Problems and Solutions to the ThirdPhase Malaysian IChO Selection Camp

Prepared by

Yau Ching Koon

Version March 23, 2015

Copyright note: the author does not own the copyright to the problems, whichshould belong to the following problem authors. However, all typesetting, minorcorrections and artworks accompanying the problems and all of the solutions dobelong to the author.

Problem authors:

Prof. Noorsaadah Abdul Rahman (carbonyl compounds) Prof. Wan Jeffrey Basirun (electrochemistry) Assoc. Prof. Lo Kong Mun (coordination compounds) Dr. Azizah Mainal (thermodynamics) Dr. Desmond Ang (reaction kinetics) Dr. Noordini Mohamad Salleh (alcohols)

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PROBLEM 1. Nitrogen pentoxide, N2O5 is an unstable and potentially dangerousoxidizer. Upon decomposition, it produces nitrogen dioxide and oxygen:

2 N2O5 4 NO2 + O2

Following is the proposed mechanism for the above reaction:

N2O5kf

kbNO2 + NO3

NO3 + NO2k2

NO + NO2 + O2

NO3 + NOk3

2 NO2

Apply steady state approximation to determine the rate law. [15 marks]

PROBLEM 2. Dilatometry technique can be employed to study chemical kinetics ofreaction that shows change in effective volume. Example is the reaction betweenethylene oxide and perchloric acid. Followings are the data obtained from anexperiment conducted at constant temperature:

Reaction time (min) Dilatometer reading

0 18.4830 18.0560 17.62

135 16.71300 15.22 12.29

Assuming it is a first order reaction, determine the average rate constant.[10 marks]

PROBLEM 3. The rate law describing the decomposition of hydrogen iodide todihydrogen and diiodine is

rate = k [HI]2

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If 0.01 mol of HI(g) is placed in a container with capacity of 1.0 L at 25 C, deter-mine the time needed for the concentration of HI(g) in the container to decreaseby 10%. Given the rate constant of the decomposition is 2.4 1031 L mol1 s1.

[5 marks]

PROBLEM 4. Which of the following entities are state functions, and which arenot?

1. work

2. heat

3. temperature

4. enthalpy

5. internal energy

6. volume

[6 marks]

PROBLEM 5. An electric motor produces a mechanical work of 30.0 kJ energy persecond. The motor also loses 4.5 kJ per second of heat to the surrounding. Whatis the change in internal energy, U, of this motor? [3 marks]

PROBLEM 6. We are often concerned with the work done on or by a system asit contracts or expands. The work done by this system depends on the volumechange and the external pressure.

1. What does the term free expansion mean?

2. What is the amount of work done in a closed system?

[2 marks]

PROBLEM 7. Sketch a pV plot for an isochoric process. Write the appropriateexpression for the internal energy, U based on this process. [4 marks]

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PROBLEM 8. Explain why the heat capacity for a gas at constant pressure, Cp islarger than the heat capacity at constant volume, CV . [4 marks]

PROBLEM 9. Given the thermochemical equation below, estimate the change inthe internal energy, fUm for the formation of ammonia, NH3 at 25 C.1

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H2(g) +12

N2(g) NH3(g) fHm = 46.1 kJ mol1

(R = 8.314 J mol1 K1) [5 marks]

PROBLEM 10. Name two situations in which the change in internal energy, U isequal to the change in enthalpy, H, i.e. U = H. [4 marks]

PROBLEM 11. Given the following heat of reaction:

H2(g) + Cl2(g) 2 HCl(g) H = 184.6 kJ

Calculate the standard enthalpy of formation, fH for HCl.2 [3 marks]

PROBLEM 12. What does it mean when the value of Gibbs free energy change, Gfor a chemical reaction is

1. Positive?

2. Negative?

3. Zero?

[3 marks]

PROBLEM 13. The latent heat of vaporization of water is 40.656 kJ mol1, and itsmelting point is 273.15 K at 1 atm pressure.

1The original question does not specify the amount of ammonia formed, hence any value forfU is technically correct.

2We shall assume that the given heat of reaction was determined at standard conditions. Oth-erwise any values of fH are correct.

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1. Calculate the molar entropy change of vaporisation of water. (Note: the unitfor molar entropy change, Sm is J K1 mol1.)

2. Would the change in entropy of the melting of ice be larger or smaller thanthat of the entropy of vaporization of water? Explain.

[6 marks]

PROBLEM 14. Provide a reasonable mechanism for the following reaction.

OH

H2SO4,

PROBLEM 15. Predict the products from each of the following reactions.

1. (1) BH3THF(2) NaOH, H2O2

2.

OH

conc. H2SO4

3.

OH

(1) H2SO4(2) BH3THF(3) NaOH, H2O2

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PROBLEM 16. Predict the organic product for each of the following reactions.

1.

O

(1) CH3MgBr, Et2O

(2) H3O+

2.

O

(1) CH3CH2Li, Et2O

(2) H3O+

3.

O

O(1) CH3MgBr (excess), Et2O

(2) H3O+

PROBLEM 17. Provide three methods that employ Grignard reagents to synthesizethe following compound.

OH

PROBLEM 18. Provide the reagents needed to achieve each of the following trans-formations.

1.OH Br

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2.

OH

O

O

Cl

O

O

3.

OHBr

ClBr

4.

HO

OH

Br

Br

5.

H

OH

H

H

H

PROBLEM 19. The reaction of CoCl2 with NH3 in the molar ratio of 1 : 4, 1 : 5and 1 : 6 yielded four types of cobalt complexes with different colours: yellow(max = 435 nm), purple (max = 495 nm), green (max = 700 nm) and violet(max = 810 nm), respectively.

1. Draw the molecular structures of the four cobalt complexes.

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2. Propose an experiment to differentiate these four types of cobalt complexes.

3. By using the ligand field splitting parameter, determine whether Cl or NH3is a strong field ligand.

PROBLEM 20. Draw the orbital energy level diagram for a d9 metal complex. Cal-culate its ligand field stabilization energy in unit of o. Draw the orbital energylevel diagram to show the effect of tetragonal distortion to the complex. Drawalso the orbital energy level diagram if the tetragonal distortion goes as far asthe total loss of the ligands along the z-axis and the formation of a square planarcomplex.

PROBLEM 21. Persulfate is a very strong oxidizing agent. If HCl is added intoa solution containing ammonium persulfate, explain (by calculations), whetherchlorine gas will be liberated or otherwise.

S2O28 + 2 e 2 SO24 E

= 2.01 V vs. SHE

Cl2 + 2 e 2 Cl E = 1.36 V vs. SHE

PROBLEM 22. Permanganate ion, MnO4 is also a very strong oxidizing agent,but unlike persulfate, it needs acidic conditions to realize its full capacity as anoxidizing agent.

Calculate at which pH, the permanganate ion will start oxidizing the chlorideions in HCl into chlorine gas.

Assume, the[MnO4

]and

[Mn2+

]are 1 mol dm3 and Cl2 pressure is 1 atm.

MnO4 + 8 H+ + 5 e Mn2+ + 4 H2O E = 1.50 V vs. SHE

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Solutions to Problems

Problem 1. The rate of this reaction can be measured by

rate = 12

d [N2O5]dt

=14

d [NO2]dt

=d [O2]

dt

Consider three possible rate determining steps.

If kf is the rate determining step, then the rate of reaction is simply

rate = kf [N2O5]

A plausible scenario of this case is illustrated by the graph below.

blue: N2O5, green: NO2, black: O2, yellow: NO3, red: NO

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If k3 is rate determining, then the faster k2 would have decomposed all of thenitrogen trioxide and is inconsistent with the stoichiometry observed. Hence, k3cannot be the rate determining step. A plausible scenario of this case is illustratedby the graph below and take note of the rapid diminishing of the yellow line.

blue: N2O5, green: NO2, black: O2, yellow: NO3, red: NO

If k2 is the rate determining step, we have

rate =d [O2]

dt= k2 [NO3] [NO2]

Also, after some finite time, the kf and kb would have achieve equilibrium. We

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may therefore write

kf [N2O5] = kb [NO2] [NO3]

K kfkr=[NO2] [NO3][N2O5]

Hencerate = Kk2 [N2O5] = kobs [N2O5]

A plausible scenario of this case is illustrated by the graph below and take note ofdifferent curvatures of the blue vs. green or black line.

blue: N2O5, green: NO2, black: O2, yellow: NO3, red: NO

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Problem 2. For first order reactions, dilatometry utilizes the equation

lnVt VV0 V = kt+ const.

Calculating the LHS term gives the following table.

t/min ln [(Vt V) / (V0 V)]30 0.072060 0.1496135 0.3368300 0.7479

The gradient, using the regression mode on a calculator, gives the average rateconstant as

k = 2.499 103 min1 = 4.165 105 s1

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Problem 3. A plausible elementary step of the reaction consistent with the ratelaw is given by

2 HI H2 + I2

Therefore,

rate = 12

d [HI]dt

= k [HI]2

where the integrated form reads

[HI]t[HI]0

d [HI]

[HI]2= 2k

t0

dt

1[HI]t

+1

[HI]0= 2kt

Given that [HI]t = 0.9 [HI]0 and [HI]0 = 0.01/1 = 0.01 mol L1, we have

t =1

2 2.4 1031(

10.9 0.01

10.01

)= 2.31 1031 s

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Problem 4.

1. w is not a state function

2. q is not a state function

3. T is a state function

4. H is a state function

5. U is a state function

6. V is a state function

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Problem 5. The first law of thermodynamics gives

dUdt

=dqdt

+dwdt

= 30 4.5 = 34.5 kJ s1

The negative sign for both the work and heat indicates that the work and energyproduced are at the expense of the internal energy of the motor.

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Problem 6.

1. Free expansion is the expansion of a matter into vacuum.

2. A closed system forbids the exchange of particles but allows the exchangeof energy. The exchange of energy obeys the first law of thermodynamics,

w = U qIt should worth noting that this work w includes p-V work, E-q (electro-chemistry) work, -A (surface) work, etc.

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Problem 7. The work done by a gas for an isochoric process is w = 0, the heat

V

p

V

p

or

absorbed or released is q = nCV,mT while the change in internal energy is

U = q+ w = nCV,mT

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Problem 8. Consider two separate gases, one heated at constant pressure andone heated at constant volume. The gas heated at constant pressure stores partof the heat absorbed into its internal energy and converts another part of the heatinto expansion-work. For the same amount of heat supplied, the gas heated atconstant volume does not need to (or rather by definition could not) expand,and hence all the heat absorbed is stored as internal energy. Hence for the sameamount of heat supplied, the temperature rises more for the case where a gas isheated at constant volume compared to heating at constant pressure. Thus, theheat capacity of a gas at constant volume is smaller than that at constant pres-sure.

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Problem 9. The enthalpy is

H = U + (pV) = U + pV +Vp

The definition for fHm implies p = 0, while V for the formation is

V =RTp[n (NH3) n (N2) n (H2)]

assuming that H2, N2 and NH3 all behaves ideally. Since for every mole of NH3formed, half mole of N2 and three halves of H2 is consumed, we have

n (N2) =12n (NH3) , n (H2) =

32n (NH3)

or

V = RTn (NH3)p

, Vm = RTpHence,

fHm = fUm RT,fUm = 46.1 103 + (8.314) (25+ 273.15) = 43.6 kJ mol1

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Problem 10. Since dU = S dT p dV and dH = S dT + V dP, the conditionU = H implies that

p dV = V dpor

d (pV) = 0

Either of these two equations can be satisfied in several ways. (1) An expansionat which p = 0 or an expansion into vacuum. (2) A change in pressure at V = 0.(3) Any process where d (pV) = 0 such as isothermal compression or expansionfor an ideal gas since d (pV) = nR dT = 0 implies dT = 0.

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Problem 11. By definition, the standard enthalpy formation refers to the forma-tion of one mole of a particular substance, here being HCl. Thus, for

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H2(g) +12

Cl2(g) HCl(g)

the standard enthalpy of formation is

fHm = 184.6

2= 92.3 kJ mol1

Due to the definition, the standard enthalpy of formation, fH and standardmolar enthalpy of formation, fHm are equivalent.

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Problem 12.

1. The forward reaction is thermodynamically non-spontaneous.

2. The forward reaction is thermodynamically spontaneous.

3. The system is in equilibrium.

Note that Gibbs energy could, in general, describes a system and not limited to areaction only.

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Problem 13.

1. 40.656 kJ of energy is required to convert one mole of water to its vapour.Hence the molar entropy of vaporisation of water is

Sm =qT=

40.656373.15

= 108.95 J mol1 K1

2. The entropy of fusion of ice is expected to be smaller than the entropy ofvaporisation of water. During fusion, the heat are absorbed to break thehydrogen bond network partially, giving some translational motion to thewater molecules. However, during vaporisation, the heat are absorbed tobreak all the hydrogen bonds between the molecules and hence is larger inmagnitude compared to the heat absorbed during fusion.

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Problem 14.

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Problem 15.

1. (1) BH3THF(2) NaOH, H2O2

OH

2.

OH

conc. H2SO4 O

3.

OH

(1) H2SO4(2) BH3THF(3) NaOH, H2O2

OH

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Problem 16.

1.

O

(1) CH3MgBr, Et2O

(2) H3O+

OH

2.

O

(1) CH3CH2Li, Et2O

(2) H3O+

OH

3.

O

O(1) CH3MgBr (excess), Et2O

(2) H3O+

OH

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Problem 17.

O

(1) MeMgBr, THF

(2) H3O+

OH

O

(1) nPrMgBr, THF

(2) H3O+

OH

O (1) PhMgBr, THF

(2) H3O+

OH

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Problem 18.

1.OH

PBr3Br

2.

OH

O

O

SOCl2

Cl

O

O

3.

OHBr

PCl5Cl

Br

4.

HO

OH

2 eqv. PBr3

Br

Br

5.

H

OH

Hconc. H2SO4

H

H

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For part (5.), the combination (1) TsCl, pyridine (2) DBU can also be used (answerdue to Chan Jer Yong). DBU = 1,8-diazabicyclo[5.4.0]undec-7-ene.

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Problem 19.

1. The structure of the four complexes of cobalt are shown below, (I)(IV). (I)and (II) are 1 : 4 adduct, (III) is a 1 : 5 adduct and (IV) is a 1 : 6 adduct.

CoNH3NH3

NH3NH3

Cl

ClI

CoNH3NH3

NH3Cl

Cl

NH3II Co

NH3NH3

NH3NH3

Cl

NH3

+

Cl

III

CoNH3NH3

NH3NH3

NH3

NH3

2+

2 Cl

IV

2. The solution may be best described in James E. Huheey, Ellen A. Keiter andRichard L. Keiter (1993), Inorganic Chemistry: Principles of structure and reac-tivity 4th edn., HarperCollins College Publisher, New York, USA, pp. 387389. These pages are reproduced here.

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3. Nothing can be inferred because all four compounds belong to four separatesymmetry point group, viz. D4h for (I), C2v for (II), C4h for (III) and Oh for(IV). The separation between the highest fully-filled d-orbital and lowestempty or partially-filled d-orbital, is dependent on the point group.

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Problem 20. The orbital energy level diagram for a d9 complex is shown below.Also shown are the energy levels due to tetragonal distortion and that of squareplanar.

e

t2g

g

D o

tetragonal distortion

xz, yz

xyz

x y

square planar

The ligand field stabilization energy is

E = 6(2

5o

)+ 3

(35o

)= 3

5o

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Problem 21. The e.m.f. for the reaction

S2O28 + 2 Cl 2 SO24 + Cl2

isE = 2.01 1.36 = 0.65 V

Since E > 0 V, the forward reaction is spontaneous at 298 K and 105 Pa. Chlorideions will be oxidized by persulfate ions to form chlorine gas.

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Problem 22. The e.m.f. for the reaction

2 MnO4 + 16 H+ + 10 Cl 2 Mn2+ + 8 H2O + 5 Cl2

isE = 1.50 1.36 = 0.14 V

Under non-standard conditions, the e.m.f. of the reaction is given by Nernst equa-tion

E = E RTnF

lna(Mn2+

)2 f (Cl2)5a(MnO4

)2 a (H+)16 a (Cl)10For the forward reaction to proceed spontaneously, E > 0 or (although not given,we shall assume that a (Cl) = 1 for simplicity)

RTnF

ln1

a (H+)16< E

16 ln {a (H+)} < 0.14 10 964858.3145 298.15 = 54.49

pH lg {a (H+)} < 54.49 lg e16

= 1.48

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Solutions to Problems