Physical Chemistry past year

8/10/2019 Physical Chemistry past year

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U n i v e r s i t ialaysia

P H N GEngineering TooInnoloQy CMy

FACULTY OF CHEMICAL NATURAL RESOURCES ENGINEERING

FINAL EXAMINATION

COURSE PHYSICAL CHEM ISTRYCOURSE CODE

BKF1253

LECTURERS

SURIATI GHAZALI

NORHAYATI ABDULLAH

DATE 8 JANUARY 2014DURATION 3 HOURSSESSION/SEMESTER

SESSION 2013/2014 SEMESTER I

PROGRAMME CODE

BKC/ BKG

INSTRUCTIONS TO CANDIDATE:

This question paper consists of FOUR (4) questions Answer ALL questions2 All answers to a new question should start on new page3 All calculations and assum ptions must be clearly stated4 Candidates are not allowed to b ring any material other than those allowed by the

invigilator into the examination room

EXAMINATION REQUIREMENT:

APPENDICESFormulas

2 Tables

DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO

This exam ination paper co nsists ofTEN (10) printed pages including front page


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CONFIDENTIAL KC BKG 131411BKF1253QUESTION 1

a) The following reaction releases the heat to the surrounding at 26.9 C . Determine the

standard entropy for this reaction.

g) + 31 -1 g) 2NH g) \H = 35.7 kJ 3 Marks)b) By using the van d er Waals equation, calculate the pressure of nitrogen gas at 2 73 .15 K

that have a molar volume of 22.41 4 L/mol. Com pare with the pressure of an ideal gas atthe same temperature and m olar volume.

7 Marks)

c) Consider 1 mo le of an ideal gas at initial pressure of 1.00 atm and initial temperature of

273 .15 K. Assume it expands adibatically against a pressure of 0.43 5 atm until its volume

doubled. Calculate:

i) Work, w

8 M arks)

ii) The final tem perature,

4 Marks)

iii) The internal change of the p rocess,

3 Marks)

Given that Cv = 1 2.47 J/mol.K.

Q U E S T I O N

a) Iron oxide is reduced to iron by hydrogen gas according to the following reaction. By

using the given information, determine w hether this reduction is spontaneous.

Fe 0 3 s) + H g) ? 2Fe s) + 3H 0 g)

12 Marks)

2


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CONFIDENTIAL KC BKG 131411BKF1253b) A stockroom attendant was asked to prepare the following solution:

1.63 g anhydrous sodium carbonate, NaCO dissolved in water and diluted to 200 mL.

Calculate the molarity of this solution.

4 Marks)

c) The ad dition of 0.24 g of sulphur to 100 g of carbon tetrachloride lowe rs the freezing point

by 0.28 C. Determ ine the molar mass of sulphur?

9 Marks)

QUESTION 3

a) The production of benzene through dealkylation process from toluene is an established

industrial activity. In this study, a 150 g of solution is prepared by mixing 40 wt

isopropanol (C H 80) and 60 wt benzene (C ) at a temperature of 100C. Given also,

at the same temperature the vapor pressure of pure benzene is 180.9 kPa and of pure

isopropanol is 74.4 kPa.

b)

i) A m ixture can occu r in either ideal or ideal-dilute conditions. Describe the differences

between ideal and ideal-dilute solutions.

4 Marks)

ii ) For an ideal solution cases, measure the Gibbs free energy (AG,,,,), entropy (AS,,,) and

enthalpy changes A H m .

8 Marks)

b) Magnesium reacts with dilute hydrochloric acid and gives off hydrogen gas. Mg ribbon

with 5cm lengths is added to excess hydrochloric acid and the hydrog en is collected in a

gas syringe. The time taken to collect 10 cm of gas is measured for different

concentrations of the acid. The results are shown in theTable 1.


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CONFIDENTIAL KCIBKG 13141 BKF1253Table 1: The concentration of hydrochloric acid varied with time

Conce ntration of acid mol.dm ) 0.2 0.4 0.6 0.8 1 0

Time to collect 10cm hydrogen s) 60 30 20 15 12

i) State the rate expression of the reaction process.

5 M arks)

ii ) Plot the data to determine the order of rea ction and the rate constant.

8 M arks)

QUESTION 4

a) The nitrogen oxide production through an oxidation process of ammonia occurs

efficiently at 298 C. The che mical reaction of the process is as follows:

02 g) + 4 NH g) -) 6 H 0 I) + 4 NO g)

i) Calculate the standard Gibbs energy change for the oxidation process at 298C.

12 M arks)

ii) Estimate the equilibrium constant, Kofthe process.

5 M arks)

b) One of the criteria that offer response to the equilibrium conditions is temperature.

Interpret the relationship betwee n the equation below with the equilibrium constant,K a t

two different temperature.

G = A H - TArS

8 M arks)

END OF QUESTION PAPER

4


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CONFIDENTIAL KCIBKG 131411BKF1253PPENDIX

Formulas

1 PV=nRT/ nRT I f l2 P=i a l V n b V )

4. T =T I)VJ

5 .\S=nR1nI( V_ L6

surr = qS Urr

sur r

7 AS =C 1n-T i

V8 W = nRT1n-----

i

9 A T = K B b B

1 W=PdV

11 qCAT

12 t\rG

A G products) VL\G reactants)

13 A G n i x =nRT [x 1flX X3 lnxB]

14 PJ=KH[J]

15 AS r n i x

nR [x lflX X8 1 f lX B ]

16 L rG = A r H - ThrS

17. A rG1?T1T

5


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ONFIDENTI L

KCIBKG 131411BKF1253

Tables 1: van der W aals parameter of the gases

Substances a(Pam3moi2) b(105m3mol)Nitrogen, N 0.1408 3 9 1

Argon, Ar 1.337 3.20Xenon, Xe 4.137 5 16Oxygen, 1.364 3.19

Table 2: The ga s constant in various units

R

.314 Jmol K1

8.314 Pam3K1moL

8.314 dm3kPa K moF1

83 45 cm3barK mol

8.206x10 dm3atmK mol

62.364 dm3TorrK mo1

1.987 calK mo11

Table 3: Molecular weight of elements

Element Molecular weight (gmor )

H 1 0 1

16.00

C 12.01Na 22.99

Cl 35 45

N 1 4 0 1

I 126.90

S 32.064

R


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CONFIDENTIAL KCIBKG/13 141JBKF1253Table 4: Thermodyn amic data for organic compound s at 298.15 K

Substance AfH(kJ mor) Sm(JK m o 1) Cp,m (J K mo1

W ater, H 0 (1) -285.83 69.91 75 29 1

H20 (9) -241.82 188.83 33.58

Carbon m onoxide,

Co (g) -110.53 197.67 29.14

Carbon dioxide,

CO2 (9) -393.51 213.74 37.11

Hydrogen,

1 4 (g ) 0 130.684 28.824

H g) 217.97 114.71 20.784

H aq) 0 0

Oxygen

02 g) 0 205.1 29.4

Acetic ac id,

CH3COOH (1) -484.5 159.8 875

CH3COO H (aq) 485 76 178.7

Nitrogen

N2 ( g ) 472 65 153.3 2 9

Nitrogen O xide

N O ( g ) 90.3 210.8 29.8

Ammonia

N}{3 ( g ) -46.1 192.5 35.1

7


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CONFiDENTIAL KCIBKG 13141JBKF1253Table : Factors for unit conversions

Mass 1 kg = 1 000 g = 0 .001 metric ton = 2.20462 ibm

5.27392 oz

1 ibm = 16 oz = 5 x 10 ton = 453 593 g = 0 453593 k g

Length 1 m = 100 cm 000 mm = 106 microns = 1010 angstroms= 39.37 in = 3.2808 ft= 1.0936 yd = 0.0006214 mile1 ft= 12 in

/3 yd = 0.3048 m = 30.48 cm

Volume 1 =1000 liters = 10 6 cm = 106 ml

= 35 3 145 ft3 = 220.83 imperial gallons = 264.17 gal =1056 68 q t1 f

=

1728 in = 7.4805 gal = 0.0283 17 m

= 28.317 liters = 28.317 cm3Force 1 N = 1 kg-mIs =

ynes = 105 g-cmIs = 0.22481 lbf

1 lbf 2.174 ibm-ft/s .4482 N .4482 x 105 dynesPressure 1 atm= 1.01325x 10 5 N/m (Pa)= 101.325kPa = 1.01325 bars=.01325 x106 dynes/cm2= 760 mm H g at00 C (torr) = 10.333 m 1-120 at 4 C= 14 .696 lbf in (psi) = 33 .9ft HO at 4 C

=29.921 in Hg at 0C

Energy 1J = 1 N-rn = 107 ergs = 107 dyne-cm

= 2.778 x 1 W -h = 0.23901 cal

= 0.7376 ft lbf= 9.486 x 10 tUPower 1W = 1 J/s = 2.3901 cal/s .7376 ft-lbffs = 9.486 x 10-4 B tU /s

= 1.341 x 10-3 hp


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CONFIDENTIAL KC/BKGI1314IJBKF1253Table : Integrated rate laws

Order Reaction Rate w t / 20 A * P v=k [A]0/2k

kt = x for O A ]0 1 A P v = k [A ] In 2)/kin[A]

kt[A} x

2 A * P v = k [A ]2 1/k[A]0

kt =

[A] [A] - x )

A+B-*P v = k[A][B]

kr=

[B] 0 [A ]0 [A] 0 x [B]0A + 2B P v = k[A][B]kt= [A]0 [B]0 2 x )[B] 0 - 2[A] 0 [A] 0 - x)[B]0A P v = k [ A ] [ P ]With autocatalysis

kt =

n

[A] 0 [P] 0 x )

[A ]0 [A] 0 x [P]0

3 A + 3B P v = k[A][B]2

kt=

x

2[B] - [A ]0 ) [B] 0 2x)[B]0

2 [ A ] , [B] ) [A] 0_ x [B]n? A * P v=k[A]

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

21_1

kt1

(n-1)k[A]ni l [ A ] 0 - x) [A]

*notes:

= [P] and v

x/dt


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CONFIDENTIAL KCIBKG/13141/BKJF1253Table 7: Henry s Law constant for gases dissolved in water at 25C.

KH / kPa m mol )

Ammonia NH 5 69

Carbon dioxide CO 2.937

Helium He 282.7

Hydrogen H 1 2 8

Methane CH4 75 5

Nitrogen N 1 5 6

Oxygen 02 79.2

Table 8: Boiling-point and freezing-point constants

Solvent Freezingpoint, C

K f

K.kg.mol1Boiling point

K b

K.kg.mol

acetone 95.32 2.40 56.2 1 . 7 1benzene 5 5 5 12 80.1 2 53carbon -2 3 29.8 7 6 5 4.95

tetrachloridephenol 43 7.27 18.2 3.04water 0 1.86 100.0 0.51

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Physical Chemistry past year