, because the solubility product con-stant is not exceeded correct

4. no, because the solubility product con-stant is exceeded

Explanation:

006 10.0 points1.56 moles of a weak electrolyte is dissolvedinto 814 grams of water. The freezing pointof the solution is −5.31128◦C. What is thepercent ionization for this substance in thissolution? Kf = 1.86◦C/m for water. Assumethat the electrolyte is a simple 1:1 ratio.

1. 23.02. 48.03. 41.04. 52.05. 35.06. 49.07. 32.08. 45.09. 39.010. 25.0

Correct answer: 49%.

Explanation:

Tf = −5.31128◦C T 0f = 0◦C

mH2O = 814 g n = 1.56 molKf = 1.86◦C/m

∆Tf = T 0f − Tf

= 0.0◦C − (−5.31128◦C) = 5.31128◦C

∆Tf = Kf · meff

meff =∆Tf

Kf=

5.31128◦C

1.86◦C/m= 2.85553 m

initial molality of solution =1.56mol MX

0.814 kg water

= 1.91646m

MX ⇀↽ M+ + X−

ini, m 1.91646 0 0∆, m −x x xfin, m 1.91646 − x x x

Total molality = 1.91646m − x + x + x

= 1.91646m + x

meff = 1.91646m + x = 2.85553 m

x = 0.939066 m

% ionization =x

mini× 100%

=0.939066 m × 100%

1.91646m= 49%

007 10.0 pointsThe energy (voltage) available in a galvaniccell depends upon the , whereas the cur-rent depends upon the

1. differences in the chemical potential en-ergy of the reactants; size of the cell. correct

2. Neither of these; voltage and current areinter-related and depend upon the details ofthe redox reaction as well as the cell design.

3. size of the cell; differences in the chemicalpotential energy of the reactants.

Explanation:

The chemical potential energy of the twohalf reactions (reduction and oxidation) setthe overall potential energy or voltage of thecell and are not affected by the physical sizeof the electrodes; however, the physical sizedoes affect how many electrons are flowing ina given time; i.e., the current.

008 10.0 points

Consider the reaction mechanism below:


Step Reaction

1 Cl2 + Pt −→ 2 Cl + Pt2 Cl + CO + Pt −→ ClCO + Pt3 Cl + ClCO −→ Cl2CO

overall Cl2 + CO −→ Cl2CO

Which species is/are intermediates?

1. Pt, ClCO

2. Cl

3. ClCO

4. Pt, Cl

5. Cl, ClCO correct

6. Pt

Explanation:

Both Cl and ClO are produced in earlysteps and stiochiometrically consumed in sub-sequent steps and neither appear in the overallreaction.

009 10.0 pointsThe graph describes the energy profile of a

reaction.

A

B

Time

Ener

gy

(kJ)

50

300

400

What are the values for ∆H and Ea, respec-tively, for the reaction in the direction writ-ten?

1. 250 kJ, 350 kJ correct

2. −250 kJ, 100 kJ

3. 250 kJ, 100 kJ

4. −250 kJ, −100 kJ

5. −250 kJ, 350 kJ

Explanation:

Ea∆H

A

B

Time

Ener

gy

(kJ)

50

300

400

∆H = 300 kJ − 50 kJ = 250 kJ

Ea = 400 kJ − 50 kJ = 350 kJ

010 10.0 pointsWhat is the molal concentration m of NaCl,a strong electrolyte in water, if the ob-served boiling temperature of the solution is100.361◦C? (Kb = 0.515◦C/m for water.)

1. 0.350 m correct

2. 0.175 m

3. 0.701 m

4. 0.186 m

5. 0.372 m

Explanation:

Tf = 100◦ C T 0f = 100.361◦ C

∆T = 100.361◦C − 100◦C = 0.361◦C

∆T = Kb m

mtotal =∆T

Kb=

0.361◦C

0.515◦C/m= 0.700971 m

mtotal = mNa + mCl− , so

mNaCl =1

2mtotal = 0.350485 m

011 10.0 pointsHow many seconds are required to produce4.94 mg of chromium metal from an acidic so-lution of potassium dichromate, using a cur-rent of 0.234 A?


1. 78 s

2. 274 s

3. 118 s

4. 196 s

5. 235 s correct

6. 157 s

Explanation:

I * t / n * F = molesmoles = 0.00494/52 = 9.5×10−5 mol Crn = 6 because the Cr is in the +6 ox state

in dichromate, Cr2O2−7 .

9.5 × 10−5(96485)(6)/0.234 = 235 s

012 10.0 pointsA buffer solution is prepared by dissolving0.50 mol HC2H3O2 and 1.00 mol NaC2H3O2

in enough water to make 1.00 L solution.What is the pH of the solution? (Ka =1.8 × 10−5.)

1. 7.00

2. 4.74

3. 5.04 correct

4. 4.44

5. 5.24

Explanation:

[C2H3O−

2 ] = 1.00 M [HC2H3O2] = 0.50 M

Ka = 1.8 × 10−5

pH = pKa + log

(

[CH3COO−]

[CH3COOH]

)

= − log(1.8 × 10−5) + log

(

1.0

0.5

)

= 5.04576

013 10.0 points

A reaction mixture that consisted of 0.2 molN2 and 0.2 mol H2 was introduced into a 26 Lreactor and heated. At equilibrium, 8% of thenitrogen gas had reacted. What is the value ofthe equilibrium constant Kc for the reaction

N2(g) + 3 H2(g) ⇀↽ 2 NH3(g)

at this temperature?1. 1128.752. 213.2393. 4429.764. 385.6585. 189.9816. 44.57267. 240.4448. 1567.879. 1344.7410. 1071.27

Correct answer: 1071.27.

Explanation:

Initially,

[N2] = [H2] =0.2 mol

26 L= 0.00769231.

Analyzing the reaction using concentra-tions,

N2(g) + 3 H2(g) ⇀↽ 2 NH3(g)ini 0.00769231 0.00769231 0∆ −x −3 x +2 xfin 0.00769231− x 0.00769231− 3 x +2 x

At equilibrium, 8% of the N2 had reacted,so 92% of the N2 remains:[N2] = (0.92) (0.00769231 mol/L) = 0.00707692 mol/

This can be set equal to the term for thefinal concentration of N2(g) in the table:

0.00769231− x = 0.00707692

x = 0.000615385 mol/L

At equilibrium,[N2] = 0.00769231 mol/L− 0.000615385 mol/L

= 0.00707692 mol/L

[H2] = 0.00769231 mol/L− 3 (0.000615385 mol/L)

= 0.00584615 mol/L


[NH3] = 2 × 0.000615385 mol/L

= 0.00123077 mol/LThe expression for the equilibrium constant

Kc is

Kc =[NH3]

2

[N2] [H2]3

=(0.00123077 mol/L)2

(0.00707692) (0.00584615 mol/L)3

= 1071.27

014 10.0 pointsYou place two beakers into an evacuatedchamber. One beaker has 100 mL of a 0.1M solution of NaCl in water and the other has100 mL of a 1.0 M sugar solution. When thesystem comes to equilibrium,

1. the beaker with the sugar solution willhave a larger volume. correct

2. the beaker with the salt solution will havea larger volume.

3. the two beakers will have identical vol-umes.

4. all of the water will be in the sugar solu-tion.

5. all of the water will be in the salt solu-tion.

Explanation:

Both liquids will evaporate and condensefrom each beaker. The chamber will cometo equilibrium when the two beakers havethe same concentration (same vapor pres-sure). The salt solution will become moreconcentrated (lower volume) and the sugar so-lution will become less concentrated (highervolume).

015 10.0 pointsConsider the following reaction and its rateconstant.

A → B k = 0.103 M −1· min−1

What will be the concentration of A after1 hour if the reaction started with a concen-tration of 0.400 M ?

1. 0.361 M

2. 0.236 M

3. 0.0843 M

4. 0.115 M correct

5. 0.384 M

6. 0.152 M

7. 8.28 × 10−4 M

8. 0.308 M

Explanation:

1

[A]t− 1

[A]0= a k t

1

[A]t=

1

[A]0+ a k t

=1

0.400+

(0.103 M−1 · min−1) (60 min)

= 8.68

[A]t = 0.115 M

016 10.0 pointsA solution prepared by adding 0.49 g of apolymer to 0.4 L of toluene (methylbenzene, acommon solvent) showed an osmotic pressureof 0.543 Torr at 19◦C. What is the molar massof the polymer?

1. 24169.72. 22284.33. 11390.24. 5626.975. 42825.66. 25592.37. 7356.68. 15782.59. 41083.210. 39850.7


Correct answer: 41083.2 g/mol.

Explanation:

m = 0.49 g V = 0.4 LT = 19◦C + 273 = 292 K P = 0.543 Torr

R = 0.08206L · atm

K · molWe assume the polymer to be a non elec-

trolyte, so i = 1, and

Π = i R T M1

M=

V MWx

m=

i R T

P

MWx =i RT m

P V

=1 (0.08206 L·atm

K·mol) (292 K) (0.49 g)

(0.543 Torr) (0.4 L)

× 760 Torr

1 atm= 41083.2 g/mol .

017 10.0 pointsThe lattice energy for MX is−505 kJ/mol andit’s heat of hydration is −345 kJ/mol. Whatis the heat of solution for MX?

1. +148 kJ/mol

2. −137 kJ/mol

3. +160 kJ/mol correct

4. −160 kJ/mol

5. −153 kJ/mol

6. +139 kJ/mol

7. −850 kJ/mol

Explanation:

∆Hsoln = ∆Hhyd − ∆Hcryst

∆Hsoln = −345 − (−505)

∆Hsoln = +160 kJ/mol

018 10.0 points

The decomposition of cyclobutane is a first-order reaction. At a certain temperature, thehalf-life for this reaction is 137 seconds. Whatfraction of a sample of cyclobutane would beleft after 685 seconds at this temperature?

1. 0.125

2. 0.0312 correct

3. 0.25

4. 0.0156

5. 0.0625

Explanation:

t1/2 = 137 s t = 685 sa = 1

k =ln 2

t1/2

ln[cyclobutane]0[cyclobutane]t

= a k t = a tln 2

t1/2

=(685 s) ln 2

137 s= 3.4657

[cyclobutane]0[cyclobutane]t

= e3.4657

= 32[cyclobutane]t[cyclobutane]0

=1

32= 0.03125

019 10.0 pointsA certain reaction has a Kc = 9.0 at 35◦C.

A(g) + B(g) ⇀↽ 2 C(g)

We have a mixture in which [A] = 2 M, [B] =4 M, and [C] = 1.5 M. Which of the followingis true of the mixture?

1. The mixture is at equilibrium.

2. More A and B will be formed to achieveequilibrium.

3. The reaction will go to the right. correct


4. The reaction will go to the left.

Explanation:

Kc = 9.0 [A] = 2 M[B] = 4 M [C] = 1.5 M

Q =[C]2

[A] [B]=

(1.5 M)2

(2 M) (4 M)= 0.28

< Kc = 9.0

Therefore, equilibrium will shift to theright.

020 10.0 pointsWrite the equilibrium expression for the reac-tion

2 NO(g) + O2(g) ⇀↽ 2 NO2(g)

1. K =P 2

NO PO2

P 2NO2

2. K =PNO

PNO2PO2

3. K =P 2

NO2

P 2NO PO2

correct

4. K =2 PNO2

2 PNO PO2

5. K =PNO2

P 2NO PO2

Explanation:

Since this is a gas phase reaction, K is theratio of the partial equilibrium pressures ofthe products to the partial equilibrium pres-sures of the reactants.

021 10.0 pointsConsider the half-reactionsMn2+ + 2 e− → Mn E0 = −1.029 VGa3+ + 3 e− → Ga E0 = −0.560 VFe2+ + 2 e− → Fe E0 = −0.409 VSn2+ + 2 e− → Sn E0 = −0.136 V

Of the species listed, the strongest oxidizingagent is

1. Sn2+ correct

2. Mn2+

3. Sn

4. Ga+3

5. Mn

Explanation:

Oxidizing agents get reduced. As E0 in-creases, the easier it is for the species to bereduced. Since Sn2+ has the biggest E0, it isreduced the easiest, making it the strongestoxidizing agent.

022 10.0 pointsWhat is the pH at the half-stoichiometricpoint for the titration of 0.88 M HNO2(aq)with 0.10 M KOH(aq)? For HNO2, Ka =4.3 × 10−4.

1. 1.86

2. 2.01

3. 7.00

4. 3.37 correct

5. 1.71

Explanation:

023 10.0 pointsA 100 ml sample of 0.100 M NH3 solution istitrated to the equivalence point with 50 mlof 0.200 M HCl. What is the final [H3O

+]?The ionization constant of NH3 is 1.8 × 10−5.

1. 6.09 × 10−6 M correct

2. 1.10 × 10−3 M

3. 1.00 × 10−7 M

4. 3.70 × 10−11 M

5. 8.61 × 10−6 M

Explanation:


[NH3] = 0.1 M [HCl] = 0.2 MInitially,

for NH3, (0.1 M)(100 mL) = 10 mmolfor HCl, (0.2 M)(50 mL) = 10 mmol

Neutralization:NH3 + HCl → NH4Cl

ini 10 mmol 10 mmol∆ −10 mmol −10 mmol 10 mmolfin 0 mmol 0 mmol 10 mmol

[NH+4 ] =

10 mmol

150 mL= 0.067 M

Equilibria re-established:

Ka

NH+4

⇀↽ NH3 + H+

0.067 M − −0.067 − x x x

Kw = 10−14 Kb = 1.8 × 10−5

Ka =Kw

Kb=

10−14

1.8 × 10−5= 5.56 × 10−10

Ka =x2

0.067 − x

Assumption:

x = [H+] =√

Ka × 0.067

=√

(5.56 × 10−10)(0.067)

= 6.09 × 10−6

024 10.0 pointsWhich of the following statements describe abuffered solution?

1. a solution containing a weak acid

2. a solution containing a conjugate base

3. a solution that resists pH change correct

4. a solution with pH = 7

Explanation:

A buffer contains comparative amounts ofa weak acid and its conjugate base or a weakbase and its conjugate acid, and is able toprevent large changes of pH. The pH of thebuffer is determined by its composition.

025 10.0 pointsThe Ksp of Cd3(PO4)2 at 18 oC is 1.08 ×10−33. What is its molar solubility at thistemperature?

1. 2.5 × 10−9 M

2. 6.5 × 10−11 M

3. 1.0 × 10−7 M correct

4. 3.3 × 10−17 M

Explanation:

The equation for Ksp = (2x)2(3x)3 =

108x5

x =

(

1.08 × 10−33

108

)1/5

= 1 × 10−7M

026 10.0 pointsThe reaction

N2O4(g) ⇀↽ 2 NO2(g)

is at equilibrium. Increasing the pressure willhave what effect on the equilibrium?

1. The equilibrium is shifted to the right.

2. The equilibrium remains unchanged.

3. The equilibrium is shifted to the left.correct

Explanation:

Increasing pressure shifts the equilibrium inthe direction that produces fewer molecules ofgas.

027 10.0 pointsWhich of the following alcohols would be theleast miscible with water?

1. pentanol (CH3CH2CH2CH2CH2OH)

2. propanol (CH3CH2CH2OH)

3. methanol (CH3OH)


4. hexanol (CH3CH2CH2CH2CH2CH2OH)correct

5. ethanol (CH3CH2OH)

Explanation:

The polar OH group is miscible with wa-ter but as the nonpolar hydrocarbon chainlengthens, solubility decreases.

028 10.0 pointsConsider an insulated system containing 150 gof liquid water and 150 g of ice at equilibriumunder atmospheric pressure. A total of 17 gof steam at 110◦C is admitted to the system.What is the phase composition and tempera-ture when equilibrium is reestablished?

1. 304 g of water and 13 g of ice at 0◦Ccorrect

2. 300 g of water and 17 g of ice at 0◦C

3. 317 g of water at 6.8◦C



6. 310 g of water and 7 g of ice at 0◦C

Explanation:

029 10.0 pointsKc = 58 at some temperature for the reaction

H2(g) + I2(g) ⇀↽ 2 HI(g) .

If 30.4 mol of HI are introduced into a 10.0liter vessel, how many moles of I2 are presentat equilibrium?

1. 1.58074 mol

2. 4.74221 mol

3. 6.32294 mol

4. 3.79377 mol

5. 3.16147 mol correct

6. 31.6147 mol

Explanation:

Kc = 58 Vvessel = 10.0 L

nHI = 30.4 [HI]ini =30.4 mol

10 L= 3.04 M

H2 (g) + I2 (g) ⇀↽ 2 HI (g)ini, M 0 0 3.04∆, M x x −2xeq, M x x 3.04 − 2x

Kc =[HI]2

[H2] [I2]= 58

(3.04 − 2x)2

x2= 58

3.04 − 2x

x=

√58

3.04 − 2 x =√

58x

x = 0.316147

nI2 = (10.0 L) [I2]

= (10.0 L)

(

0.316147mol

L

)

= 3.16147 mol

030 10.0 pointsWhat is the molar solubility of Zn(OH)2 in asolution buffered at pH of 8.0?(Zn(OH)2, Ksp = 3.0 × 10−17)

1. 5.0 × 10−6

2. 2.0 × 10−6

3. 4.0 × 10−9

4. 8.0 × 10−4

5. 3.0 × 10−5 correct

Explanation:

pH=8 means that pOH=6 and [OH−]=10−6

Ksp = [Zn2+][OH−]2 (solve for Zn2+)

[Zn2+] = 3.0 × 10−17/(10−6)2

[Zn2+] = 3.0 × 10−5 M


031 10.0 pointsA 0.20 M solution of Cu(NH3)4SO4 contains[Cu+2] = 9.2 × 10−4 M, [NH3] = 3.7 × 10−3

M, [Cu(NH3)2+4 ] = 0.20 M, and [SO2−

4 ] = 0.20M. What is the dissociation constant for theCu(NH3)

2+4 complex ion?

1. 3.4 × 10−14

2. None of the other answers is correct

3. 4.4 × 10−11

4. 1.7 × 10−13

5. 8.6 × 10−13 correct

Explanation:

032 10.0 pointsThe hydronium ion concentration of 0.05 Macetic acid is 9.4×10−4 mol/L. What is of thepH of the solution?

1. 11.45

2. 1.30

3. 3.00

4. 4.97

5. 1.12

6. 3.03 correct

7. 1.00

Explanation:

By definition,pH = −log [hydrogen ion molarity].

= −log(

9.4 × 10−4)

= 3.03

033 10.0 pointsA saturated solution of BiI3(s) is found tohave an I− concentration of 3.96 × 10−5 M.What is the value of Ksp for BiI3?

1. 6.6 × 10−17

2. 8.2 × 10−19 correct

3. 2.7 × 10−19

4. 9.1 × 10−20

5. 3.0 × 10−20

6. 4.2 × 10−25

7. 5.9 × 10−18

Explanation:

034 10.0 pointsFour of the following factors can affect theforward rate of a chemical reaction. Whichone cannot affect this rate?

1. concentration of reactants

2. particle size of solid reactants

3. temperature

4. removal of some of the products correct

5. presence of a catalyst

Explanation:

As long as reactants are present, the for-ward reaction can occur; the higher the re-actant concentration, the faster the reactioncan proceed. Other factors which increasereaction rates are increased temperature, in-crease in surface area of solid reactants andthe presence of a catalyst.

035 10.0 pointsCalculate the potential for the cell indicated:

Fe | Fe2+ (10−2 M) || Cu+ (10−2 M) | CuCu+ + e

− → Cu E0 = +0.210 V

Fe2+ + 2 e− → Fe E0 = −0.440 V

1. 0.650 V

2. 0.591 V correct

3. 0.709 V


4. 0.620 V

5. 0.000 V

Explanation:

The overall reaction is

Fe + 2Cu+ → Fe2+ + 2Cu

Please notice that since the concentrations arenot 1 M, the Nernst equation must be used.

In this cell notation, the anode is located onthe left of the salt bridge || and the cathodeon the right. So first calculate

E0cell = Ecathode − E0

anode

= +0.210 V − (−0.440) V = 0.65 V

Using the Nernst Equation

Ecell = E0cell −

0.05916

nlog Q

= 0.65 V − 0.05916

2log

(

[Fe2+]

[Cu+]2

)

= 0.65 V − 0.05916

2log

(

10−2

10−4

)

= 0.59084 V

036 10.0 pointsThe titration curve for the titration of 0.5

M Na2CO3(aq) with 0.5 M HClO4(aq) is givenbelow.

0

2

4

6

8

10

12

14

0 20 40 60 80 100 120 140

Volume of acid (mL)

pH

What are the main species in the solutionafter the addition of 35 mL of HClO4?

1. HCO−

3 , Na+, and ClO−

4 .

2. H2 CO3, Na+, and ClO−

4 .

3. CO2−3 , Na+, and ClO−

4 .

4. HCO−

3 , H2 CO3, Na+, and ClO−

4 .

5. CO2−3 , HCO3−, Na+, and ClO−

4 . cor-

rect

Explanation:

037 10.0 pointsA reaction

3 A2+ + 2 B → 3 A + 2 B3+

is used to make a battery. The standard Gibbsfree energy change for the given reaction is−655 kJ. What is the standard potential(E0) for the battery?

1. 1.5322. 1.473. 1.1074. 1.3135. 1.1316. 1.3637. 1.6628. 1.0549. 1.69610. 1.401

Correct answer: 1.131 volts.

Explanation:

038 10.0 pointsA solution with a higher pH is more acidic.

1. False correct

2. True

3. The answer cannot be determined with-out additional information.

Explanation:


pH = − log [H3O+], so as [H3O] increases,

pH decreases.

039 10.0 pointsThe phase diagram for a pure substance is

given below.

Solid

Liquid

Vapor50

100

150

200

250

300

100 200 300 400Temperature, K

Pre

ssure

,atm

The substance is stored in a container at 150atm at 25◦C. Describe what happens if thecontainer is opened at 25◦C.

1. The liquid in the container freezes.

2. The vapor in the container escapes.

3. The solid in the container melts.

4. The solid in the container sublimes.

5. The liquid in the container vaporizes.correct

Explanation:

040 10.0 pointsAn indicator changes from yellow to blue go-ing from its acidic (nonionized) form to itsbasic (ionized) form. The indicator has a Ka

of 2.5×10−8. At pH 8.1, this indicator’s colorwould best be discribed as which color?

1. blue

2. green

3. yellow

4. yellow-green

5. blue-green correct

Explanation:

The pKa is 7.60 so that is where you wouldsee green (50/50 blend of yellow and blue).You would see all yellow for pH 6.60 andlower, and all blue for pH 8.60 and above.The pH 8.1 given is on the basic side of themiddle so it would have a blue-green color.

041 10.0 pointsWhat is the conjugate base of ammonia?

1. NH+4

2. NH3

3. NH2OH

4. OH−

5. NH−

2 correct

Explanation:

042 10.0 pointsThe vapor pressure of water at 37◦C is 47.1torr and its enthalpy of vaporization is 44.0kJ·mol−1. Estimate the vapor pressure of wa-ter at 87◦C. Assume the enthalpy of vaporiza-tion of water is independent of temperature.

1. 112 torr

2. 713 torr

3. 503 torr correct

4. 256 torr

5. 52 torr

Explanation:

T1 = 37◦C + 273.15 = 310.15 KT2 = 87◦C + 273.15 = 360.15 K∆H◦

vap = 44.0 kJ · mol−1 P1 = 47.1 torrUsing the Clausius-Clapeyron equation,


ln

(

P2

P1

)

=∆H◦

vap

R

(

1

T1− 1

T2

)

=44 kJ · mol−1

8.314 J/mol−1 · K−1· 1000 J

1 kJ

×(

1

310.15 K− 1

360.15 K

)

= 2.36896P2

P1= e2.36896

P2 = P1 e2.36896

= (47.1 torr) e2.36896

= 503.323 torr

043 10.0 pointsWhat is the molarity of 555 L of a Ba(OH)2solution if the pH is 10.20?

1. 7.92 × 10−5 M correct

2. 6.31 × 10−11 M

3. 2.26 × 10−5 M

4. 1.58 × 10−4 M

5. 3.15 × 10−11 M

6. 4.40 × 10−2 M

7. 3.14 × 10−4 M

8. 5.15 × 10−7 M

Explanation:

VBa(OH)2 = 555 L pH = 10.2First we need to calculate the concentration

of OH−:

pOH = 14 − pH

= 14 − 10.2 = 3.8

pOH = −log[OH−]

[OH−] = 0.000158489 M OH−

The Ba(OH)2 delivers 2 hydroxides for ev-ery 1 Ba(OH)2 so the concentration of theBa(OH)2 is 1/2 the overall hydroxide concen-tration.

[Ba(OH)2] = 0.000158489/2

= 7.92447 × 10−5M

Note that the volume of 555 L is just someextra information not needed in the calcula-tion.

044 10.0 pointsOxidation occurs

1. at both anode and cathode.

2. in the electrolyte.

3. at the cathode.

4. at either, depending on whether the cellis electrochemical or electrolytic.

5. at the anode. correct

Explanation:

045 10.0 pointsConsider the reaction

2 Fe2O3(s) + 3 C(s) → 4 Fe(s) + 3 CO2(g) ,

∆H◦ = 462 kJ, ∆S◦ = 558 J · K−1 . Calcu-late the equilibrium constant for this reactionat 525◦C.

1. 2.18 × 10−2

2. 3.04 × 10−3

3. 1.9 × 106

4. 5.20 × 10−7

5. 8.07 × 10−2 correct

Explanation:

046 10.0 pointsThe rate of formation of oxygen in the follow-ing reaction is 2.28 M/s.

2 N2O5(g) → 4 NO2(g) + O2(g)


What is the rate of formation of NO2?

1. 1.14 M/s

2. 4.56 M/s

3. 0.57 M/s

4. 9.12 M/s correct

5. 2.28 M/s

Explanation:

The NO2 forms at 4× the rate of the O2.

047 10.0 points

If one added 130 mL of 2 M NaOH to 1 Lof a buffer composed of 2.3 M NH3 and 3.2 MNH4Cl, what would be the pH of the resultingsolution? The Kb of NH3 is 1.8 × 10−5.

1. 2.2

2. 9.2 correct

3. 11.8

4. 10.0

5. 4.8

Explanation:

The general reaction that takes place is:BH+ + OH− ↔ B + H2OInitial amounts of each species are 3.2, 0.26and 2.3 moles respectively. After the reactiongoes to completion, the equilibrium amountsare 2.94, 0 and 2.56 moles respectively. Thereis no need to calculate the final concentra-tions. For a buffer composed of a weak baseand its conjugate acid,[OH−] = Kb(Cb/Ca)

=(

1.8 × 10−5)

(2.56 moles/2.94 moles)

= 1.57 × 10−5

pOH = 4.8

pH = 9.2

048 10.0 pointsSuppose we put 1.0 mol of HI(g), 1.0 mol

of H2(g), and 1 mol of I2(g) in a 2.0 literreaction vessel and the following equilibriumis established:

2 HI(g) ⇀↽ H2(g) + I2(g)

If Kc = 10 for this reaction at the tempera-ture of the equilibrium mixture, compute theequilibrium concentration of HI.

1. 0.240 M

2. 0.071 M

3. 0.295 M

4. 0.260 M

5. 0.145 M

6. 0.205 M correct

7. 0.429 M

8. 0.102 M

Explanation:

[HI]ini =1 mol

2 L= 0.5 M

[H2]ini =1 mol

2 L= 0.5 M

[I2]ini =1 mol

2 L= 0.5 M

Q =[H2] [I2]

[HI]2

=(0.5) (0.5)

(0.5)2

= 1 < Kc = 10

Therefore equilibrium moves to the right.2 HI ⇀↽ H2 (g) + I2 (g)

ini, M 0.5 0.5 0.5∆, M −2x x xeq, M 0.5 − 2x 0.5 + x 0.5 + x

(0.50 + x) (0.50 + x)

(0.50 − 2x)2= 10

0.5 + x

0.50 − 2x=

√10

0.5 + x = 0.50√

10 − 2√

10 x

x = 0.148


[HI] = 0.5 − 2 x = 0.205 M

049 10.0 pointsWhat is the activation energy for a reactionif its rate constant is found to triple when thetemperature is raised from 300 K to 310 K?

1. 84,900 J/mol correct

2. 20,300 J/mol

3. 195,600 J/mol

4. 418,400 J/mol

5. No other choice is within 3 percent.

Explanation:T1 = 300 K

k2 = 3 k1

T2 = 310 K

ln

(

k2

k1

)

=Ea

R

(

1

T1− 1

T2

)

Ea =

R ln

(

k2

k1

)

1

T1+

1

T2

The rate constant tripled, so k2 = 3 k1:

Ea =(8.314 J/mol · K) ln 3

1

300 K− 1

310 K= 84944 J/mol

050 10.0 pointsChoose the effective pH range of a pyridine-pyridinium chloride buffer. For pyridine, thevalue of Kb is 1.8 × 10−9.

1. 10.3 to 12.3

2. 9.1 to 11.1

3. 1.4 to 3.4

4. 4.3 to 6.3 correct

5. 7.7 to 9.7

Explanation:

051 10.0 pointsWhat is the E◦

cell of

Zn(s) | Zn2+(aq) || Ce4+(aq) | Ce3+(aq)

Zn2+ + 2 e− → Zn E◦

red = −0.76

Ce4+ + e− → Ce3+ E◦

red = +1.61

1. +2.37 correct

2. +0.85

3. −2.37

4. −0.85

5. +1.61

Explanation:

052 10.0 pointsThe following figure represents the progress ofa given reaction at 298 K.

b

b

b

b

b

G

AB

C D

E

rxn progress

At point B on this figure, what is the relation-ship of Q to K?

1. Q < K correct

2. Cannot be determined

3. Q = K

4. Q > K

Explanation:

Point B is on the reactants-heavy side ofequilibrium, so Q is less than K. Note also


that dG (slope) is negative here which meansthe reaction would be spontaneous in the for-ward direction. Spontaneous in a forwarddirection corresponds to Q being less than K.

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