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CHE2612 Examination

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UNIVERSITY EXAMINATIONS
January/February 2023
CHE2612
Physical Chemistry II (Theory)
100 Marks
Duration 2.00 hours
First Examiner: Dr BS Dladla
Moderator:
Dr MG Smith
Instructions:
This is a closed book examination.
Use a ball point pen, preferably black.
This question paper consists of 9 pages.
A Resource section appears on page 7.
Name all symbols used.
Use ONLY SI units unless instructed otherwise.
Answer ALL questions. Each question counts 25 Marks.
The use of non-programmable calculator is permissible.
Guidelines for uploading the answer file on myExams can be found on page 9.
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Upon submission remember to tick the Honesty Declaration.
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Question 1 [25]
1.1
Consider the following pressure-temperature phase diagram of a pure substance and identify
by name:
i.
1.2
The intersection at 1.
[1]
ii. The regions 2, 3, and 4.
[3]
iii. The endpoint at 5.
[1]
iv. The lines 6, 7, and 8.
[3]
Show by calculating Gibbs energy of reaction whether evaporation or condensation occurs
spontaneously at 298 K where standard molar Gibbs energy of formation of water vapour is 229 kJ/mol and that of liquid water is -237 kJ/mol.
1.3
1.4
[4]
Consider a gas with a molar volume 1.5 times that of a perfect gas.
a.
Calculate the compression factor.
[3]
b.
Interpret the value obtained in a.
[2]
Explain the following terms:
i.
Incongruent melting.
[2]
ii.
Enthalpy of formation
[2]
iii. Osmosis
[2]
iv. Partial molar volume
[2]
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Question 2 [25]
2.1
Consider two initially separate metal blocks of aluminium (Mr = 26.98 g/mol) and lead
(Mr = 207.19 g/mol) in a chamber with adiabatic walls. The lead block of mass 3 g is initially
at 1000C and that of aluminium of mass 4 g is initially at 250C. The blocks are then placed in
contact and allowed to thermally equilibrate. The molar heat capacities of the metals are
considered constant over the temperature of experiment. At 1 bar, Cp,m(Al) = 24.35 JK-1mol-1
and Cp,m(Pb) = 26.44 JK-1mol-1
Calculate
2.2
a.
the final temperature.
[4]
b.
the change in the entropy of the blocks in contact.
[4]
c.
identify evidence that the process occurred spontaneously or not.
[2]
Consider the following reactions at 298 K.
2 O3(g) → 3 O2(g)
O2(g) → 2 O(g)
NO(g) + O3(g) → NO2(g) + O2(g)
βˆ†π‘Ÿπ‘Ÿ 𝐻𝐻 ⦡ = −427.0 π‘˜π‘˜π‘˜π‘˜/π‘šπ‘šπ‘šπ‘šπ‘šπ‘š
βˆ†π‘Žπ‘Žπ‘Žπ‘Ž 𝐻𝐻 ⦡ = 495.0 π‘˜π‘˜π‘˜π‘˜/π‘šπ‘šπ‘šπ‘šπ‘šπ‘š
βˆ†π‘Ÿπ‘Ÿ 𝐻𝐻 ⦡ = −199.0 π‘˜π‘˜π‘˜π‘˜/π‘šπ‘šπ‘šπ‘šπ‘šπ‘š
Apply Hess’s law to calculate the standard reaction enthalpy for the reaction
2.3
NO(g) + O(g) → NO2(g)
[14]
Choose one of the words in bold to give the sentence the correct scientific meaning.
At constant temperature and pressure, spontaneous change in a process tends to occur in the
direction of increasing/decreasing Gibbs energy.
[1]
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Question 3 [25]
3.1
Consider a state for which the molar volume is 4.00 × 10-4 m3/mol at 288 K. If the van der
Waals parameters a and b are 0.76 Pa m6 mol-2 and 1.26 × 10-4 m3/mol respectively, calculate
the gas pressure in Pa, using i) the perfect gas law and ii) van der Waals equations of state.
3.2
Express your answers in kPa and in 5 significant figures
[4]
Explain the difference in the values obtained in 3.1.
[2]
3.3
The graph in the figure above depicts a temperature-composition solid-liquid phase diagram.
1.
Name the phases in the regions 1-8
[8]
2.
At what temperatures do the pure components solidify.
[2]
3.
Name the line curve marked 9.
[1]
4.
Write the probable formula for the incongruent-melting compound.
[2]
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5.
Write the probable formula for the congruent-melting compound.
[2]
6.
Identify the lines marked 4, 5, and 10 as either peritectic or eutectic.
[3]
7.
Name the phase above line curve marked 9.
[1]
Question 4 [25]
4.1
Calculate total vapour pressure above an ideal liquid mixture for which the vapour
composition of one component is 5.32. The vapour pressures of the pure components at 298 K
are 75 Torr and 21 Torr.
4.2
[7]
Consider 1.5 mol of methane gas in a 32.414 dm3 vessel at 298 K. The van der Waals constants
for methane gas are a /atm L2 mol-2 = 2.273, and b /10-2 L mol-1 = 4.31. The universal gas
constant is R /J mol-1 K-1 = 8.3145.
i. Express each van der Waals constant in SI units.
[2]
ii. Calculate perfect gas pressure of sample in kPa. Express your result in SI units.
[2]
iii. Calculate as van der Waals gas the pressure. Express your result in SI units.
[4]
Express all your answers in 5 significant figures
4.3
A 125 mg sample of butane, (Mr/g mol-1 = 58.13) was burnt in a bomb calorimeter and an
increase in temperature of 3.25 K was observed. βˆ†cH⦡(C4H10, g) = –2878 kJ/mol,
βˆ†cH⦡(C5H12, g) = –3537 kJ/mol at 298,15 K.
Calculate
i. The calorimeter constant.
[6]
ii. The change in temperature upon combustion in the same bomb calorimeter of 125 mg
pentane (Mr/g mol-1 = 72.15)
[4]
Express your answers in 5 significant figures.
END
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RESOURCES SECTION
Van der Waals equation
𝑛𝑛𝑛𝑛𝑛𝑛
𝑉𝑉−𝑛𝑛𝑛𝑛
−
𝑛𝑛2 π‘Žπ‘Ž
𝑉𝑉 2
First law of Thermodynamics
𝑝𝑝 =
Internal energy change
ΔU = n CV,m ΔT
Standard reaction enthalpy
βˆ†π‘Ÿπ‘Ÿ 𝐻𝐻 βŠ– = ∑ πœˆπœˆβˆ†π‘“π‘“ 𝐻𝐻 βŠ– (𝑃𝑃𝑃𝑃) − ∑ πœˆπœˆβˆ†π‘“π‘“ 𝐻𝐻 βŠ– (𝑅𝑅𝑅𝑅)
βˆ†U = q + w
Dalton’s law
βˆ†π‘Ÿπ‘Ÿ 𝐺𝐺 βŠ– = ∑ πœˆπœˆβˆ†π‘“π‘“ 𝐺𝐺 βŠ– (𝑃𝑃𝑃𝑃) − ∑ πœˆπœˆβˆ†π‘“π‘“ 𝐺𝐺 βŠ– (𝑅𝑅𝑅𝑅)
Partial pressure
pi = xi ptot
Standard reaction Gibbs energy of reaction
Compression factor
Reversible work
Irreversible work
Entropy of mixing
Partial molar volume definition
Ideal gas law
𝑝𝑝 = ∑ 𝑝𝑝𝑖𝑖
𝑍𝑍 =
π‘‰π‘‰π‘šπ‘š
π‘œπ‘œ
π‘‰π‘‰π‘šπ‘š
𝑀𝑀 = −𝑝𝑝𝑒𝑒𝑒𝑒𝑒𝑒 (𝑉𝑉2 − 𝑉𝑉1 )
βˆ†π‘šπ‘šπ‘šπ‘šπ‘šπ‘š 𝑆𝑆 = −
βˆ†π‘šπ‘šπ‘šπ‘šπ‘šπ‘š 𝐺𝐺
𝑇𝑇
πœ•πœ•πœ•πœ•
οΏ½
πœ•πœ•π‘›π‘›π‘‹π‘‹ 𝑝𝑝,𝑇𝑇,𝑛𝑛
𝑉𝑉𝑋𝑋 = οΏ½
pV = nRT
Maximum work at constant temperature
βˆ†A = βˆ†U - Tβˆ†S
Reversible expansion entropy
βˆ†π‘†π‘† = 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑉𝑉𝑓𝑓
Entropy variation with temperature
at constant pressure
Temperature dependence of enthalpy
Universal gas constant, R
Critical compression factor
Reduced variables
Osmometry
Total volume
Composition of vapour in equilibrium
with an ideal liquid mixture
𝑉𝑉2
𝑉𝑉1
𝑀𝑀 = −𝑛𝑛𝑛𝑛𝑛𝑛 ln
𝑉𝑉
π‘Œπ‘Œ
𝑖𝑖
𝑇𝑇 𝐢𝐢𝑝𝑝
𝑆𝑆�𝑇𝑇𝑓𝑓 οΏ½ = 𝑆𝑆(𝑇𝑇𝑖𝑖 ) + ∫𝑇𝑇 𝑓𝑓
𝑖𝑖
𝑇𝑇
𝑑𝑑𝑑𝑑
𝑇𝑇
βˆ†π»π» = ∫𝑇𝑇 𝑓𝑓 𝑛𝑛 𝐢𝐢𝑝𝑝,π‘šπ‘š 𝑑𝑑𝑑𝑑
𝑖𝑖
8.3145 J/K mol
𝑍𝑍𝑐𝑐 =
π‘‰π‘‰π‘Ÿπ‘Ÿ =
Π
π‘π‘π‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘š
𝑝𝑝𝑐𝑐 𝑉𝑉𝑐𝑐
𝑅𝑅𝑇𝑇𝑐𝑐
π‘‰π‘‰π‘šπ‘š
𝑉𝑉𝑐𝑐
=
=
𝑅𝑅𝑅𝑅
𝑀𝑀
3
8
+
π‘π‘π‘Ÿπ‘Ÿ =
𝑝𝑝
𝑝𝑝𝑐𝑐
𝐡𝐡𝐡𝐡𝐡𝐡
𝑐𝑐
𝑀𝑀 2 π‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘šπ‘š
π‘‡π‘‡π‘Ÿπ‘Ÿ =
𝑇𝑇
𝑇𝑇𝑐𝑐
𝑉𝑉 = 𝑛𝑛𝑀𝑀 𝑉𝑉�𝑀𝑀 + 𝑛𝑛𝑁𝑁 𝑉𝑉�𝑁𝑁
π‘₯π‘₯ 𝑝𝑝∗
𝐴𝐴
𝑦𝑦𝐴𝐴 = 𝑝𝑝∗ +�𝑝𝑝𝐴𝐴∗ −𝑝𝑝
∗
𝐡𝐡
𝐴𝐴
𝐡𝐡 οΏ½π‘₯π‘₯𝐴𝐴
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Conversion factors
1 atm
101.325 kPa
1 litre
1 dm3
1 bar
105 Pa
Thermodynamic data
Gravitational acceleration for free fall, g
9.80665 m/s2
Liquid water density at 293 K
0.997 g/cm3
At 298 K
H2(g)
βˆ†π‘“π‘“ 𝐻𝐻 βŠ– /kJ mol-1
0
βˆ†π‘“π‘“ 𝐺𝐺 βŠ– /kJ mol-1
O2(g)
0
0
H2O(l)
-285.83
-237.13
0
Temperature variation of molar heat capacities,
𝐢𝐢𝑝𝑝,π‘šπ‘š
(𝐽𝐽𝐾𝐾−1 π‘šπ‘šπ‘šπ‘šπ‘šπ‘š −1 )
a
N2(g)
28.58
𝑐𝑐
= π‘Žπ‘Ž + 𝑏𝑏𝑏𝑏 + 𝑇𝑇 2
b / (10-3 K-1)
c / (105 K2)
3.77
– 0.50
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