physical chemistry past year
TRANSCRIPT
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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
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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
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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
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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
1