answer chm520-apr2011.pdf

CONFIDENTIAL AS/APR 2011/CHM520

Hak Cipta Universiti Teknologi MARA CONFIDENTIAL

2

Avogradros number (NA) = 6.02214 1023 mol-1 Gas constant (R) = 8.314 J K-1 mol-1 or 0.0821 dm3 atm K-1 mol-1 or 0.0831 dm3 bar K-1 mol-1 1 Pa = 1 N m-2, 1 bar = 105 Pa, 1 atm = 101.325 kPa RT/F at 25 oC = 0.02569 V mol ln a = 2.303 log a Faraday constant (F) = 96485 C mol-1 QUESTION 1 a) For one mole of an ideal gas, external pressure (Pexternal) = pressure (P) = 200 103 Pa.

The temperature is changed from 100 C to 25.0 C at constant pressure. Molar heat capacity at constant volume (CV,m) = 3/2 R. Calculate (in unit J) i) change of internal energy (DU), ii) change of enthalpy (DH), iii) heat (q) and iv) work done (w).

(12 marks)

b) A pellet of Zn of mass 10.0 g is dropped into a flask containing dilute H2SO4 at a pressure (P) = 1.00 bar and temperature (T) = 298 K. i) Write the balanced chemical equation for the reaction that occurs. ii) Calculate the work done (w) for the process (in unit J).

(8 marks)

QUESTION 2 Calculate the change of entropy of system (DSsys), the change of entropy of surroundings (DSsurr) and the total change of entropy (DStot), when the volume of 85.0 g of CO initially at 298 K and 1.00 bar increases by a factor of three in

a) an adiabatic reversible expansion, b) an expansion against external pressure (Pexternal) = 0, and c) an isothermal reversible expansion.

State whether each process (a to c) is spontaneous.

Given that the molar heat capacity at constant pressure (Cp,m) to be constant at the value 29.14 J mol1 K1 and assume ideal gas behavior.

(20 marks)



3

QUESTION 3 a) Express mean activity coefficient ( g ) in terms of activity coefficient for cation (g + ) and

activity coefficient for anion (g - ) for i) SrSO4, ii) MgBr2, iii) K3PO4, and iv) Ca(NO3)2.

Assume complete dissociation.

(8 marks)

b) Using the DebyeHckel limiting law, calculate the value of mean activity coefficient ( g ) in 5.0 103 molal solutions of

i) KCl, ii) Ca(NO3)2, and iii) ZnSO4.

Assume complete dissociation. Given that: A = 0.509 for aqueous solution at 25 oC.

(12 marks)

QUESTION 4 a) For the half cell reaction Hg2Cl2 (s) + 2 e- 2 Hg (l) + 2 Cl- (aq), standard cell potential

(E) = +0.26808 V. Determine the standard molar Gibbs energy of formation (Gf) (Cl-, aq). Given that: (Gf) (Hg2Cl2, s) = 210.7 kJ mol-1

(5 marks)

b) Reaction of 4 NO3- + 4 H+ 4 NO + 2 H2O + 3 O2 is referred.

i) Show the electrode half-reactions and standard cell potential (E). ii) Calculate the equilibrium constant (K) at 298.15 K for the reaction.

(7 marks)

c) The cell potential (E) = +0.260 V for the cell

Pt (s) H2 (g,

2Ha = 1) H+ (aq, +Ha = 1 NaCl (aq, m = 0.300) AgCl (s) Ag (s)

i) Show the electrode half-reaction, the overall reaction and standard cell potential (E),

reaction quotient (Q) and the Nernst Equation of each cell. ii) Determine -Clg assuming that -+ == ClNa ggg .



4

Quantities a, g and m denote activity, activity coefficients and concentration. (8 marks)

QUESTION 5 a) Radium-224 decays by -particle emission.

i) Write the balanced nuclear equation. ii) Calculate the mass defect (m) (in unit kg). iii) Determine the energy (in unit J) associated with this decay. Given that:

1 amu = 1.6606 10-27 kg 1 J = 1 kg m2 s-2 Speed of light (c) = 2.9979 108 m s-1

Table 1 Masses of fundamental particles of matter.

Particle Mass (amu) Electron 0.00054858 proton 1.007825

neutron 1.008665

Table 2 Nuclide and its nuclear mass Particle Nuclear mass (amu)

224Ra 223.9719 220Rn 219.9642 4He 4.00150

(8 marks) b) Copper consists of several radioactive isotopes, some of which are listed in Table 3.

i) Which of these isotopes are stable and which isotope is less stable? Give your explanation.

ii) Justify the observation that Cu-64 can undergo emission, positron emission, or electron capture by referring to Figure 1. Write the balanced nuclear equation for each case.

Table 3 Isotopes of copper

Isotope Cu-63 Cu-64 Cu-65



5

Figure 1 Neutron-proton ratio and the belt of stability of nuclides (dots in a region).

(12 marks)

END OF QUESTION PAPER

of 9

Faculty of Applied Science, UiTM Final Exam Apr 2011

Course : Physical Chemistry II, CHM 520 First Examiner : Dr. Chan Chin Han Second Examiner : Mr Borhanuddin Answer : 5 questions only

of 9

QUESTION 1 Solution: Mark [Sub

total] a) i)

( )-1 -1,3 8.314 J mol K 298 K 373 K 935 J2V m

U nC T = = =

ii) and iii)

RCC = mp,mV,

( )( )

, ,

-1 -1

3

5 8.314 J mol K 298 K 373 K2

1.56 10 J =

P m V m

P

H nC T n C R T

q

= = +

=

=

iv) U = q + w

3935 J 1.56 10 J 624 JPw U q= = + =

3 1 1 2 2 1 2

[3]

[4]

[2]

[3]

b) i) Zn (s) + H2SO4 (aq) Zn2+ (aq) + SO42- (aq) + H2 (g) ii) w = -PexV and Vf >>> Vi, so dV ~ Vf

exPnRTV =

exex P

nRTPw =

w = - nRT So, n = amount of H2 produced. Due to, 1 mol of H2 1 mol Zn, So, n = amount of Zn consumed.

2 1 1

[2]

of 9

1mol g39.65g10

massmolar mass

==n

K15.298mol K J315.8mol g39.65g10 11

1 =

w

w = -379 J

2 2

[6]

Total: 20 QUESTION 2 Solution: Mark [Sub

total] a) an adiabatic reversible expansion Stot = Ssys + Ssur

surr

surrsurr T

qS =

Ssys = 0 because the process is reversible. Ssurr = 0 because q = 0 Stot = Ssys + Ssurr = 0. The process is not spontaneous. The system and surroundings are at equilibrium.

1

1

2

1

[5]

b) an expansion against Pexternal = 0 (Isothermal) w = -PexV T and w = 0. Therefore U = q = 0.

Ssys : -1 -1

-1

-1 -1 -1

85.0 gln 8.314 J mol K ln 328.01g mol

3.03 mol 9.13 J mol K = 27.7 J K

f

i

VS nR

V = =

=

Ssurr = 0 because q = 0 Stot = Ssys + Ssurr = 27.7 J K-1 + 0 = 27.17 J K-1. The process is spontaneous.

3

1

2

1

[7]

of 9

c) an isothermal reversible expansion T = 0. Therefore U = 0.

-1 -1 3ln 3.03 mol 8.314 J mol K 298 K ln 3 8.25 10 Jfi

Vw q nRT

V= = = =

Ssys : 3

-18.25 10 J = 27.7 J K298 K

reversibleqST

= =

3

-18.25 10 J 27.7 J K298 Ksurroundings

qST

= = =

Stot = Ssys + Ssurr = 27.7 J K-1 27.7 J K-1 = 0. The process is not spontaneous. The system and surroundings are at equilibrium.

2

1

1

2

1

[8]

Total: 20

of 9

PART B - QUESTION 3 Solution: Mark [Sub

total] a) To the case of a compound MpXq that dissolves to give a solution of p cations and q anions.

( ) qpqp + += /1

i) SrSO4: ( )1

2 + =

ii) MgBr2: ( )1

2 3 + =

iii) K3PO4: ( )1

3 4 + =

iv) Ca(NO3)2: ( )1

2 3 + =

2 2 2 2

[8]

b)

= i ii zmI2

m 21

mlog IAzz + =

i) KCl

( ) ( )

( )

2 2 2 2

1 1 1

-1

12 21 0.0050 mol kg 0.0050 mol kg 0.0050 mol kg2

ln 1.173 1.173 0.0050 0.08294mol kg

0.92

mI v z v z m z m z

I

Iz z

+ + + +

+

= + = +

= + =

= = =

=

2 2

[4]

of 9

ii) Ca(NO3)2

( ) ( )

( )

2 2 2 2

-1 2 -1 -1

1

12 21 0.0050 mol kg + 2 0.0050 mol kg 0.0125 mol kg2

ln 1.173 1.173 2 0.0125 0.26229mol kg

0.77

mI v z v z m z m z

I

Iz z

+ + + +

+

= + = +

= =

= = =

=

2 2

[4]

iii) ZnSO4

( ) ( )

( )

2 2 2 2

1 1 1

-1

12 21 4 0.0050 mol kg 4 0.0050 mol kg 0.020 mol kg2

ln 1.173 1.173 4 0.020 0.6635mol kg

0.52

mI v z v z m z m z

I

Iz z

+ + + +

+

= + = +

= + =

= = =

=

2 2

[4]

Total: 20

of 9

PART B - QUESTION 4 Solution: Mark a) [Sub

total] ( ) ( )

( ) ( )

( )

-2 2

2 2-

-1 -1- -1

2 Cl , Hg Cl ,

Hg Cl ,Cl ,

2210.7 kJ mol 2 mol 96485 C mol 0.26808 VCl , 131.2 kJ mol

2

R f f

ff

f

G nFE G aq G s

G s nFEG aq

G aq

= =

=

= =

o o o o

o o

o

o

2

1

2

[5]

b) i) NO3- + 4H+ +3e- NO + 2H2O E = +0.957 V 2 H2O O2 + 4H+ + 4 e- E = 1.229 V The overall reaction is 4NO3- + 4H+ 4NO + 2H2O + 3O2 E = 0.272 V

1 1

1

ii) -1

-1 -1

56

12 96485 C mol 0.272 Vln8.314 J K mol 298.15 K

127.0476.67 10

nFK ERT

K

= =

=

=

o

2

1

1

[7] c) i)

The half cell and overall reactions are AgCl + e- Ag + Cl- E = +0.22233V 1/2 H2 H+ + e- E = 0 AgCl + 1/2 H2 Ag + H+ + Cl- E = +0.22233V

2H

HCl

aaa

Q +

=

=

+

2H

HCl lna

aanFRTEE , n =1

1

[3]

of 9

ii) + == ClNa ,

mma = , m

= 1 mol/kg.

( )

( )

2

-1

-1 -1

ln ln ln

ln ln

1 96, 485 C mol 0.22233 V 0.260 Vln ln 0.30 0.2630

8.314 J mol K 298 K0.769

H Cl ClCl

H

Cl

a a mRT RT RTE E E a EnF nF nF ma

nF E E mRT m

+

= = =

=

= =

=

o o o

o

o

o

2

1

1

1

[5]


total] a) i)

HeRnRa 4222086

22488 + 2 [2]

ii) m = (mass 220Rn + 4He) mass 224Ra m = (219.9642 + 4.00150 223.9719) amu = -0.0062 amu AND

( ) kg100296.1g100296.1amu10022.6

g1amu0062.0 292623 ==

=m

2mcE =

( )( ) 22132-29 s m kg1027.9m/s10800.3kg101.0296- ==E 1 J = 1 kg m2 s-2

J1027.9 13=E

1

2

1 1

1

[3]

[3]

b) i) Cu-63: stable No. of proton = 29 (odd) No. of neutron = 34 (even) Stable nuclide with an odd number of protons and even number of

1

1

of 9

neutrons. Cu-65: stable No. of proton = 29 (odd) No. of neutron = 36 (even) Stable nuclide with an odd number of protons and even number of neutrons. Cu-64: less stable No. of proton = 29 (odd) No. of neutron = 35 (even) Stable nuclide with an odd number of protons and odd number of neutrons.

1

1

1

1

[6]

ii) emission

e016430

6429 ZnCu + Zn-64 is even-even isotope, close to the belt of stability and is likely to be stable.

1 1

positron emission

e016428

6429 NiCu ++ Ni-64 is even-even isotope, close to the belt of stability and is likely to be stable.

1 1

electron capture NiCu 6428

01-

6429 + e Ni-64 is even-even isotope, close to the belt of stability and is likely to be stable.

1

1

[6]

Total: 20

CONFIDENTIAL AS/JUL 2010/CHM520


2

Avogradros number (NA) = 6.02214 1023 mol-1 Gas constant (R) = 8.314 J K-1 mol-1 or 0.0821 dm3 atm K-1 mol-1 or 0.0831 dm3 bar K-1 mol-1 1 Pa = 1 N m-2, 1 bar = 105 Pa, 1 atm = 101.325 kPa RT/F at 25 oC = 0.02569 V mol ln a = 2.303 log a Faraday constant (F) = 96485 C mol-1 QUESTION 1 a) Calculate change of enthalpy (H) (in unit J) and the change of internal energy (U) (in

unit J) for the transformation of one mole of an ideal gas from 27.0 C and 1.00 atm to 327 C and 17.0 atm if molar heat capacity at constant pressure (Cp,m) = 20.9 + 0.042T in units of J K-1 mol-1.

(9 marks)

b) In the adiabatic expansion of one mole of an ideal gas from an initial temperature of 25 C, the work done on the surroundings is 1200 J. If molar heat capacity at constant volume (CV,m) = 3/2 R, calculate: i) heat (q) (in unit J), ii) work done (w) (in unit J), iii) final temperature (T) (in unit K) iv) change of internal energy (U) (in unit J), and v) change of enthalpy (H) (in unit J).

(11 marks)

QUESTION 2 a) Are change of entropy of fusion (Sfusion) and change of entropy of vaporization

(Svaporization) always positive or negative? Explain your answer. (4 marks)

b) Are change of enthalpy of fusion (Hfusion) and change of enthalpy of vaporization

(Hvaporization) always positive or negative? Explain your answer. (4 marks)

c) Calculate change of entropy (S) if the temperature of 1 mol of an ideal gas with molar

heat capacity at constant volume (CV,m) = 3/2 R is increased from 150 to 350 K under conditions of i) constant pressure and ii) constant volume.

(7 marks)

d) Calculate the normal boiling point (Tb) (in unit of K) of Br2, using values of standard molar enthalpy of formation (Hf), standard molar entropy (S) and standard molar Gibbs energy of formation (Gf) from Table 1.



3

Table 1 Standard molar properties at 298.15 K

Substances Hf / kJ mol-1 S / J K-1 mol-1 Gf / kJ mol-1 Br2 (g) 30.91 245.46 3.11 Br2 (l) 0 152.23 0

HBr (g) -36.40 198.70 -53.45

(5 marks)

QUESTION 3 a) Why are activity coefficients ( ) calculated using the DebyeHckel limiting law always

less than one? Explain your answer. (3 marks)

b) Why is it not possible to measure the activity coefficient ( + ) of Na

+ (aq)? Explain your answer.

(2 marks)

c) Calculate the molar solubility (S) of BaSO4 [solubility constant (Ksp) = 1.08 1010] i) in pure H2O and ii) in an aqueous solution with ionic strength (Im) = 0.0010.

hint: MX M+ + X-, + = aaKsp Given that: S = number moles of solute in 1 L of a saturated solution in units of mol L-1.

mma = , m = 1 mol kg-1. Quantities a, and m denote activity, activity

coefficients and concentration. mlog IAzz + = . Quantities zi = charge ion i and A = 0.509 for aqueous

solution at 25 oC. (15 marks)

QUESTION 4

a) Reaction of 4 NiOOH + 2 H2O 4 Ni(OH)2 + O2 is referred. i) Show the electrode half-reactions and standard cell potential (E). ii) Calculate the equilibrium constant (K) at 298.15 K for the reaction.



4

Given that: NiOOH + H2O + e- Ni(OH)2 + OH- E = +0.52 V

(7 marks)

b) Consider the Daniell cell for the indicated molalities: Zn (s) ZnSO4 (aq, 0.3000 m) CuSO4 (aq, 0.2000 m) Cu (s) The activity coefficient ( ) has the value 0.1040 for CuSO4 and 0.08350 for ZnSO4 at the indicated concentrations.

i) Show the electrode half-reactions, the overall reaction and standard cell potential

(E), reaction quotient (Q) and the Nernst Equation of each cell. ii) Calculate cell potential (E) by setting the activity (a) equal to the molality (m). iii) Calculate E by using the above values of the . iv) How large is the relative error for E calculated in ii) and iii) if the concentrations,

rather than the activities, are used? (13 marks)

QUESTION 5 a) For , and radiation, answer the following questions. Explain your answer for each

case. i) Which has the greatest ability to penetrate through matter? ii) Which has the greatest ability to ionize matter? iii) Which suffers the greatest deflection in a magnetic field? iv) How does a ray differ from and X-ray?

(8 marks)

b) Without doing detailed calculations, determine which of the radioactive nuclides in Table 2: i) has the greatest activity (A), for a 1-million-atom sample, ii) would undergo a 75 % reduction in activity in about 2 days.

Explain your answer in i) and ii).

Table 2 Nuclide and its half-life (t0.5)

Nuclide t0.5 15O 122 s 18F 109.8 min

28Mg 21 h 44Sc 3.93 h 33P 25.3 d

(6 marks)



5

c) The mass defect (m) in the formation of a nucleus of neon-20 is 0.173 amu. i) What is the nuclear mass of this nuclide (in unit of amu)? ii) What is its atomic mass (in unit of amu)? Given that: Table 3 Masses of fundamental particles of matter.


neutron 1.008665 (6 marks)


of 9

Faculty of Applied Science, UiTM Final Exam Nov 2010

Course : Physical Chemistry II, CHM 520 First Examiner : Dr. Chan Chin Han Second Examiner : Mr Borhanuddin Answer : 5 questions only

of 9


total] a)

( )

,

600

300600 K2

300K3 3

3

20.9 0.42K

20.9 600 K 300 K J 0.21 J

6.27 10 J 56.7 10 J 63.0 10 J

f

i

T

P mT

K

K

H n C dT

T dT

T

=

= +

= + = +

=

( )3 -1 1

3

63.0 10 J 8.314 J K mol 300 K 60.5 10 J

U H PV H nR T

= =

=

=

1 1 1 1 1 2 1 1

[5]

[4]

b) i) to iv) U = q + w, q = 0 for adiabatic process

( ),,

,

-1 1

-1 -1

0 because the process is adiabatic1200 J

1200 J 7.5 8.314 J mol K 298 K 1.5 8.314 J mol K

202 K

V m f i

V m if

V m

qU w

U nC T T

U nC TT

nC

= = =

=

+=

+ =

=

vi)

RCC = mp,mV,

( ) ( ) ( )( )

, ,

-1 1

3

2.5 8.314 J mol K 202 K 298 K

2.00 10 J

P m f i V m f iH n C T T n C R T T

= = +

=

=

1

1+1 1 1 1 1 2 1 1

[1] [2]

[3]

[5] Total: 20

of 9


total] a) Sfusion = positive Solid liquid, degree of randomness increases. Svaporization = positive Liquid gas, degree of randomness increases.

2

2

[4]

b) Hfusion = positive Solid liquid, attractive forces must be overcome. vaporization = positive Liquid gas, attractive forces must be overcome.

2

2

[4]

c) i) at constant pressure

RCC = mp,mV,

-1 -1 -1,

3 350 Kln 1 mol + 1 8.314 J mol K ln = 17.6 J K2 150 K

fP m

i

TS nC

T = =

ii) at constant volume

-1 -1 -1,

3 350 Kln 1 mol 8.314 J mol K ln = 10.6 J K2 150 K

fV m

i

TS nC

T = =

1

3

3

[7]

d) Br2 (l, 1 atm) Br2 (g, 1 atm)

STHG = , at equilibrium, G = 0,

b SHT

=

= )reactants()products( nHnHH kJ91.30091.30)l ,Br()g ,Br( 2

2

=== nHnHH

1

of 9

= )reactants()products( nSnSS 1

2

2 K J23.9323.15246.245)l ,Br()g ,Br( === nSnSS

K5.331K J23.93

J10091.31

4

b =

=

= SHT

1

2

[5]

Total: 20

of 9


total] a)

1 the charge on an electrolyte lowers the chemical potential of the

electrolyte (due to the charges interact attractively with the solvent, lowering the energy.)

when compared with an analogous solution of uncharged solute molecules.

2 1

[3]

b) It is impossible to create a solution of pure Na+ in water. A counter ion is always required to give a solution that is

electrically neutral, and this anion will always affect the measurement.

1 1

[2]

c) i)

mma = , m = 1 mol/kg

= i ii zmI2

m 21

mlog IAzz + =

2 2-4

2 2-4

2

2+ 2-4 4

+

2 -10

2

2

a) BaSO (s) Ba (aq) + SO (aq) 1, 1 2, 2

1.08 10

1.08

aB SOsp

Ba SO

Basp

z zc c

Kc c

c c

CK

C

+

+

+

+

= == =

= =

=

= =

o o

o

-1010

Assume: = 1 2 -5 -11.039 10 mol LBac + =

1 1 2

[4]

of 9

when = 1 2 -5 -11.039 10 mol LBac + =

( )

( )

2 2

-5-5 -1

-5

21.039 10 4 4 4.157 10 mol kg

2ln 1.173 4 4.157 10 0.03025 0.97020

mI z z

+ +

= +

= + =

= = =

1 1

[2]

when = 0.970230 2 -5 -11.0711 10 mol LBac + =

( )-5

-5 -1

-5

1.0711 10 8 4.2846 10 mol kg2

ln 1.173 4 4.2846 10 0.03071 0.9698

I

= =

= = =

1 1

[2]

when = 0.9698 2 -5 -11.0716 10 mole LBac + =

( )-5

-5 -1

-5

-5 -1

1.0716 10 8 4.2866 10 mol kg2

ln 1.173 4 4.2866 10 0.03072 0.9697

solubility 1.07 10 mol L

I

= =

= = =

=

3 iterations are sufficient.

1 1 1

[3]

ii)

( )2

2

-1

2`2 -10

-5 -1

b) 0.0010 mol kg

ln 1.173 4 0.0010 0.148374 0.8621

0.8621 1.08 10

1.21 10 mol kg

Basp

Ba

I

cK

c

c

+

+

=

= = =

= =

=

o

2 1 1

[4]

Total: 20

of 9


total] The half cell reactions are NiOOH + H2O + e- Ni(OH)2 + OH- E = +0.52V 4OH- O2 + 2H2O + 4e- E = 0.401V The overall reaction is 4 NiOOH + 2 H2O 4 Ni(OH)2 + O2 E = + 0.12V

-1

-1 -1

8

4 96485 C mol 0.12 Vln8.314 J K mol 298.15 K

18.6831.30 10

nFK ERT

K

= =

=

=

o

1

2

2

2

[7]

b) i) Oxidation: Zn (s) Zn2+ (aq) + 2 e- E = 0.7618 V Reduction: Cu2+ (aq) + 2 e- Cu (s) E = 0.337 V Overall: Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s) E = 1.099 V

+

+

= 2Cu

2Zn

aaQ

2

2lnZn

cell cellCu

aRTE EnF a

+

+

=

o o , n = 2

1

1

1

[4]

ii) Activities equal to molalities:

( ) ( )( ) ( )

-1 -1

-1

8.3145 J mol K 298.15 K 0.3001.099 ln 1.094 V0.2002 96, 485 C mol

ocellE

= =

3

[3]

iii) Using :

mma = , m

= 1 mol/kg. Quantities a, and m denote activity,

activity coefficients and concentration.

4

[4]

of 9

( ) ( )( ) ( )

-1 -1

-1

8.3145 J mol K 298.15 K 0.0835 0.3001.099 n 1.097 V0.104 0.2002 96, 485 C molcell

E = =

o l

iv) Relative error: 11.094 V 1.0971 V 100% 0.27%

1.097 V

=

2

[2]

Total: 20


total] a) i)

has the greatest penetrating power through matter.

+= 242He , e01= , = energy having neither mass nor charge. (most

energetic)

1

1

[2]

ii) has the greatest ionizing power.

+= 242He , with charge of 2+.

1 1

[2]

iii) suffers the greatest defection in magnetic field.

e01= , its mass is much lighter than that of .

1 1

[2]

iv) ray is emitted from the nucleus while X-ray is produced in the electron shell. The X-ray is usually more energetic.

1 1

[2]

of 9

b) i) 15 O

Nt

NA5.0

693.0==

The shortest t0.5 would be the greatest activity.

1

2

[3]

ii) 28Mg with t0.5 = 21 h

%25%50%100 5.05.0 21 tt 25 % left 2 t0.5 If 2 days 2 t0.5 Then, t0.5 1 day.

1

2

[3]

c) i) Mass of proton = 1.0073 amu 10 = 10.073 amu Mass of neutron = 1.0087 amu 10 = 10.087 amu m = 0.173 amu m = mass of (proton + neutron) nuclear mass nuclear mass = mass of (proton + neutron) m nuclear mass = (20.160 -0.173) amu = 19.987 amu

1

2

[3]

ii) nuclear mass = atomic mass - mass of extranuclear electrons Atomic mass = nuclear mass + mass of extranuclear electrons Atomic mass = 19.987 amu + (10 5.48610-4 amu) = 19.992 amu

1

2

[3] Total: 20



2

Avogradros number (NA) = 6.02214 1023 mol-1 Gas constant (R) = 8.314 J K-1 mol-1 or 0.0821 dm3 atm K-1 mol-1 or 0.0831 dm3 bar K-1 mol-1 1 Pa = 1 N m-2, 1 bar = 105 Pa, 1 atm = 101.325 kPa RT/F at 25 oC = 0.02569 V mol, ln a = 2.303 log a Faraday constant (F) = 96485 C mol-1 QUESTION 1 a) Two ideal gas systems undergo reversible expansion starting from the same pressure

(P) and volume (V). At the end of the expansion, the two systems have the same volume. The pressure in the system that has undergone adiabatic expansion is lower in the system that has undergone isothermal expansion. Explain this result without using equations.

(4 marks)

b) A cup of water at 278 K (the system) is placed in a microwave oven and the oven is turned on for one minute, during which it begins to boil. State positive, negative or zero for the following quantities. Explain your answer for each case. i) Heat (q) ii) Work (w) iii) Change of internal energy (U)

(6 marks) c) One mole of an ideal gas at an initial pressure (Pi) of 2.00 bar undergoes adiabatic

expansion from an initial temperature (Ti) of 450 K to a final temperature (Tf) of 300 K. i) Calculate work done (wad) (in unit J). ii) Write an expression for the work done (w) in the isothermal reversible expansion of

the gas at 300 K from an initial pressure of 2.00 bar. iii) What value of the final pressure (Pf) (in unit bar) under isothermal reversible

expansion would give the same value of wad under adiabatic expansion?

Assume that molar heat capacity at constant pressure (Cp,m) = 5/2 R. (10 marks)

QUESTION 2 a) Classify the following processes as spontaneous or not spontaneous. Explain your

answer for each case. i) The reversible isothermal expansion of an ideal gas. ii) The vaporization of superheated water at 102 C and 1 bar. iii) The constant pressure melting of ice at its normal freezing point by the addition of an

infinitesimal quantity of heat. iv) The adiabatic expansion of a gas into a vacuum.

(8 marks)



3

b) Calculate the change in entropy (S) if 1 mol of liquid water is heated from 0 to 100 C under constant pressure. Molar heat capacity under constant pressure (Cp,m) for water is 75.291 J K1 mol1. The melting point (Tm) of water at the pressure of interest is 0 C and the enthalpy of fusion (Hfus) is 6.0095 kJ mol1. The boiling point (Tb) is 100 C and the enthalpy of vaporization (Hvap) is 40.6563 kJ mol1. Calculate S for the transformation H2O (s, 0 C) H2O(g, 100 C).

(10 marks)

c) Under what conditions is 0dG ; a condition that defines the spontaneity of a process? (2 marks)

QUESTION 3 a) Calculate ionic strength (Im), mean activity coefficient ( ), and activity ( 3AlCla ) for a

0.0250 molal solution of AlCl3 at 298 K. Assume complete dissociation. Given that: A = 0.509 for aqueous solution at 25 oC.

(10 marks)

b) Conductivity measurements were one of the first methods used to determine the ionization constant of water (Kw). The ionization constant of water is given by the following equation:

[ ] [ ]

==

+

+ 1MOH

1MH

OHHwaaK

Given that: mma = , m = 1 mol/kg. Quantities a, and m denote activity, activity

coefficients and concentration. Assume that 1 and water is a weak electrolyte with degree of dissociation of . i) Using the expression provided, show that the molar conductivity of pure water

( ) can be written as ( ) ( ) ( ) ( )[ ]-oo21w2 OHHM1OH K += + . Quantity o denotes molar ionic conductivity at infinite dilution.

ii) Kohlrausch and Heydweiller measured the conductivity of water in 1894 and determined that ( )OH 2 = 5.5 106 S m1 at 298 K. Using the information provided in Table 1, determine Kw.



4

Table 1. Cation

o (-1 cm2 mol-1)

Anion o

(-1 cm2 mol-1) H3O+ or H+ 350.0 OH- 199.2 NH4+ 73.5 Br- 78.1 K+ 73.5 Cl- 76.3 Na+ 50.1 I- 76.8 Ag+ 62.1 NO3- 71.4 Ca+ 118.0 CH3COO- 40.8 Mg2+ 106.1 SO42- 159.6

(10 marks)

QUESTION 4 a) A current of 2.00 A is applied to a metal wire for 30 s. How many electrons pass through

a given point in the wire during this time?

Given that: 96500 C 1 mol e. (4 marks)

b) Reaction of Hg2Cl2 (s) 2 Hg (l) + Cl2 (g) is referred.

i) Show the electrode half-reactions and standard cell potential (E). ii) Calculate change of molar Gibbs free energy for the reaction ( )rxnG and the

equilibrium constant (K) at 298.15 K for the reaction.

Given that: Hg2Cl2 (s) + 2 e 2 Hg (l) + 2 Cl- (aq) E = + 0.26808 V (8 marks)

c) Determine solubility constant (Ksp) for AgBr at 298.15 K using the electrochemical cell

described by Ag (s) AgBr (s) Br- (aq) Ag+ (aq) Ag (s). Given that: AgBr + e- Ag + Br- E = + 0.07133 V

(8 marks)

QUESTION 5 a) Complete the following nuclear equations and identify X in each case:

i) X3 Br U 108735

10

23592 +++ nn

ii) 4XZn Sm U 723016062

10

23592 +++ n

iii) 2XTeSr U 135529940

10

23592 +++ n

(6 marks)



5

b) Nuclear waste disposal is one of the major concerns of the nuclear industry. In choosing a safe and stable environment to store nuclear wastes, consideration must be given to the heat released during nuclear decay. Consider the decay of Sr 90 (89.907738 amu):

1st decay: 01-9039

9038 YSr + half-life (t0.5) is 28.1 yr

The Y90 (89.907152 amu) futher decays as follows:

2nd decay: 01-9040

9039 Zr Y + t0.5 = 64 h

Zirconium-90 (89.904703 amu) is a stable isotope. i) By referring to Table 2, calculate the following for the 1st and the 2nd decays:

The mass defect (m) (in unit amu and kg) The energy released (in unit J) for the decays

Given that:


Table 2


neutron 1.008665

ii) Starting with one mole of Sr 90 , calculate the number of moles (n) of Sr 90 that will decay in one year.

iii) Calculate the amount of heat released (in unit kJ) corresponding to the number of moles of Sr 90 decays to Zr 90 in ii).

(14 marks)


of 11

Faculty of Applied Science, UiTM Final Exam Apr 2010

Course : Physical Chemistry II, CHM 520 First Examiner : Dr. Chan Chin Han Second Examiner : Ms Hjh Mashiah Domat Shaharudin Answer : 5 questions only

of 11


total] a) In the system undergoing adiabatic expansion, all the work done must come through the lowering of U, and therefore of the temperature. By contrast, some of the work done in the isothermal expansion can come at the expense of the heat that has flowed across the boundary between the system and surroundings.

2 2

[4]

b) i) q is positive because heat flows across the system-surrounding boundary into the system. ii) w is negative because the vaporizing water does work on the surroundings (expansion). (w = PV) iii) U is positive because the temperature increases and some of the liquid is vaporized. (U = nCp,mT)

1 1 1 1 1 1

[6]

c) i)

RCC = mp,mV, U = q + w, q = 0 for adiabatic process

J1087.1K150KmolJ314.823)( 311mp,ad ====

TRCnUw

ii) The total work of reversible expansion is

= fi dVV VPw

Isothermal reversible expansion: (of a perfect gas) PV = nRT

VnRTP = , T is constant

= fi dVV VPw

1 1 2 1

[4]

of 11

= fidV

V VVnRTw

i

flnVVnRTw =

For ideal gas:

111 nRTVP = and 222 nRTVP =

2

1

22

11

nRTnRT

VPVP

= , when T1 = T2

f

i

i

f

PP

VV

=

f

ilnPPnRTw =

iii)

nRTw

PP ad

f

iln =

( )( )( )( )300KKmol8.314J1mol

J101.87bar2ln 1-1-3

f

=P

( ) 4720bar2f

.P

=

Pf = 4.24 bar

1 1 1 1 1

[3]

[3]

Total: 20


total] a) i) is not spontaneous because the system and surroundings are in equilibrium. ii) is spontaneous because the equilibrium phase under the stated conditions is a gas. iii) is not spontaneous because the process is reversible. iv) is spontaneous because at equilibrium, the density of a gas is uniform throughout its container.

2

2

2

2

[8]

of 11

b) The heat input is the same for a reversible and an irreversible process.

: heatingS

,

,,

-1 -1

-1

ln

373.15 K 1 mol 753 J mol K ln273.15 K

23.49 J K

reversible P m

fP mP m

i

dq dq nC dTTC

S n dT nCT T

= =

= =

=

=

1

1

1 1

[4]

( )

-1

-1

-1

6009 J 22.00 J K273.15 K

40656 J 108.95 J K373.15 K

22.00 108.95 23.49 J K

15

fusionfusion

fusion

vaporizationvaporization

vaporization

total fusion vaporization heating

HS

TH

ST

S S S S

= = =

= = =

= + + = + +

= -14.4 J K

2

2

2

[6]

c) This is the case at constant T and P if no non-expansion work is possible.

2 [2]

Total: 20

of 11


total] a)

= i ii zmI2

m 21

( ) ( )[ ]{ }22m 107503025021 += ..I 150m .I =

mlog IAzz + = ( )( ) 590150509013log ... ==

2560. =


so, mma = and m

ma +++ =

To the case of a compound MpXq that dissolves to give a solution of p cations and q anions. m+ = pmi and m- = qmi , mi = electrolytes stoichiometric molality.

( )[ ]( )[ ] ( ) ( )

=

+++

qq

ppqp

mm

mmaa

( )[ ]( )[ ] ( ) ( )

= ++

qiq

pipqp

mqm

mpm

aa

( )[ ]( )[ ] ( ) ( ) ( ) ( )qp

iqpqpqp

mmqpaa

+

++

= , where ( ) ( ) ( )

qpqp+

+

( )[ ]( )[ ] ( ) ( ) ( )qp

iqpqpqp

mm

qpaa+

++

=

( )[ ]( )[ ] ( ) ( ) ( )qp

iqpqp

mm

qpaa+

+

=

( )[ ] ( )[ ] ( ) ( ) ( )( )[ ] 3131313 02500256031ClAl ++ = ..aa ( )[ ] 83 10511AlCl = .a

1 1 1 1 1 1 1 1 1 1

[3]

[3]

[4]

of 11

b) i) The conductivity of a weak electrolyte is given as

o = The degree of disassociation is given by the equilibrium expression

[ ] [ ]

==

+

+ 1MOH

1MH

OHHwaaK , [H+] = [OH-] = 1 M

( )22

w M1K =

( ) 21wM1 K =

And, o is given as ( ) ( ) ( )+ += OHHOH oo2o

Combining the above expressions yields the relationship of interest:

( ) ( ) ( ) ( )[ ]+ += OHHM1OH oo21w2 K

1 1 1 1 1

[5]

ii)

( ) 12o molmS30500H + = . , ( ) 12o molmS01990OH = . ( ) -162 mS1055OH = .

Substituting these values in the expression derived in part (a) of this question and rearranging to isolate Kw yields:

( )( ) ( ) ( )[ ]

2

oo

2w OHHM1

OH

+

= + K

( ) ( ) ( )[ ]2

oo

ow OHHM1

+

= + K

( )[ ]2

1-21-2

1-6

w mol m S 0.0199 mol m S 0350.0M1m S 105.5

+

=

K

23

62

228

w L 1000m 1

m molL mol1000.1

= K

14w 1000.1

=K

1 1 1 1 1

[5]

of 11

( ) ( ) ( )( )

( )

2

+

2? ?

2 ? 2 ?

22 2 3?

2 6

? 4

1 M H OH

5.5 10 S m 1 M 0.0350 S m mol + 0.0199 S m mol

mol L 1 m =1.00 10mol m 1000 L

1.00 10

mwK

= +

=

=

Total: 20

of 11


total] Current (I) is equal to the amount of charge (Q) that flows per unit time (t):

( )( ) 2.00 A 30 s 60.0 C

Q It==

=

Coulombs of charge can be converted to number of electrons using the charge per electron as follows: 96500 C 1 mol e. Avogradros number (NA) = 6.02214 1023 mol-1 So, 1 e = 1.60210-19 C

( ) 20? 91 60.0 C 3.75 10

1.60 10 Ce e

=

1

1

2

[4]

b) i) Hg2Cl2(s) +2e- 2Hg(l) +2Cl-(aq) E = + 0.26808 V 2Cl-(aq) Cl2(g) + 2e- E = 1.35827 V Hg2Cl2(s) 2Hg(l) +Cl2(g) E= 1.09019 V

1

1

ii) -1 -1

-1

-1 -1

37

2 96485 C mol 1.09019 V =210.4 kJ mol

2 96485 C mol 1.09019 Vln8.314 J K mol 298.15 K

85.0021.21 10

reactionG nFEnFK ERT

K

= =

= =

=

=

o o

o

3

2

1

[8] c)

The half cell and overall reactions are R: Ag Ag+ + e- E = 0.7996 V L: AgBr + e- Ag + Br- E = + 0.07133 V Overall AgBr Ag+ + Br- E = 0.72827 V

1

1

Ksp = [Ag+][Br-] ( )[ ] ( )[ ]+= BrAg aaQ , Q = Ksp

1 1

of 11

At equilibrium E = 0,

QnFRTE ln0 =

RTnFEQ

ln =

( )

13sp 10884

3528V025690

V7282701ln

==

=

=

.QK

..

.Q

1

1

1

1

[8]


total] a) i) 235 1 87 146 192 0 35 57 0U n Br La 3 n+ + + ii) 235 1 160 72 192 0 62 30 0U n Sm Zn 4 n+ + + iii) 235 1 99 13592 0 40 52U + n Sr + Te + 2 10n

2

2

2

[6] b) i) In the 90Sr decay (1st decay), the mass defect is: m = (mass 90Y + mass e) mass 90Sr = [(89.907152 amu + 5.4857 104 amu) 89.907738 amu] = 3.743 105 amu AND 5 29 32

231 g( 3.743 10 amu) 6.216 10 g 6.216 10 kg

6.022 10 amu = = =

The energy change is given by: E = (m)c2 = (6.126 1032 kg)(3.00 108 m/s)2 = 5.59 1015 kg m2/s2 = 5.59 1015 J

1

[3]

of 11

Similarly, for the 90Y decay (2nd decay), we have m = (mass 90Zr + mass e) mass 90Y = [(89.904703 amu 104 amu) 89.907152 amu] = 1.900 103 amu AND

3 27 3023

1 g( 1.900 10 amu) 3.156 10 g 3.156 10 kg6.022 10 amu

= = =

and the energy change is: E = (3.156 1030 kg)(3.00 108 m/s)2 = 2.84 1013 J

1

[3]

ii) This calculation requires that we know the rate constant for the decay. From the half-life, we can calculate .

12

10.693 0.693 0.0247 yr 28.1 yr

= = =t

To calculate the number of moles of 90Sr decaying in a year, we apply the following equation:

ntn

=dd

=tn

n tnn 0to

dd1

no = number of atoms present at time 0. nt = number of atoms present at time t. [ ] [ ]tonn tn dln to =

tnn

=

o

tln

0ln = t

Nt

N

1ln (0.0247 yr )(1.00 yr)1.00

= x

where x is the number of moles of 90Sr nuclei left over. Solving, we obtain: x = 0.976 mol 90Sr Thus the number of moles of nuclei which decay in a year is (1.00 0.976) mol = 0.024 mol

2

1

1

1

[5]

of 11

This is a reasonable number since it takes 28.1 years for 0.5 mole of 90Sr to decay. iii)

Since the halflife of 90Y is much shorter than that of 90Sr, we can safely assume that all the 90Y formed from 90Sr will be converted to 90Zr. The energy changes calculated in part (i) refer to the decay of individual nuclei. In 0.024 mole, the number of nuclei that have decayed is:

23226.022 10 nuclei0.024 mol 1.4 10 nuclei

1 mol

=

Realize that there are two decay processes occurring, so we need to add the energy released for each process calculated in part (i). Thus, the heat released from 1 mole of Sr 90 waste in a year is given by: The energy released in the above two decays is 5.59 1015 J (1st decay) and 2.84 1013 J (2nd decay). The total amount of energy released is: (5.59 1015 J) + (2.84 1013 J) = 2.90 1013 J.

1322 2.90 10 J(1.4 10 nuclei)

1 nucleus

= = =9 6heat released 4.06 10 J 4.06 10 kJ

This amount is roughly equivalent to the heat generated by burning 50 tons of coal! Although the heat is released slowly during the course of a year, effective ways must be devised to prevent heat damage to the storage containers and subsequent leakage of radioactive material to the surroundings.

1

1

1

[3]

Total: 20



2

Avogradros number (NA) = 6.02214 1023 mol-1 Gas constant (R) = 8.314 J K-1 mol-1 or 0.0821 dm3 atm K-1 mol-1 or 0.0831 dm3 bar K-1 mol-1 1 Pa = 1 N m-2, 1 atm = 101.325 kPa RT/F at 25 oC = 0.02569 V mol, ln a = 2.303 log a Faraday constant (F) = 96485 C mol-1 QUESTION 1 a) State True or False for the statement below and explain you answer.

i) The P-V work in a mechanically reversible process in a closed system always equals

-PV. Quantities P and V denote pressure and volume, respectively. ii) For every cyclic process, the final state of the system is the same of the initial state.

(6 marks) b) A strip of magnesium of mass 15 g is dropped into a beaker of dilute hydrochloric acid.

The atmospheric pressure is 1.0 atm and the temperature 25 C. i) Write the balanced chemical equation for the above reaction. ii) Calculate the work done (w) (in unit J) by the system as a result of the reaction.

(6 marks) c) The constant-pressure heat capacity (Cp) of a sample of a perfect gas was found to vary

with temperature according to the expression Cp/(J K1) = 20.17 + 0.3665(T/K). When the temperature (T) is raised from 25 to 200 C at constant pressure Calculate (in unit kJ): i) Heat (q) ii) Work (w) iii) Change of internal energy (U) iv) Change of enthalpy (H)

(8 marks)

QUESTION 2 Consider a system consisting of 2.0 mol CO2 (g), initially at 25 C and 10 atm and confined to a cylinder of cross-section 10.0 cm2. It is allowed to expand adiabatically against an external pressure of 1.0 atm until the piston has moved outwards through 20 cm. Assume that carbon dioxide may be considered a perfect gas with constant-volume molar heat capacity (CV,m) = 28.8 J K1 mol1. Calculate:

a) Heat (q) (in unit J) b) Work (w) (in unit J) c) Change of internal energy (U) (in unit J) d) Change of temperature (T) and the final temperature (Tf) (in unit K) e) Initial volume (Vi) and final volume (Vf) (in unit dm3) f) Change of entropy for the system (Ssys) (hint: constant volume cooling followed by

isothermal expansion)

(20 marks)



3

QUESTION 3 a) The mobility () of a chloride ion in aqueous solution at 25 C is 7.91 10-8 m2 s-1 V-1.

Calculate the molar ionic conductivity (). (3 marks)

b) The mobilities () of H+ and Cl- at 25 C in water are 3.623 10-7 m2 s-1 V-1 and 7.91 10-

8 m2 s-1 V-1, respectively. i) What proportion of the current [or transport number (I)] is carried by the protons in 10-

3 M HCl (aq)? ii) What fraction of the current do protons carry when the NaCl is added to 10-3 M HCl?

The final concentration of NaCl in the solution is 1.0 mol dm-3. The concentration of 10-3 M HCl does not vary after addition of NaCl is assumed. (hint: concentration (c) as well as mobility governs the transport of current) Given that of Na+ is 5.19 10-8 m2 s-1 V-1.

(7 marks) c) The molar conductivity () of KBr solutions as a function of concentration (c) at 25 oC is

given in Table 1.

Table 1 c /10-3 M / S cm2 mol-1

0.25 150.16 0.36 149.87 0.50 149.55 0.75 149.12 1.00 148.78 1.60 148.02 2.00 147.64 5.00 145.47

10.00 143.15

i) Write the relationship between the cell constant (Kcell) and the electrolytic conductivity ().

ii) Write the relationship between and the .

iii) Verify that the follows the Kohlrausch law

= 2

1

o c and determine

graphically the molar conductivity at infinite dilution (o) for KBr (in unit S cm2 mol-1).

iv) Determine the coefficient . v) Use scientific calculator to calculate the slope, intercept and r2 for the figure

presented in iii). (10 marks)



4

QUESTION 4 a) The cell of Ag | AgBr (s) | KBr (aq, 0.050 mol kg1) || Cd(NO3)2 (aq, 0.010 mol kg1) | Cd is

referred. i) Show the electrode half-reactions, the overall reaction and standard cell potential

(E), reaction quotient (Q) and the Nernst Equation of the cell. ii) Use the DebyeHckel limiting law to calculate activity coefficient ( ) of anion and

cation; and mean activity coefficient ( ) of anion and cation. Given that: A = 0.509 for aqueous solution at 25 oC.

iii) Use the Nernst equation to estimate the cell potential (E) at 25C of the cell which is referred.

Given that: AgBr (s) + e Ag (s) + Br- (aq), E = 0.07 V.

(14 marks) b) State True or False for the statement below and explain you answer.

i) Doubling the coefficients in a chemical reaction will square the value of the

equilibrium constant (Kc). ii) Doubling the coefficients in a chemical reaction will double the change in standard

Gibbs free energy (G). iii) Doubling the coefficients in a chemical reaction will not change the standard cell

potential (E) (6 marks)

QUESTION 5 a) Complete the following nuclear equations and identify X in each case:

i) X3 Ba U 1014056

10

23592 +++ nn

ii) 2XRb Cs U 9037144

5510

23592 +++ n

(4 marks) b) For He42 (4.0026 amu), calculate the following by referring to Table 2:

i) The mass of total proton and neutron (in unit amu) ii) The mass defect (m) iii) The nuclear binding energy (BE) (in unit J) and the binding energy per nucleon

Given that:


(10 marks)



5

Table 2


neutron 1.008665 c) Tritium ( )H31 is radioactive and decays by electron emission. Its half-life (t0.5) is 12.5 yr.

In ordinary water, the ratio of H11 to H31 atoms is 1.0 10

17 to 1. i) Write a balanced nuclear equation for tritium decay. ii) How many tritium atoms and disintegrations per minute will be observed in a 1.00

kg sample of water? (6 marks)


of 11

Faculty of Applied Science, UiTM Final Exam Nov 2009

Course : Physical Chemistry II, CHM 520 First Examiner : Dr. Chan Chin Han Second Examiner : Ms Hjh Mashiah Domat Shaharudin Answer : 5 questions only

of 11

QUESTION 1 Solution: Mark a) [Sub

total] i) False 1 In equilibrium, Pex = P, so dw = -PexdV; dw = -PdV The total work of reversible expansion is

= fi dVV VPw

Isothermal reversible expansion: (of a perfect gas) PV = nRT


= fi dVV VPw

= fidV

V VVnRTw

i

flnVVnRTw =

1 1

ii) True

In a thermodynamic cycle, the overall change in a state function (from the initial state to the final state and then back to the initial state again) is zero.

1 1 1

[6]

of 11

b) i) Mg (s) + 2 HCl (aq) H2 (g) + MgCl2 (aq)

2

ii) The external, opposing pressure is 1.0 atm and Vi 0,

Molar mass for Mg = 24.31 g mol-1

mol6170molg 24.31

g 151 .n =

=

w = PV , Pf = Pex

ff P

nRTV =

nRTP

nRTPVP)VV(Pw ====f

ffexifex

( )( )( )K15298molKJ3148mol6170 11 ...w = J1529=w

1 1

1

[6]

c) i)

= fi pdTT TCH

( )[ ] ( ) += 473298 /Kd/K0.365520.17 TTH

( )( ) [ ]47329822366502984731720 T..H

+=

kJ328 .H =

1 1 1

[3]

ii) q = H = 28.3 kJ

1

[1]

iii) w = PV , Pf = Pex w = (PV) , PV = nRT w = nRT w = -(1 mol) (8.314 J K-1 mol-1) (473 K 298 K) w = -1.45 kJ

1

[2] iv) U = q + w U = 28.3 -1.45 = 26.8 kJ

1 1

[2]

Total: 20

of 11


total] a) q = 0 for adiabatic process

1

[1]

b) w = PexV

( )( )( ) J20cm

m10cm10cm20Pa10011 336

25 =

=

.w

Note: 1 Pa m3 = 1 J

1

[2]

c) U = q + w U = 0 - 20 = -20 J

1

[1]

d) U = nCV,mT

( )( ) K3470molKJ828mol2J20 11 ..

T =

=

Tf = Ti + T = 298.15 0.347 = 297.803 K

1

1

[3] e)

ii P

nRTV =

( )( )( ) 3113i dm8934atm10

K15298molKatmdm082060mol02 ....V ==

Vf = Vi + V = 4.892 + 0.2 = 5.093 dm3

1

1

[3]

f) 21sys SSS +=

Cooling the system from 298.15 K to 297.803 K at 4.893 dm3 (constant volume cooling)

=fi

rev2 T

dqS , dqrev = CV,mdT

= fimV,

2d

TT TTC

nS

+

of 11

=

i

fmV,2 ln T

TnCS

( ) 12 KJ06710K15298K803297ln

molKJ828mol2 =

= .

...S

1

[3]

Isothermal reversible compression of an ideal gas from 10 to 1 atm at 297.803 K PV = nRT


= fiVV PdVw

= fiVV V

dVnRTw

i

f

VVlnnRTw =

U = 0, qrev = -wrev

i

frev V

VlnnRTq =

TqS rev1 = , when T = constant

i

f1 V

VlnnRS =

( ) 133

1 KJ6660dm8934dm0935ln

molKJ3148mol2 =

= .

.

..S

1

1

[5]

21sys SSS +=

( ) 1sys KJ666006710 += ..S 1

sys KJ600= .S

1

1

[2]

Total: 20

of 11


total] a)

Fz = ( )( )( ) 12311128 molmS10637molC96485Vsm109171 == ..

1 2

[3]

b) i) I+ = z++pcFEA for I+ and the equivalent term for I-.

++

+++ +

=qzpz

pzt ++

+

=qzpz

qzt

In the special case where z+ = z- and necessarily, p = q,

+

++ +

=

t +

+

=

t

82079106233

6233H ...

.t =+

=+

1 1 1

[3]

ii) ( )

( ) ( )++++

++ ++

=ClNaH

HII

It ; I+ = z++pcFE, so I+ = constant +c

( )( ) ( )++++

+++

+ ++=

ClNaHH

ccct

( )( )( )( ) ( )( ) ( )( ) 002807910001151900162331001

623310013

3

.......

..t =++

=

+

1 1 2

[4]

c) i)

RK cell= , R = resistance

1

ii)

c

= 1

of 11

iii)

c /10-3 M c (mol L-1)

c1/2 (mol1/2 L-1/2)

/ S cm2 mol-1

0.25 0.00025 0.01581 150.16 0.36 0.00036 0.01897 149.87 0.50 0.00050 0.02236 149.55 0.75 0.00075 0.02739 149.12 1.00 0.00100 0.03162 148.78 1.60 0.00160 0.04000 148.02 2.00 0.00200 0.04472 147.64 5.00 0.00500 0.07071 145.47 10.00 0.01000 0.10000 143.15

2

For strong electrolyte in dilute solution;

= 2

1

o c (Kohlrauschs law)

o = molar conductivity at infinite dilution (extrapolation at c = 0 of a plot

vs c .

1

c1/2 ( mol1/2 L-1/2)0.00 0.02 0.04 0.06 0.08 0.10 0.12

(S

cm

2 mol

-1)

142

144

146

148

150

152Interceipt=151.41slope =-83.30r =0.9996

o = 151.41 S cm2 mol-1

2 1

iv) K = -83.30 1 v) From the calculation of scientific calculator. Interceipt=151.41slope =-83.30r =0.9996

1

[10]

Total: 20

of 11


total] a) i) R: Cd2+ (aq) + 2e Cd (s) E = -0.40 V L: 2AgBr (s) + 2e 2Ag (s) + 2Br- (aq), E = 0.07 V Overall, R L: Cd2+ (aq) + 2Ag (s) + 2Br- (aq) Cd (s) + 2AgBr (s) E = -0.47 V (R-L)

1

1

[3]

( )[ ] ( )[ ]+= BrCd1

22 aaQ

( )[ ] ( )[ ]

= + BrCd

1ln2 22

aaFRTEE

( )[ ] ( )[ ]{ }++= BrCdln2

22 aaF

RTEE

1

1

[2]

ii)


so, mma = and m

ma +++ =

( ) = ma Br and ( ) +++ = ma 2Cd

( ) ( )KBrBr = and ( ) [ ]22 Cd(NO)Cd ++ =

For anion Br-:

= i ii zmI2

m 21

( ) ( )[ ]{ }22m 105001050021 += ..I 050m .I =

mlog IAzz + = ( )( ) 110050509011log ... ==

7690. === + ; ( ) ( )KBrBr =

[3]

of 11

For anion Cd2+:

= i ii zmI2

m 21

( ) ( )[ ]{ }22m 102002010021 += ..I 030m .I =

mlog IAzz + = ( )( ) 1760030509012log ... ==

8380. === + ; ( ) [ ]22 Cd(NO)Cd ++ =

[2]

iii)

( )[ ] ( )[ ]{ }++= BrCdln2

22 aaF

RTEE

( ) = ma Br and ( ) +++ = ma 2Cd 7690. === + ; ( ) ( )KBrBr =

8380. === + ; ( ) [ ]22 Cd(NO)Cd ++ =

[ ][ ]{ }2 ln2 ++

+= mmF

RTEE

( ) ( )( )[ ]( ) ( )[ ]{ }22 05076900108380lnmol2

mol V 0.02569V470 .....E

+=

( ) ( ) V620V1450V470 ...E =+=

1

2

1

[4]

b) i) True. A + B C

[ ][ ][ ]BA

Cc =K

When 2A + 2B 2C

[ ][ ] [ ] c

222

2

c BAC K"K ==

1

1

ii) True A + B C

xG = When 2A + 2B 2C

22 Gx"G ==

1

1

iii) True E is an intensive property. They do not depend on how many times the reaction occurs.

1

1

[6]

of 11

As long as the concentrations of ions and gases are 1 M or 1 atm, E is a constant and not dependent on the coefficients in the balanced equation.


total] a) i) 235 1 140 1 9392 0 56 0 36U n Ba 3 n Kr+ + +

ii) 235 1 144 90 192 0 55 37 0U n Cs Rb 2 n+ + +

2

2

[4] b) i) 4 1 12 1 0He 2 p + 2 n (There are 2 protons and 2 neutrons in the helium nucleus.) The mass of 2 protons is: (2)(1.007825 amu) = 2.015650 amu the mass of 2 neutrons is: (2)(1.008665 amu) = 2.017330 amu The predicted mass of 42He is 2.015650 + 2.017330 = 4.032980 amu

1

1

1

ii) the mass defect is: m = 4.032980 amu 4.0026 amu = 0.0304 amu

1

iii) The energy change (E) for the process is E = (m)c2 = (0.0304 amu)(3.00 108 m/s)2

2

152

amu m2.74 10s

=

Lets convert to more familiar energy units (J/He atom).

15 2

2 23 2

2

2.74 10 amu m 1.00 g 1 kg 1 J1000 g1 s 6.022 10 amu 1 kg m

s

=

124.55 10 J

1

1

1

of 11

nuclear binding energynuclear binding energy per nucleon

number of nucleons=

For the helium-4 nucleus,

124.55 10 J/He atomnuclear binding energy per nucleon 4 nucleons/He atom

= = 121.14 10 J/nucleon

1

2

[10]

c) i) 3 3 01 2 1H He +

1

The number of tritium (T) atoms in 1.00 kg of water is:

233 2 2

2 172 2 2

1 mol H O 6.022 10 molecules H O 2 H atoms 1 T atom(1.00 10 g H O)18.02 g H O 1 mol H O 1 H O 1.0 10 H atoms

= 6.68 108 T atoms The number of disintegrations per minute will be:

12

0.693rate (number of T atoms)= = =N Nt

( )80.693 1 yr 1 day 1 hrate 6.68 10 T atoms12.5 yr 365 day 24 h 60 min

=

rate = 70.5 T atoms/min = 70.5 disintegrations/min

1

1

1

1

1

[6]

Total: 20

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