Physical Chemistry I
Working Book
for the first-year bachelor course Physical Chemistry I
Albert P. Philipse, Sofie Ferwerda, Matthijs Alting, Marlous Kamp
Edition November 2024
Contents
Exercises
4
Chapter 1
A First Encounter with the First and Second Law
5
Chapter 2
Thermodynamic States and Ideal Molecules
7
Chapter 3
The First Law: the Total Energy Cannot Change
10
Chapter 4
The Second Law: the Total Entropy Cannot Decrease
15
Chapter 5
The Gibbs Energy
21
Chapter 6
Liquid-Gas Equilibria
27
Chapter 7
Osmosis and Osmotic Pressure
32
Chapter 8
Redox Equilibria
37
Chapter 9
The Chemical Potential
42
Chapter 10
Electrolyte Solutions
45
Chapter 11
Diffusion in Liquids and Gases
48
Chapter 12
Kinetic Theory
51
Answers
55
Chapter 1
A First Encounter with the First and Second Law
56
Chapter 2
Thermodynamic States and Ideal Molecules
59
Chapter 3
The First Law: the Total Energy Cannot Change
62
Chapter 4
The Second Law: the Total Entropy Cannot Decrease
66
Chapter 5
The Gibbs Energy
73
Chapter 6
Liquid-Gas Equilibria
77
Chapter 7
Osmosis and Osmotic Pressure
83
Chapter 8
Redox Equilibria
89
Chapter 9
The Chemical Potential
92
Chapter 10
Electrolyte Solutions
96
Chapter 11
Diffusion in Liquids and Gases
98
Chapter 12
Kinetic Theory
101
2
Preface
In the following chapters, several exercises are given per subject. Each chapter in this
working book corresponds to the same chapter in the Lecture Notes.
Some of the chapters consists of exercises of series A and B, relating to the first and second
lecture on that topic, respectively.
The exercises of each chapter are divided into three categories: theoretical exercises,
math-related exercises and review Exercises. The theoretical exercises cover the contents
of the corresponding lecture, and chapter from the lecture notes. Math-related exercises
involve computations, derivations of equations, numerical calculations etcetera. Review
exercises, addressing topics treated in previous lectures and literature, are provided to
further increase your understanding of them.
In the Answers Section you will find either brief answers or worked-out solutions to all
exercises. To get the most out of the Physical Chemistry tutorials, you are strongly
recommended to first tackle exercises yourself before taking a look at answers.
We would like to thank Dr Ben Erné for sharing his database of exercises and answers with
us.
We cannot hope to have escaped errors in a document so packed with formulas, numbers,
calculations etcetera – though we have done our best to avoid them. For any corrections
or suggestions, we will be very grateful.
The authors
Utrecht, October 2023
3
Chapter 1 A First Encounter with the First and Second Law
Theoretical Exercises
1.1
a) What is – in thermodynamics- the meaning of a spontaneous reaction?
b) Explain the difference between an exothermic and an endothermic reaction.
1.2
a) Explain the difference between mass and weight.
b) Explain why weight cannot be a conserved quantity.
1.3
a) Give the reaction equation for combustion of methane.
b) The reaction leads to an energy release of 604.5 kJ per mole of methane. Calculate
the decrease in mass (in nano-gram) due to this release. Hint: see Lecture Notes
(LN) page 15.
1.4
a) What is the difference between a perpetuum mobile of the first and second kind?
b) Take a look at LN Figure 1.6; can you think of a reason why the boat indeed cannot
cycle by heat exchange with the quay, as described in the legend to Figure 1.6?
1.5
The sun (see also LN Figure 1.2) is a gigantic ball of hydrogen and helium gases. The
surface temperature of the gas ball is about 6000 K; the core temperature is 15 · 106
K. Its large mass compresses the gas sphere so much that a chain of nuclear-fusion
reactions is triggered, in which four H-atoms fuse to one He-atom. Hydrogen and
helium nuclei have a molar mass of, respectively, MH = 1.0078 g mol-1 and MHe =
4.0026 g mol-1.
a) Write down the net reaction equation for this fusion.
b) Calculate how much energy is released by the fusion of four moles of H.
c) Argue whether or not the energy production in b. violates the First Law.
1.6
Let us consider a few standard properties of water:
a) What is the molar mass of H2O?
b) What is the concentration of H2O (in units of mol dm-3) in water?
1.7
A certain amount of liquid has a mass m of 40 g and a volume V of 0.050 dm3.
Calculate the density ρ of this liquid. Express your answer in SI-units.
1.8
Argue if the molar weight of ATP (adenosine triphosphate, a molecule used for storage
and transport of energy in the cell) is similar at the moon and on earth.
4
Math-related Exercises
1.9
a) Evaluate:
1
1 x2 dx
21
2. 
dx
1 x
2 dV
3. 
1 V
1.
2
4.
5.
1
a
0
 e dx
− ax
4
 cdy ; c = constant
2
𝑉+∆𝑉
6. ∫𝑉
𝑝𝑑𝑉′ ; 𝑝 = constant
b) Simplify:
1. ln A + ln B
5. ln e-2
2. ln A – ln B
6. ea·eb
B

 A
−1
3. − ln 
7.√e2
4. ln e
8. √e6
3
𝜕𝑓
1.10 Calculate ( ) for the following functions:
𝜕𝑥 𝑦
a) 𝑓(𝑥) = 𝑥 4
f) 𝑓(𝑥) = ln (𝑥)
b) 𝑓(𝑥) = 𝑥 𝑦
g) 𝑓(𝑥) =
c) 𝑓(𝑥) = 𝑒 𝑥
h) 𝑓(𝑥) = 𝑥𝑦
d) 𝑓(𝑥) = 𝑒
i) 𝑓(𝑥) = 𝑥 2 𝑦 2
𝑔(𝑥)
4
e) 𝑓(𝑥) = 𝑒 𝑥 +𝑥+𝑦
j) 𝑓(𝑥) =
1
𝑥
𝑦
𝑥
1.11 Evaluate the indefinite integral ∫ 𝑓(𝑥)d𝑥 for the following functions:
a) 𝑓(𝑥) = 𝑥 𝑛
b) 𝑓(𝑥) = 4𝑥 3
c) 𝑓(𝑥) = 𝑒 𝑥
d) 𝑓(𝑥) = 𝑒 𝑔(𝑥) 𝑔′(𝑥)
3
e) 𝑓(𝑥) = 3𝑥 2 𝑒 𝑥 +𝑎 with 𝑎 is a constant
f) 𝑓(𝑥) =
1
𝑥
g) Verify that ∫ 𝑢(𝑥)𝑣 ′ (𝑥)d𝑥 = 𝑢(𝑥)𝑣(𝑥) − ∫ 𝑢′ (𝑥)𝑣(𝑥)d𝑥 for 𝑢(𝑥) = 4𝑥 and
𝑣(𝑥) = 𝑥 2
1.12 Consider the constants 6.02214 · 1023 mol-1 and 1.38066 · 10-23 J K-1. How are they
usually called? Evaluate their product; how is that product usually called?
1.13 Which manipulations with a force F (in Newton, N) and a distance s (in meter, m) are
not allowed? (NB: ds represents an infinitesimal change in s)
a) F + s
d) ln(F/s)
g) ∫ 𝐹 𝑑𝑠
j) √𝐹/𝑠
b) F · s
e) F/s
h) F/ds
k) sin(2F/s)
c) exp(F · S)
f) (F + 2S)2
i) FdS
l)
5
 F dS
2
1.14 For the permitted cases in Exercise 1.13, specify the resulting units. For which case(s)
is the result energy in Joule?
1.15 Convert the following units:
1Pa = ..Nm-2
1mL = ..L
1atm = ..Pa
1J = ..Nm
1cm = ..L
1atm = ..bar
1mm = ..km
1bar = ..Pa
3
1 lightyear = ..km
1 nanoton = .. mg (nano = 10-9)
1.16 An ideal gas with initial pressure p expands form volume V1 to a final volume V2.
a) Evaluate the integral ∫ 𝑝𝑉 where pV=C
The product pV for the gas remains a constant C
b) What is the unit of the constant C?
c) Calculate C for one mole of ideal gas at 25 °C.
1.17 Given is the quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0. What are the formulas for the two
solutions for x?
1.18 One heartbeat requires an amount of energy or 1 J. Suppose the hydrolysis of
adenosine triphosphate (ATP) generates 31 kJ mol-1, calculate the amount of ATP
needed to make your heart beat for one day. Assume a constant heartbeat of 60
beats per minute.
1.19 Calculate the pressure exerted on a surface by a person (who has a mass of 65 kg),
by standing on
a) shoes with a total surface area of 250 cm 2;
b) ice skates with a total surface area of 2.0 cm 2.
Express your answers in units of atm, Nm-2 and Pa.
1.20 Calculate the pressure (both in units of Pa and atm) exerted by a cylindrical column
with height h = 10 m of:
a) water, mass density 1.0 g mL-1 and
b) mercury, mass density 13.5 g mL-1
6
Chapter 2 Thermodynamic States and Ideal Molecules
Theoretical Exercises
2.1
Are you breathing an ideal gaseous mixture at the moment?
2.2
Calculate the molar volume of an ideal gas at temperature T=298 K and pressure
p=1 bar
2.3
Assume atmospheric air is an ideal gas mixture of 80 vol% N2 and 20 vol% O2.
a) Calculate the mass of one liter of air at T=298 K and a total pressure of p=1 bar.
Assume density of N2 and O2 gas are, respectively, 1.17 kg m-3 and 1.29 kg m-3.
b) Now assume you do not know the mass densities of N2 and O2 gas, but that the
ideal gas law applies. Calculate the mass of one liter of air by using Dalton’s law.
What do you conclude regarding the validity of the ideal gas law?
2.4
Calculate the molar volume of methane at 330 K and 100 bar.
2.5
One mole of ideal gas occupies 22.414 L at 1 atm and 273.15 K. Evaluate the gas
constant R (in units of J mol-1 K-1).
2.6
A glucose solution contains one weight percent of glucose (M=180 g mol-1).
Calculate the osmotic pressure the solution exerts against pure water (T=25 oC).
2.7
A solution of 20 g hemoglobin in 1 L water exerts an osmotic pressure of π=763 Nm2
2.8
at 298 K. What is the molar mass of hemoglobin?
In an isolated box at fixed temperature T a movable piston separates a gas with
pressure p1=1 bar and volume V from a gas with pressure p2=2 bar and equal
volume V. Calculate the pressure at mechanical equilibrium.
2.9
Round marbles that have been poured in a vessel form a “random sphere
packing” in which the marbles occupy a fraction φ= 0.64 of the vessel volume.
a) Calculate the molar volume (in km3 mol-1) of randomly packed marbles with a
diameter of d=1 cm.
b) A cube with side length L contains a random packing of one pico (10-12) mol of the
marbles from a). Calculate L in m.
Math-related Exercises
7
2.10 A certain amount of gas has a volume of 100 cm3 with a pressure 80 kPa and a
temperature of 310 K. Calculate the volume of the gas at a temperature of 0 °C and
1 bar pressure.
2.11 A gaseous mixture consists of 2 g hydrogen (H2), 2 g helium (He) and 16 g oxygen
(O2). The total pressure equals 12.0 kPa. Calculate the partial pressure of hydrogen
for this mixture. Give your answer in units of kPa.
Data: Mw(H2) = 2.0 g mol-1, Mw(He) = 4.0 g mol-1 and Mw(O2) = 32.0 g mol-1
2.12 A certain gaseous mixture consists of methane (CH 4) and oxygen (O2) at standard
temperature (298.15 K) has a total pressure of 1 bar and has the same density as
pure nitrogen (N2) at standard temperature and at 1 bar. Assume that CH4 and O2 do
not react with each other. Calculate the mole fraction of O2 in the mixture.
2.13 Calculate the air pressure needed to burn 1.5 g liquified ethanol (C 2H5OH) with the
exact amount of oxygen necessary in a vessel of 1 dm3 at T=303 K. Neglect the
volume of the liquid ethanol and assume that air consists of 21vol% of oxygen.
Data: Mw(C2H5OH) = 46.0 g mol-1, Mw(O2) = 32.0 g mol-1
2.14 One mixes bacteria species Micrococcus denitrificans in water without oxygen, where
0.100 mol KNO3 and an excess of an oxidation organic compound is dissolved. This
mixture (which has a volume of 200 cm3) is transferred to a 2.000 dm3 flask. This
flask is filled with pure nitrogen gas and sealed. Under these circumstances (without
oxygen and in presence of an oxidation agent), these bacteria react with nitrogen
according to:
2 H2O + 4 NO3- → 4 OH- + 5 O2 + 2 N2
(2.1)
Assume that all nitrogen has been converted according to reaction (2.1). Calculate
the partial pressure of nitrogen after this reaction. Given are the following data: the
initial partial pressure of nitrogen is 101 kPa; the temperature has kept constant at
303 K. Neglect the change in volume and the solubility of nitrogen.
8
Review Exercises
2.15 Balloon flight
a) Your spherical balloon has a radius of r = 3.0 m. How much H2 gas (in moles) is
needed to blow up a balloon to a pressure of 1.0 bar, at sea level at a temperature
of 25 °C?
b) How large is the mass that the balloon can lift at sea level? (air mass density at
sea level: ρ = 1.22 kg m-3).
c) How large is this mass when helium (He) gas is used instead of hydrogen?
d) A lecturer claims that for the balloon volume and temperature in exercise a), it is
possible to achieve a balloon pressure of 1.0 bar, with less hydrogen than you have
calculated in a). He argues that it is just a matter of finding a more efficient way to
fill the balloon. Why do you (dis)agree with this claim?
9
Chapter 3 The First Law: the Total Energy Cannot Change
Theoretical Exercises
3.1
The total energy of an isolated system
A is always constant
B always increases
C always decreases
D may increase as well as decrease
3.2
A spontaneous reaction
A is always exothermic
B is always endothermic
C occurs without any external assistance
D is by definition a fast reaction
3.3
Water is evaporating from an open vessel at room temperature. This evaporation is
A exothermic and spontaneous
B endothermic and spontaneous
C exothermic and not spontaneous
D endothermic and not spontaneous
3.4
A non-spontaneous reaction
A is always endothermic
B needs a catalyst to occur
C is by definition a slow reaction
D can only occur with external assistance (work)
3.5
What are reversible processes? Do they occur in nature or technology? Why are these
processes nevertheless very important in thermodynamics?
3.6
Formulate the First Law for a closed system; explain the terms in it.
3.7
Explain the difference between heat and temperature.
3.8
What are the three types of thermodynamic systems? To which type belong:
a) a perfect thermos bottle;
b) a human being;
c) the earth;
d) a closed reaction tube?
3.9
Why is heat no state function?
10
3.10 Somebody claims that by driving a car reversibly, one can extract energy from
gasoline with maximum efficiency. Argue whether you agree with this person.
3.11 The equilibrium 3 NO ® N2O + NO2 has an equilibrium constant K = exp(42). What
can you conclude about the equilibrium composition?
3.12 Give at least two examples of:
a) an endothermic, spontaneous reaction
b) an exothermic, spontaneous reaction
c) an endothermic, non-spontaneous reaction
d) an exothermic non-spontaneous reaction
e) a spontaneous process that is neither exothermic nor endothermic
f) a non-spontaneous process that is neither exothermic nor endothermic
3.13 A nail of zinc is immersed in acetic acid. What will happen? Give the reaction equation.
How could we extract work from this reaction? What type of work?
3.14 A teacher claims that a suitable equation of state for a gas is p = nRT/V2, for n moles
of gas. Can this equation be correct? Why?
3.15 Specify in the following cases the thermodynamic system. Specify the nature of the
energy exchange with the surroundings and explain the sign of q or w.
a) A discharging battery that heats up an external resistance.
b) A ball that bounces on the floor and gradually comes to rest.
c) A balloon filled with ideal gas that reversibly expands in open air at constant
temperature.
3.16 a) Show that for the reversible, isothermal expansion of 𝑛 mol of ideal gas from
volume V1 to V2:
𝑉
𝑤𝑟𝑒𝑣 = −𝑛𝑅𝑇𝑙𝑛 1
𝑉2
b) Calculate wrev for the case V2 = 2V1, n=1 mol and T=298 K (or to be precise 1.0000
mol and 298.15 K). What is the dimension of wrev and what is its sign, and why?
c) How large is the heat exchange q? Why?
11
3.18 One liter of liquid water at 25 °C is converted to steam of 500 oC at p=1 bar.
Data: Cp (steam, at 1 bar and 100 °C)1 = 2000 J kg-1 K-1 ;
∆Hvap at 1 bar = +2258 kJ kg-1
Cp (liquid water) = 4180 J kg-1 K-1
a) What is the volume of the steam?
b) How much energy is released in the form of heat when the steam is allowed to
condense on a surface with a temperature of 25 °C?
3.19 The external pressure on an ideal gas with volume V=1 L under a movable piston is
increased from 1 to 2 bar at a constant temperature of T=298 K. Calculate ∆H for
the gas due to the pressure increase. Conclusion?
3.20 An example of a strongly exothermic process is the so-called thermite reaction:
2Al + Fe2O3 → Al2O3 + 2Fe
;
∆H = –851 kJ mol-1
The temperature due to this reaction may rise to about 2000 °C: enough to melt the
reaction product iron which has a melting temperature of 1500 °C. Because of this
enormous heat production the thermite reaction is used for the welding of steel and
–
unfortunately
–
also
for
the
fabrication
of
incendiary
bombs.
a) Calculate how much heat is produced when 125 gram of iron oxide powder reacts
−1
with excess aluminum powder. (MFe2O3 = 159.68 g mol ). Which assumption do you
have to make?
b) How many liters of water can be heated from 25 to 100 °C from the conversion of
one kilo of iron oxide in the thermite reaction? Assume that the heat capacity of liquid
water, Cp = 4.18 kJ kg-1 K-1, is constant.
3.21 For the dissociation of acetic acid:
CH3COOH → CH3COO- + H+
; ∆H = –0.25 kJ mol-1
Assuming the reaction has come to equilibrium, argue how the pH of a dilute acetic
acid solution will change upon increasing the solution’s temperature.
3.22 a) Calculate the isochoric heat capacity of one mole of ideal monatomic gas at
T=298 K.
b) Experimental values are CV = 12.47 J K-1 for argon, and CV = 21.05 J K-1 for
oxygen. What do you infer from these values?
1
This is an estimate: Cp depends on temperature, which we ignore here.
12
𝜕𝑉
3.23 a) Calculate for one mole ideal gas the partial derivative (
) for p=1, 2 and 3 bar.
𝜕𝑇 𝑝
b) What is the physical meaning of this derivative?
Math-related Exercises
3.24 Calculate the work w needed to lift your body to a height of h=10 m.
3.25 Show that for an ideal gas molecules and ideal solute molecules in solution:
æ ¶U ö
çè ÷ø = 0
¶V T
3.26 Show that ∆H = 0 J for any isothermal change of n moles of an ideal gas.
3.27 Differentiate the following 2:
a)
(1- 4x ) cos(x)
2
d) x ln(x)
g) ln(2x
3.28 Find
2
- 3x +1)
1+ 2x + 3x 2
b)
3 + x3
ln(x)
e)
x
c)
( 2 + 3x ) e x
f)
2 + x2
2 2 x 2 +3
h) x e
dV
at constant T and n for the following equations of state (assume a, b and B
dp
are constants)2.
æ
è
b) pV = nRT ç 1+
a) pV=nRT
(
)
c) p V - nb - nRT = 0
d)
nB ö
÷
Vø
æ
n2a ö
p
+
(V - nb) = nRT
çè
V 2 ÷ø
2
Adapted from E. Steiner (2008), The Chemistry Maths Book, Oxford University Press, USA,
second edition.
13
Review Exercises
3.29 The pressure p above the earth surface (z = 0 m) decreases by increasing z. Assume
that this decrease is described by the first-order differential equation:
d𝑝
d𝑧
=−
𝑀𝑔𝑝
(3.1)
𝑅𝑇
Here, g is the gravitation acceleration, M the molar mass of the gas, T the
temperature and R the ideal gas constant.
a) Evaluate (1) to obtain an expression for p as function of z. Use the boundary
condition p(z = 0) = p0.
b) What is the potential energy U for a gas molecule at height z?
3.30 The reversible work wrev by changing the volume of a gas is given by:
(3.2)
𝑤𝑟𝑒𝑣 = − ∫ 𝑝d𝑉
a) Calculate wrev for doubling the volume of ideal gas with an initial volume of 2.0
dm3 at constant temperature. (You can set nRT in the expression). What is the sign
of wrev and why?
b) Suppose that the gas in a) does not behave as an ideal gas and it complies instead
with the state function:
𝑉
(3.3)
𝑝 ( − 𝑏) = 𝑅𝑇
𝑛
where b is a constant with the value b = 0.2 dm3 mol-1. You may further assume n=1
mol. Calculate wrev for doubling the volume of this gas.
3.31 Hundred liters of water are stored at a height of h = 10 m; an empty bathtub is
located on the floor at h = 0 m.
a) Calculate the potential energy of the water due to its height.
b) Calculate the kinetic energy of the falling water at the moment it enters the
bathtub.
c) Calculate the total amount of molecular kinetic energy in 100 L of water at T =
298 K.
d) Calculate the height h from which 100 L of water has to fall such that it enters the
tub with a kinetic energy that equals its total molecular kinetic energy.
14
Chapter 4 The Second Law: the Total Entropy Cannot Decrease
A) Theoretical Exercises
4.1
A reversible reaction
A always increases the total entropy
B can never deliver heat
C does not change the total entropy
D can change the total energy
4.2
A system or reaction in equilibrium
A cannot deliver any type of work
B slowly increases the total entropy
C may deliver non-volume work
D slowly decreases the total entropy
4.3
A refrigerator
A cools spontaneously
B always operates in a reversible manner
C is in equilibrium with its surroundings
D increases the entropy of its surroundings
4.4. In the formula ΔS = q/T , q must be the
A heat–exchange in a spontaneous process.
B reaction heat at constant pressure
C heat exchange in a reversible process
D heat production by an isolated system
4.5. The entropy of an isolated system
A always increases
B always decreases
C always remains constant
D cannot decrease
4.6. An irreversible reaction
A occurs spontaneously
B is always exothermic
C reduces the total entropy
D is a sequence of equilibrium states
4.7. True or false? Briefly motivate your choice; remember that statements of the form
´it
is
always
the
case
that
x’
are
refuted
by
one
counter
example.
a) In an ideal gas the molecules do not exert a force on the wall of the gas container.
b) In an isolated system the entropy is always constant.
c) Work performed by a gas on the surroundings is always negative.
15
d) The enthalpy is always a state function.
e) Biking is a reversible process.
f) Water can never freeze spontaneously because the entropy of ice is lower than
that of liquid water.
g) A spontaneous process is always exotherm.
h) Avogadro’s number has no unit.
i) A mole fraction can never be larger than one.
A) Math-Related Exercises
4.8
In an isolated system n moles ideal gas expands at constant temperature from
volume V1 to volume V2 (see figure 4.1).
a) Show that the total change in entropy for this expansion is given by
V
DS = nR ln 2
V1
(4.1)
Figure 4.1
b) What is the sign of ∆S? What is its physical meaning?
c) Show that for an ideal gas, equation (4.1) is equal to:
DS = -nR ln
p2
p1
(4.2)
d) A teacher claims for an gas expansion in an isolated system that:
“By definition, a change in entropy is given by
dS =
dqrev
T
(4.3)
An isolated system cannot exchange any heat with its surroundings so at every
moment during the expansion dS = 0. Hence, the total change in entropy should be
equal to ∆S = 0, so equation (4.3) is wrong.”
Comment on this statement. Show which parts are correct and which are wrong.
4.9
3
æ ¶U ö
= nR
÷
è ¶T ø V 2
a) Show that for an ideal gas: CV = ç
16
b) Show that for an ideal gas: Cp = CV + nR
c) Why does the isobaric heat capacity exceed the isochoric heat capacity?
4.10 Derive Carnot’s theorem from DS =
qrev
T
4.11 Calculate the volume of 1.000 mole ideal gas at 1.000 bar and 10.00 °C.
4.12 Calculate the change in entropy in which 100.0 J of heat is added reversibly to a
system at 20.00 °C
4.13 An ideal gas has been put into a balloon at 20.00 °C and 1.000 atm external pressure.
The volume increases from 1.000 L to 3.000 L. Calculate the absolute value of the
volume-work.
4.14 Calculate the change in entropy for a system with heat capacity of 12.0 J K -1 which
is cooled down from 20.00 °C to 10.00 °C.
B) Theoretical Exercises
4.15 Explain why ΔU = 0 J for the isothermal expansion of an ideal gas. Does it make any
difference whether the expansion is reversible or spontaneous?
4.16 For the irreversible, isothermal expansion of an ideal gas against vacuum:
A ΔU < 0 J
B ΔU > 0 J
C ΔU = 0 J because q = 0 J and w = 0 J
D ΔU = 0 J because w = -q
4.17 For a reversible, isothermal expansion of an ideal gas, which statement is true for
the enthalpy change ΔH ?
A ΔH < 0 J
B ΔH > 0 J
C ΔH = 0 J because ΔU = 0 J and Δ(pV) = 0 J
D ΔH = 0 J because enthalpy is a state function
4.18 The entropy production σ in the Clausius inequality
A is zero for a system at equilibrium
B is negative for a reversible process
C equals the reversible heat exchange
D is positive only for an exothermic process
4.19 If heat is added to a system
A the temperature always increases
B the temperature sometimes stays constant
17
C the systems entropy must decrease
D this heat must be counted negative
4.20 Indicate for the following parameters if they are a state variable or not:
a) Heat
b) Temperature
c) Pressure
d) Work
e) Volume
4.21 An “erg’’ is an old-fashioned unit for energy (1 erg = 1 g cm2 s-2). Convert 1.0 J into
erg.
B) Math-Related Exercises
4.23 A balloon has at T=20.00 °C a volume of 1.000 L and a pressure of 1.000 atm. By
pressing the balloon against a wall the volume will be halved reversibly at 20.00 °C.
a) Calculate the work which will be performed on the balloon.
b) Calculate the change in entropy.
c) Calculate the change in enthalpy.
4.24 a) Calculate the changes in entropy of a block of iron when 25 kJ as heat is added to
it via a reversible, isothermal process at 1) 0 °C and 2) 100 °C.
b) How are the outcomes for a block of copper?
c) A lecturer claims that according to the Second Law “∆S = 0 for any reversible
process so your answers are incorrect”. What is wrong with the lecturer’s
assessment?
4.25 a) Calculate ΔU, q, w, ΔH and ΔS for an isolated vessel containing two blocks of
copper (one with temperature 100 °C and the other one with temperature 0 °C),
which are brought in close contact. The heat capacity of copper is 0.385 J K-1 g-1.
Assume that this value does not depend on temperature and that CV = Cp. The blocks
have a mass of 10 kg each.
b) What can one conclude from the sign of ΔS?
4.26 The reversible volume work wv,rev could be evaluated by a volume change of the
system from V1 to V2 against the total external pressure pex, which equals the gas
pressure p.
a) Show that an expression for the reversible volume work for this volume change
is given by:
wv, rev = nRT ln(3)
(4.4)
for the case V1 = 3V2.
Another situation is described by the transfer of n moles of an ideal gas from pressure
p to pressure p*. In this situation, we consider two liquid samples, one of a pure
18
solvent with a vapor pressure of pure solvent p* and that of a solution, containing
that solvent, at the same temperature with vapor pressure p. This situation is
described in figure 4.2.
Figure 4.2 Left: a pure solvent in equilibrium with its vapor pressure p*, Right: a solution
generates a lower equilibrium vapor pressure p < p*. The work done in step D to transfer one
mole of solvent to a solution is the opposite of the work done to increase in step I in which the
pressure of one mole of vapor is raised from p to p*.3
In this question, we are only interested in the transfer of one mole of ideal gas from
the solution with vapor pressure p to the pure solvent with vapor pressure p*. Assume
equal temperature in both systems.
b) Show that the reversible work is given by:
æ p* ö
w = nRT ln ç ÷
è pø
(4.5)
[Hint: a standard integral is given by: ∫ 𝑑𝑥 −1 = ∫ −𝑥 −2 𝑑𝑥
]
c) If p* > p, should work be performed on or by the system? Explain.
4.27 Given two vessels with volumes VA and VB containing two ideal gases A and B,
respectively. When mixing two gases, the total entropy increases due to the mixing
of the gases. The absolute value of the mixing entropy depends on the ratio between
the volumes VA and VB. Assume xA = xB.
a) Derive equation (4.29), starting from (4.21) in the Lecture Notes.
b) Assume VA = VB and calculate ∆Smix
b) Assume VA = 10VB and calculate ∆Smix
c) Comment on the difference in ∆Smix in a) and b).
4.28 Given three vessels with volumes VA, VB and VC, containing three ideal gases A, B and
C, respectively, separately. Suppose VA = V, VB = 2V and VC = 3V and
nA = nC =
1 mol and nB = 2 mol, derive an expression for ΔSmix by mixing these gases.
3 Adapted from A. P. Philipse, Brownian Motion: Elements of Colloid Dynamics, Springer Nature,
2018
19
Review Exercises
4.29 Given are the heat capacities for ice and water [2.2 and 4.18 J K-1 g-1 respectively;
and they do not depend on temperature] and the melting enthalpy of ice ΔHmelt =
332 J g-1.
Calculate the change in entropy ΔS by freezing water at –10 °C of 1 mole cooled
water of –10 °C.
HINT: This is an irreversible process, which can be calculated via a reversible path
for the freezing of water at 0 °C, followed by changing the temperature from 0 to –
10 °C reversibly.)
Data: Mw(H2O) = 18.0 g mol-1
4.30 For an ideal gas, the ideal gas law is
(4.6)
𝑝𝑉 = 𝑛𝑅𝑇
Here n is the number of moles and R the gas constant, 8.314 J mol
-1
K . This law is
-1
valid for a certain amount of gas per mole. However, there is an analogous expression
for the ideal gas law per number of molecules, which is
(4.7)
𝑝𝑉 = 𝑁𝑘𝐵 𝑇
Here N is the number of molecules and kB Boltzmann’s constant, 1.381 · 10-23 J K-1.
Derive expression (4.7) from expression (4.6). Do you have to make any additional
assumptions in comparison to the derivation of (4.6) as done in the lectures?
20
Chapter 5 The Gibbs Energy
5A) Theoretical Exercises
5.1
Which statement is true. The Gibbs energy change ΔG
A always refers to the system
B always refers to the surroundings
C always refers to the total change of G in system plus surroundings
D always increases
5.2. The Gibbs energy is defined as G = H – TS. Show that for an isothermal process dG
= dH –T dS and ΔG = ΔH – T ΔS. What is the difference between ‘d’ and Δ?
5.3
The following reaction takes place at constant pressure and T=298 K:
SO2 + 1/2 O2  SO3 ; H = ?
(5.1)
For the same p and T the following enthalpies of formation are known:
S + O2  SO2 ; H SO = −17 kJ mol-1
(5.2)
S + 3/2 O2  SO3 ; H SO3 = −23 kJ mol-1
(5.3)
2
a) Calculate H for reaction (5.1). Clearly explain your method.
b) Enthalpy values are given in J mol-1. Per mole of what?
c) Suppose reaction (5.1) has reached equilibrium. How large is in that case S of
the reaction?
d) How large is Stot for this equilibrium? Explain any difference with S from c).
5.4
a) Calculate S and G for the evaporation of 3.00 mol benzene at its boiling point of
Tb=80.1 °C at p=1 atm. The molar enthalpy for this evaporation at the boiling point
is +30.80 kJ mol-1.
b) Comment on the sign of the entropy change.
c) Do you think it is relevant that the pressure is specified to be p=1 atm?
5.5 A galvanic cell operates at T=298 K and delivers electrical work
we
= -118.3 kJ to the surroundings. The cell produces 24.6 kJ of heat while the
surroundings perform 1.9 kJ of volume work on the cell.
a) What is a galvanic cell?
b) Assume that all mentioned processes are reversible, isothermal and isobaric.
Briefly explain what these terms mean.
c) Calculate the changes in U, H, S and G of the cell.
d) Compare sign and magnitude of G to we. Conclusion?
5.6
True or false:
21
a) Gibbs energy has the same unit as heat.
b) ΔG < 0 J is the criterion for spontaneity at given p, T.
c) The total Gibbs energy is not always constant.
d) ΔG = 0 J for every reversible process.
e) ΔG > 0 J applies to every non-spontaneous reaction at given p, T.
f) ΔG > 0 J implies that the total entropy decreases.
5A) Math-related Exercises
5.7. One mole ideal gas expands at 27 °C isothermal and reversible from p=10 bar to p=1
bar. Calculate U, q, w, H, G and S due to this gas expansion. Carefully check
signs and units.
5.8. An ideal gas is compressed from the standard pressure p0 to p=2 bar at
T=25
°C.
a) Calculate the Gibbs energy change for one mole of gas.
b) Do the same for two moles of gas.
c) Comment on the sign of the Gibbs energy change.
5.9
Prove that for an ideal gas at constant T:
G( p2 ) − G( p1 ) = nRT ln
p2
p1
(5.4)
5B) Theoretical Exercises
5.10 Argue whether the following statements are true or false:
a) The equilibrium constant does not depend on the total pressure.
b) The standard Gibbs energy change always relates to T=298 K.
c) The total pressure has no effect on the equilibrium composition for gas phase
reactions.
d) In equilibrium molecules do not move.
e) The value of the equilibrium constant does not depend on the choice of the
standard pressure (see Lecture Notes section 5.7).
f) For an exothermic reaction, the equilibrium constant decreases when the
temperature is increased (see Lecture notes section 5.7).
5.11 True or false:
a) When ∆G < 0 J it is useful to look for a catalyst.
b) ∆G° always refers to a reaction at p=1 atm and T=298 K.
c) A reaction is always in equilibrium when ∆G°= 0 J.
d) ΔG > 0 J for the isothermal compression of an ideal gas.
e) When ΔG = 0 J the reaction quotient Q equals the equilibrium constant K.
22
5.12 For the oxidation of glucose:
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
we have: ΔSo=+182.4 J K-1 and ΔHo=–2808 kJ.
a) Calculate the maximal amount of non-volume work that can be obtained per mole
of glucose at T = 37 °C.
In the biological degradation of glucose one of the reactions steps is:
with Go = +1.7 kJ mol-1
glucose 6-phosphate → fructose 6-phosphate
This step is catalyzed by the enzyme glucose-phosphate isomerase. In an in vitro
experiment at 298 K this enzyme is added to an aqueous solution of 1.2 mmol glucose
6-phosphate.
b) How many mmol of glucose 6-phosphate and fructose 6-phosphate are present in
equilibrium?
c) What will be the eventual equilibrium composition in absence of the catalyst?
d) In your body the above-mentioned reaction occurs takes place. How is that
possible if Go is positive?
5.13 A mixture of CO(g), H2(g) and CH3OH (g) with partial pressures 𝑝𝐶𝑂 = 10 𝑏𝑎𝑟, 𝑝𝐻2 =
1 bar and 𝑝𝐶𝐻3𝑂𝐻 = 0.1 𝑏𝑎𝑟, is studied at a temperature of T=500 K. For the formation
of methanol from CO and H2 it is known that DG 0 = +21.21 kJ mol-1 for T = 500 K.
a) Does it make sense to add a catalyst to the mixture to speed up the rate of
formation of methanol?
b) And if pressures of CO and H2 are adjusted to pCO = 1 atm, pH2 = 10 atm?
c) Calculate the equilibrium constant at 500 K for the formation of methanol from CO
and H2.
d) Calculate the partial hydrogen pressure at equilibrium, assuming the partial
pressures of CO and methanol are equal.
e) How large are the equilibrium constant and Go for the decomposition of methanol
tot CO and H2 at 500 K?
f) Suppose the equilibrium constant in e) decreases at decreasing temperature. Is in
that case the decomposition exothermic or endothermic?
5.14 Ammonia gas is prepared by the gaseous reaction:
3H2 + N2 ® 2NH3
The
standard
(5.5)
change
in
Gibbs
energy
at
T=298
∆G° = -33.2 kJ mol .
-1
a) What does ‘standard change in Gibbs energy’ mean?
23
K
for
this
reaction
is
b) Give the expression for the equilibrium constant K in terms of partial pressures
pH2 , pN2 , pNH3 and standard pressure p0.
c) Calculate ln K (K = equilibrium constant) for this ammonia synthesis at T=298 K.
What can you conclude about the equilibrium composition?
d) Argue how the equilibrium composition will be affected by increasing the total
pressure.
e) Suppose that all pressures are equal with respect to each other, so
pH2 = pN2 = pNH3 = x atm .
Will there be any value of x for which the reaction is in equilibrium at
T=298 K? If yes, which value?
5B) Math-related Exercises
5.15 For the thermal decomposition of NO2 under standard conditions at T=295 K:
2NO2 (g) → 2O2 (g)+N2 (g)
;
H0 = + 66 kJ mol−1
;
S0 = +122 J mol−1 K−1
a) What does ‘under standard conditions’ mean?
b) Interpret the signs of the standard enthalpy and entropy change.
c) How large is the standard Gibbs energy change?
d) At which temperature would the decomposition be in equilibrium?
5.16 For the evaporation of liquid water under a pressure of p0= 1 bar:
H2O(l) ® H2O(g) ; DH 0 = +40.6 kJ mol-1 ; DS 0 = +108.7 J mol-1 K -1
a) Calculate the standard Gibbs energy change for the evaporation at 298 K.
b) The same for T=373.2 K; conclusion?
5.17 When the pressure is increased isothermally from 1.0 to 2.0 atm, calculate the
change in Gibbs energy per mole of (1) liquid (18 mL mol-1) which behaves as an
incompressible liquid and (2) vapor which behaves like an ideal gas.
5.18
Ag2CO3 (s) ® Ag2O(s) + CO2 (g)
T/Kelvin
350
K
3.98x10
400
-4
1.41x10
-2
450
500
0.186
1.48
Calculate the enthalpy change ΔH0 for the decomposition of silver carbonate under
standard conditions.
5.19 Investigate whether the reaction
2 Fe3+ + 2 I- → 2 Fe2+ + I2
;
∆G° = -45,3 kJ mol-1
24
will occur spontaneously at 25 °C in an aqueous solution with the following
concentrations:
[Fe3+]
= 1.0 · 10-3 mol dm-3
[I-]
= 1.0 · 10-2 mol dm-3
[Fe2+]
= 1.0 · 10-2 mol dm-3
[I2]
= 1.0 · 10-3 mol dm-3
5.20 The first step in the degradation of alcohol in the liver (37 oC) is
C2H5OH + NAD+ → C2H4O + NADH + H+
ethanol
; ∆G° = +63.7 kJ mol-1
acetaldehyde
Suppose [C2H5OH] = 1.0 · 10-5 mol dm-3 ; [C2H4O] = 1.0 · 10-8 mol dm-3; pH =
7.0. What is the maximal concentration ratio [NADH]/[NAD+] for this reaction to
occur spontaneously?
5.21. Consider the equilibrium constant:
K=
pB
p 2A
(5.6)
of the gas phase reaction 2A → B. Can you express 𝑝𝐴 as well as 𝑝𝐵 in terms of K
and the total pressure ptot = pA + pB ?
5.22. In a closed vessel, a gaseous mixture consisting of oxygen and hydrogen is present
at constant temperature T=298 K. Consider the formation of liquid water via:
1
H 2 (g) + O2 ® H 2O(l)
2
(5.7)
Data:
∆fG0 = -237.129 kJ mol-1 and ∆fH0 = -285.83 kJ mol-1 at p0 and T = 298 K.
a) Calculate the equilibrium constant K for this so-called blast gas reaction and
comment on the equilibrium composition.
b) Suppose that at the beginning the gas pressures are equal at 𝑝H2 =2 atm and 𝑝O2 =1
atm. Calculate ∆G for this blast gas reaction at these pressures. Assume that all gases
behave ideally and set the contribution of liquid water in the concentration quotient
equal to 1.
c) Argue whether it is useful to add a catalyst to the mixture in b).
d) Calculate the standard change in entropy ∆S0 for the blast gas reaction and explain
the sign of it.
e) Calculate the average Kinetic energy per molecule for both H2 and O2 molecules in
the mixture.
f) Argue the effect of increasing the temperature on the equilibrium composition.
25
5.23 Given the following reaction:
C6H12O6 + O2 → 6 CO2 + 6 H2O
(5.8)
For these reaction, the following conditions apply at T=298.15 K:
ΔH° = –2816 kJ mol-1; ΔG° = –2879 kJ mol-1
a) Calculate the change in standard entropy (ΔS°) for reaction (5.8).
b) Does the number of microstates for reaction (5.8) increase or decrease under
standard conditions? Explain your answer.
c) Calculate ΔG° for the following reaction:
6 CO2 + 6 H2O → C6H12O6 + O2
(5.9)
d) Reaction (5.9) does not occur spontaneously, even under biological circumstances
However, this reaction does occur (like via photosynthesis). How could you explain
this observation?
Review Exercises
5.24 For an isothermal process, equation (4.4) can be integrated easily to obtain
DSsys =
qrev
T
a) Explain what is meant by ‘isothermal process’ and give an example of such a
process.
b) For a non-isothermal process, equation (4.4) cannot be integrated directly. The
change in heat can be expressed as a function of temperature by using the heat
capacity of the system. For a system containing n moles of a solid, derive an
expression like LN equation (4.12) for the change in entropy due to changing the
temperature from T1 to T2.
5.25 Given equation (5.8) in the Lecture Notes for a closed system: DSsurr = -
0
DH sys
T
a) Derive this equation. Indicate which assumption(s) you have to make.
b)
What
is
the
difference
with
an
equilibrium
∆G°=0 J mol ), where under standard pressure DS
0
sys
-1
difference in sign.
26
=+
process
0
DH sys
T
(for
which
, i.e.: explain the
Chapter 6 Liquid-Gas Equilibria
Theoretical Exercises
6.1
Which statement is correct? A liquid boils when
A its vapor pressure equals the atmospheric pressure.
B its vapor pressure exceeds the atmospheric pressure
C air bubbles in the liquid collapse
D when atmospheric pressure exceeds the vapor pressure.
6.2
Which statement is correct? (hint: browse Lecture Notes sections 6.1 and 6.2)
A A solid has no vapor pressure.
B Vapor pressure is independent of temperature.
C For an endothermic vaporization the vapor pressure increases with temperature.
D A vapor pressure must exceed the standard pressure.
6.3
On a mountain top water boils at a lower temperature than at sea level. From this
we can conclude that the vaporization of water
A must be exothermic
B must be endothermic
C can either be endothermic or exothermic
D must be a non-spontaneous process.
6.4
Which statement is correct:
A The Clausius-Clapeyron equation holds for all liquid-vapor equilibria.
B The Van’t Hoff equation assumes that gases are ideal.
C The Clausius-Clapeyron equation assumes that vapors obey the ideal gas law.
D The ideal gas law assumes that gas molecules do not collide with the wall.
6.5
For the condensation of water vapor (g) onto liquid water (L) we can write the
equilibrium as:
H2O (g) → H2O (l)
(6.1)
At a certain temperature T the vapor pressure equals 0.5 bar. Assuming the vapor is
ideal, the thermodynamic equilibrium constant K of the water condensation equals:
A 2 bar
B 0.5
C2
D 0.5 bar
6.6
At constant temperature, a solute distributes itself between two immiscible solvents
only in a particular ratio. This is known as the Nernst Distribution law, which describes
the relative distribution of a component that is soluble in two liquids, assuming that
27
these liquids do not mix. This law also states that, at equilibrium, the ratio of the
concentrations of a third component in two liquid phases is constant, i.e.:
c1
= KD
c2
(6.2)
where c1 and c2 are the molar concentrations of the third component in the first and
second liquid phase, respectively, KD is the distribution coefficient, which is
temperature dependent.
a) Assume a solute has a KD of 5.00 between water and chloroform. In which liquid
phase will the solute be dissolved mostly? Explain.
b) Argue whether a solute can have a negative value for KD.
c) Suppose you have a solute dissolved in hexane. Knowing that it has a KD of 0.13
between hexane and water, propose a method to extract this solute.
6.7
a) What is Raoult’s law for a component of a liquid mixture, if this component behaves
ideally?
b) If both components of a binary liquid mixture are very similar, such as benzene
and toluene, then Raoult’s law can be used for each component. Derive a relationship
between the mole fraction yB for benzene in the gas phase and the mole fraction xB
for benzene in the liquid mixture of benzene and toluene.
c) The vapour pressures of pure benzene and toluene are 0.513 and 0.185 atm at 60
C, respectively. Calculate yB at this temperature, if xB = 0.5. What is the partial
pressure of toluene in this case?
6.8
The HCl fountain
A closed, cubicle vessel is at time t=0 filled with pure hydrochloric acid gas at a
pressure of 1 bar. The vessel is connected to a water bath via a capillary as indicated
in par. 5.5 in the Lecture Notes. Then HCl starts to dissolve in water, a dissolution
that can be represented by:
HCl(g) ® HCl(aq)
(6.3)
The temperature is everywhere T=25 °C ; the atmospheric pressure is equals the
standard pressure p0=1 bar.
a) Calculate the height z (in cm) of the cubic vessel assuming that at t = 0 the
vessel contains one mole HCl vapor that behaves as an ideal gas.
b) Explain what will happen when HCl vapor dissolves in water.
c) Calculate ΔU, ΔH, ΔS and ΔG (all per mole HCl) of the dissolution reaction.
Comment on their signs.
d) How much heat is released in this set-up, per mole HCl, upon dissolution?
e) What would be the heat release if HCl dissolves in water, without delivering any
work?
28
f) What is the total entropy increase ΔStot due to the dissolution? (HINT: see Lecture
Notes, section 5.1). Compare TΔStot to ΔG; conclusion?
g) Calculate the thermodynamic equilibrium constant for reaction (6.3). Conclusion?
h) Calculate the maximal mass m of water, per mole HCl, that could be lifted to a
height of h=10 cm via the maximal amount of work that the dissolution reaction can
deliver.
6.9
p0= 1 bar ; T = 298.15 K
f H0 / kJ mol−1
S 0 / J mol−1 K −1
HCl (g)
− 92.3
186.9
HCl (aq)
− 167.2
56.5
Combination of Raoul´s law and the Clausius-Clapeyron equation (see also lecture
notes Section 6.3) provides a prediction for the vapor pressure of a solution.
a) Estimate from this prediction how many moles (nNaCl) of sodium chloride should be
added to nH O moles of water such that the salt solution boils on top of Mount Everest
2
at 100 °C. Assume the pressure on the mountaintop is p ≈ 0.28 bar. Discuss whether
your estimate is realistic or not.
b) The salt concentration in the Dead Sea is about 360 g L-1; estimate the boiling
point of Dead Sea water at atmospheric pressure. The change in standard molar
enthalpy equals +40 kJ mol-1.
c) In a closed box, a bucket full of Dead Sea water is placed next to a bucket full of
tap water; what will happen?
Math-related Exercises
6.10 Water is distilled at a reduced pressure of p=0.38 atm and boils at T=75 °C
a) Calculate the molar enthalpy change of water evaporation. What does the sign
mean?
b) Calculate the molar entropy change for the evaporation of water at 75 °C and 100
°C. Explain the difference; what does the sign of the entropy change mean?
6.11 Consider the gas phase reaction 2I →I2.
Data:
Iodine
M/g mol-1
ΔHf0/kJ mol-1
ΔGf0/kJ mol-1
Sm0/J mol-1 K-1
I2(s)
253.81
0
0
+116.35
I2(g)
253.81
+62.44
+19.33
+260.69
I(g)
126.90
+106.84
+70.25
+180.79
a) Calculate ΔG° for this reaction.
b) Calculate the equilibrium constant at T=298 K.
29
c) Calculate ΔS° for this reaction and explain the sign.
d) Verify that ΔG° = ΔH° – TΔS°.
6.12 At a temperature of 273.16 K the constant-pressure heat of melting (fusion) of ice is
qp=6.0 kJ mol-1. Calculate the entropy of melting of ice at 273.16 K.
6.13 Perform the integration of equation (6.13) to obtain equation (6.14) in the Lecture
Notes. What is the physical meaning of the integration constant C?
6.14 H2 gas and I2 gas react quickly at 327 K to form HI gas according to an equilibrium
reaction:
H2 (g) + I2 (g) ® 2HI(g)
(6.4)
The equilibrium constant for this reaction is known to be 92.6. If the initial partial
pressures of H2 and I2 are 1.980 atm and 1.710 atm, respectively, what are the
equilibrium partial pressures of each gas?
6.15 The partition coefficient of a substance X between benzene and water is 5.0 (X
dissolves best in benzene). A 100 cm3 solution of X in water, with a concentration [X]
of 0.10 mol dm-3 is obtained. With how much cm-3 benzene should the aqueous
solution be shaken to extract 7.5 mmol of X from the aqueous solution? (Dissociation
or association of X does not occur).
A 20
C 60
E 100
B 40
D 80
F 120
30
Review Exercises
6.16 Given two isotherms for an ideal gas in the figure depicted below. In this graph, three
trajectories A, B and C are indicated.
a) What is ΔU for trajectory A? Explain.
b) Formulate ΔU for trajectories B and C. Assume that Cp and CV do not depend on
temperature.
c) Show that for an ideal gas the following relation holds:
𝐶𝑝 = 𝐶𝑣 + 𝑛𝑅
d) Suppose that a certain amount of heat q will be added to an ideal gas. Argue for
which process the temperature will increase the most: if the pressure has been kept
constant or when the volume has been kept constant?
6.17 Ethanol can be prepared from ethane via:
C2H4 + H2O → CH3CH2OH
The compounds in the reaction are all contained in a sealed container. Assume that
all compounds are in the gas phase and behave ideally. The temperature is 25 C and
the partial pressures of all compounds are equal to 1 bar (standard conditions).
Assume that DH f and S
0
0
are independent of T.
Data: At T = 25C:
Compound
∆𝐻𝑓0 [kJ mol-1]
𝑆 0 [J mol-1 K-1]
∆𝐺𝑓0 [kJ mol-1]
C2H5OH (g)
–235.10
+282.70
–168.49
C2H4 (g)
+52.26
+219.56
+68.15
H2O (g)
–241.82
+188.83
–228.57
H2O (l)
–258.8
+69.9
–237.13
a) What are the molar volumes of each of the three substances in the contained for
the abovementioned reaction?
b) What is the change in enthalpy for the reaction?
c) Calculate the temperature at which the reaction is in equilibrium, under
standardconditions.
d) The volume V of the reaction mixture in equilibrium is reduced isothermally.
Discuss whether the amount of ethanol in mixture will increase or decrease.
31
Chapter 7 Osmosis and Osmotic Pressure
Theoretical Exercises
7.1
What would be the effects of long-term drinking of very clean distilled water?
7.2
Arrhenius investigated the dissociation of an electrolyte in 1884 (Uppsala). He got
into a disagreement with his contemporaries, who saw little in his idea. They said,
for example, that is was very unlikely that NaCl would split in water to Na + and Cl-,
because these ions have a strong Coulomb attraction, so they would stick together
a) Explain why you agree or disagree with this argument.
b) How can it be demonstrated experimentally that NaCl does indeed dissociate in
water?
7.3
The surrounding atmosphere has a total pressure of p=1 atm and has a mole fraction
of 0.21 of O2. Suppose that the amount of water vapor in this atmosphere is exactly
the same as the vapor pressure of water at 37 °C, namely 0.06 atm.
Data:
𝑀CO2 = 44.0 g mol-1, 𝑀N2 = 28.0 g mol-1,
Henry’s law: 𝑃O2 = 𝐾𝑥O2 ; K = 52.8 · 103 atm
a) Explain the term ‘vapor pressure’ and how this can be measured for water at 37
°C.
b) Calculate the partial pressure of oxygen in aforementioned atmosphere.
c) Suppose that blood is in equilibrium with the inhaled atmosphere in the lungs.
Calculate the mole fraction of oxygen in the blood. Assume that blood mainly consists
of water.
d) Convert the answer of c) into the concentration of oxygen [O2] in units of mol L-1.
Indicate which assumptions you need to make in order to perform this calculation.
e) Suppose that blood contains 0.5 mol L-1 of cations. Assume that blood only consists
of 1:1 electrolyte, calculate the osmotic pressure of blood exerts against pure water.
(T=37°C).
7.4
Consider a glucose solution in water (T=298 K) with a concentration of 1 g glucose
per liter (𝑀glucose =180 g mol-1). Assume glucose molecules behave ideally.
a) What does ‘ideal behavior’ mean?
b) Calculate the average kinetic energy for these glucose molecules and for water
molecules in solution in units of J mol-1.
c) Calculate the osmotic pressure this sugar solution exerts against pure water, for
which water of the solution is separated by a water permeable membrane.
32
d) The vapor above the sugar solution is in contact with water vapor above pure
water, as sketched in figure 7.1. Argue if something will happen in this setup. If yes,
explain what will happen and why.
Figure 7.1 A closed system in which the vapors above a sugar solution and pure
water are connected.
7.5
Does Raoult’s Law allows a perpetuum mobile?
Consider the following situation in an isolated system: water vapor from a pure
solvent (right) migrates spontaneously to the vapor with lower pressure (left) above
a solution. Next water vapor condenses on the solution; as a result, the liquid column
left becomes higher than to the right, so liquid water will be pushed through a porous
membrane to the right, where it evaporates, thus closing the water cycle. We could
place a rudder in the vapor or liquid which would be rotated and could perform work,
extracted from the water cycle.
Do you think this device is going to work? Does it contradict the First or Second Law?
If so, explain why.
What will happen in reality in the set-up sketched in the figure above?
33
Math-related Exercises
7.6
a) Calculate the osmotic pressure of a solution of 1 gram table salt in 1 litre water
(T=25 °C).
b) What is the height of a water column that exhibits the same pressure?
c) Calculate the osmotic pressure of a solution of 1 gram protein
(molecular
weight 6 · 104 g mol-1) in 1 litre water. Explain the difference in the answer obtained
at a).
7.7
Human blood has an osmotic pressure of 7.2 bar. A solution with a higher salt
concentration than blood will, when mixed with blood, lead to shrinkage of blood cells.
An isotonic NaCl solution will cause neither shrinkage nor swelling of blood cells; what
is the NaCl concentration in an aqueous isotonic solution? Assume T=37 °C.
7.8
Sea water contains dissolved salt with a total ion concentration of about
1.13
mol L-1. Estimate the minimal pressure needed (at T=25 °C) in a reverse osmosis
set-up to induce a flow of pure water through a membrane that is impermeable for
salt molecules. What is the essential assumption in your calculation?
7.9
Osmotic pressure measurements can be employed to determine the molecular weight
of dissolved solutes. The osmotic pressure of a solution with 5.0 g of polystyrene per
liter is measured to be 0.0100 bar. Calculate the molecular weight of the polystyrene.
Assume T=298 K.
7.10 Decompression sickness
In caissons submerged under water the air pressure may be equal to the atmospheric
pressure plus the hydrostatic pressure of the water. Someone who works for some
time in such a caisson has blood plasma with an increased nitrogen concentration
due to the increased partial pressure of nitrogen in inhaled air. Upon a too rapid
return to the standard atmospheric pressure, disease symptoms arise because
gaseous nitrogen is released into the blood – the disease is also known as divers
disease or caisson disease.
Calculate the maximum volume of nitrogen gas that may be released into the blood,
using the suggestions and data given below:
- Consult table 7.1 on the next page.
- The mole fraction of nitrogen in the air is always 0.8
- Set the mass densities of water and blood plasma both equal to 1.0 kg dm -3 and
assume a temperature of 37 °C.
- Volume blood plasma is 3.2 dm3
34
- The extra hydrostatic pressure is ρgh, where ρ = density of water (see above), g =
acceleration of gravity (9.81 m s-2) and h = depth below the water level (assume that
h= 100 m)
- Consider the evolved nitrogen gas as an ideal gas with a pressure of 1 bar and a
temperature of 37 °C.
- Assume that blood mainly consists of water.
Table 7.1: Equilibrium constants for Henry's Law
Gas
T (°C)
K (103 atm)
N2
25
85.7
N2
37
107
O2
25
43.4
O2
37
52.8
CO2
25
1.64
CO2
37
2.3
7.11 Blue energy
Salt solutions and pure water mix spontaneously - which may be exploited to extract
non-volume work. Rivers of fresh water that end up into the salty North Sea are,
consequently, a potential source of useful work (also referred to as ‘blue energy’).
Concentration of salt (most of which is NaCl, 𝑀NaCl =58.44 g mol-1) in the North Sea
is about 35 g L-1. Assume salt molecules behave as ideal solutes.
a) Suppose 1 m3 of fresh river water, with virtually no dissolved salt, at
25
°C merges with 1 m3 of salty seawater. What is the drop in osmotic pressure of the
salt in the seawater?
b) Estimate how much non-volume work can be maximally extracted from this
mixing.
Hint: look at the entropy increase of an ideal gas that spontaneously expands from
volume V1 to V2.
35
Review Exercises
7.12
In an isolated system, n mole of ideal gas at constant temperature, expands
spontaneously against a vacuum from its initial volume V1 to a final volume of V2 =
2 V1.
a) What is the change in energy ΔU of the gas due to the expansion? Why?
b) What is the work wirr performed by the gas for the irreversible expansion?
c) Show, using the ideal gas law that the volume work for a reversible expansion of
V1 to V2 is equal to: wrev = -nRT ln(2) .
Explain the sign.
d) Calculate ΔH, ΔS and ΔG for the gas expansion per mole of gas. R and T can
remain in the answer. Clearly explain how you achieve your answers and comment
on the signs.
7.13
To study the influence of an oil layer on water, we do the following test:
A beaker containing 200 cm3 water and 100 cm3 octanol is in equilibrium with air at
standard temperature and pressure (T* = 298.15 K and p*= 1 atm ≅
1.0 · 10
5
Pa). The density of octanol under these circumstances is smaller than that of water.
Given are:
•
the Henry constant of oxygen in octanol K2 = 1.0 · 106 atm;
•
the partition coefficient of oxygen between octanol and water K = 4.5;
•
the density of octanol ρ = 0.83 kg dm-3,
•
the molar mass of octanol M = 130 g mol-1
•
the mole fraction of oxygen in the air xO = 0.2.
2
Octanol and water do not dissolve in each other. During this exercise, make use of
Table 7.1 which is given on the previous page.
a) Calculate the amount of O2 (in mole) in octanol.
b) Do the same for water: what is the concentration?
c) Does the concentration of O2 in clean water increase or decrease when covered
with oil?
The Henry constant for oxygen in water is 43 · 103 atm.
d) What influence does the thickness of the oil layer have on the concentration of
O2 in water in the equilibrium state?
36
Chapter 8 Redox Equilibria
Theoretical Exercises
8.1
Which statement is correct:
A A reductor accepts electrons
B An oxidator donates electrons
C An oxidator accepts electrons
D A reductor is reduced
8.2
a) Give the electrode reactions and the cell reaction of the following Daniel-cell:
Zn | Zn2+ || Cu2+ | Cu
b) Do you expect that the cell reaction will take place spontaneously?
8.3
The following reaction occurs during the alcoholic fermentation reaction:
CH3CHO + NADH + H+ → CH3CH2OH + NAD+
aceetaldehyde
ethanol
Herein NAD is short for nicotinamide adenine dinucleotide (a coenzyme). The
redoxcouple of NAD has H+ + NAD+ as an oxidizer and NADH as a reducer
a) Split the abovementioned reactions into two half reactions.
b) Give the cell diagram of an electrochemical cell (with salt bridge), of which the
abovementioned reaction is de cell reaction.
8.4
Below you will find several total cell reactions. Split every reaction into two half
reactions (reactions which occur on an electrode). Indicate for each reaction the
oxidation and reduction half reaction.
Calculate the reversible cell potential under standard conditions. Indicate if the cell
reaction will occur spontaneously.
a) Au + 3H+ → Au3+ + 3/2 H2
b) Na + H+ → Na+ + ½H2
c) Na+ + ½H2 → Na + H+
Compare your answer to b). Explain.
d) 4Na + 4H+ → 4Na+ + 2H2
Compare your answer to b). Explain.
e) 2Na + 2 H2O → 2NaOH + H2
Compare your answer to b). Explain.
f) 2 H2O → 2 H2 + O2 at pH 0
g) 2 H2O → 2 H2 + O2 at pH 14
Compare your answer to f). Explain.
h) Fe2+ + H+ + 1/4 O2 → Fe3+ + ½H2O
37
i) AgCl → Ag+ + ClNext, calculate the solubility product of AgCl at T = 25.00°C.
j) Zn + Cu2+ → Zn2+ + Cu
Next, calculate the change in Gibbs free energy.
In addition, calculate E and ∆G for 10-5 M Cu2+ + 10-2 M Zn2+ (25.00 °C).
Table 8.1 Standard reduction potentials4
8.5
Battery. A battery widely applied in automobiles is sketched below. It consists of two
plates of lead in a solution of sulfuric acid. The positive electrode is covered with lead
dioxide (PbO2).
a) Formulate the reactions that occur at the two electrodes, and the cell reaction.
Hint: see question c
b) Specify in which direction electrons are migrating.
4
(Adapted from D. W. Ball, Physical Chemistry (Wadsworth, 2015).
38
c) During the spontaneous cell reaction, sulfuric acid is removed from solution and
deposited as lead sulfate on the two electrodes. When is the battery ‘empty’ and how
can the battery be re-charged?
Math-related Exercises
8.6
A new battery for a flashlight with a voltage of 1.5 V delivers a current of 1 A. Suppose
it takes 1 h for the battery to run down: how much work does the battery deliver in
this hour?
8.7
Suppose a galvanic cell at T=298 K generates a voltage of ΔV=1.56 volts, with
electrons flowing through an external resistance of R=2.0 ohms.
a) Calculate the electrical current flowing through the resistance.
b) Calculate the number of coulombs of charge that passes the resistance in 250
seconds.
c) How many milli-moles of electrons flow in 250 seconds through the resistance?
d) The power (energy per unit time) produced by a current I in a resistance R is I2R.
The released (‘dissipated’) energy in t seconds is I2Rt. What is the dissipated energy
in 250 seconds?
e) How much entropy does the electron flow for 250 seconds produce (at minimum)?
8.8
The standard potential of the Daniell cell is E0 = +1.1 V. Calculate the equilibrium
constant
K
(at
T=298
K)
for
the
reaction
Cu2+ + Zn → Cu + Zn2+. Conclusion?
8.9
A modern ‘pH electrode’, in fact, consists of two electrodes on both sides of a glass
membrane. A voltage is caused by the difference in H + concentration between an
internal compartment and the analyte. This is a concentration cell: assume that
without a difference in H+ concentration the voltage equals zero. Calculate at both
20.00°C and 25.00°C the voltage for the case the intern compartment contains 0.100
M HCl and the analyte contains 1.00 · 10-2 M HCl.
39
8.10 If a base metal is placed in a 1 M HCl solution, the metal can dissolve and the
development of hydrogen gas will take place simultaneously. Calculate if this could
be the case for:
a) Sodium
b) Gold
For the following exercises given is T=25.00 C and the standard values of H and
G apply at this temperature. (Use table 8.1.)
8.11 Given is the following electrochemical cell:
Fe3+ (10-4 mol dm-3)
Ce4+ (10-3 mol dm-3)
Pt
Pt
Fe2+ (10-3 mol dm-3)
Ce3+ (10-1 mol dm-3)
a) Calculate the reversible cell potential.
b) Give the cell reaction, calculate the change in Gibbs-energy (G) for this reaction,
and calculate the standard value for G (G).
c) Calculate the equilibrium constant of the cell reaction.
8.12 The reaction that occurs for corrosion in acidic media is:
1
Fe(s) + 2H+(aq) + O2(g) → Fe2+(aq) + H2O(l)
2
We can assume the temperature at which corrosion is happening is 273 K < T < 373
K. Is the equilibrium constant favourable for the formation of Fe2+(aq)?
8.13 Calculate the equilibrium constant for the following reaction:
2 Cu+(aq) → Cu(s) + Cu2+(aq)
at 298 K.
a) Will Cu+ (aq) be a common ion in water?
b) Sketch the galvanic cell for this redox reaction and give the cell diagram.
c) How big is the change in G and in the cell potential if 4 Cu+ react instead of 2
Cu+?
8.14 The following electrochemical cell
Pt (H2, 1 atm.) | solution A || solution B | (H2, 1 atm.) | Pt
Contains a solution A that is a watery buffer solution with a pH of 6.88 and a solution
B with a blood serum. The EMK (electromotive force = cell potential) of this cell at a
temperature of 20 C (=293.15 K) is –0.0250 V.
Calculate (to the nearest 0.01 pH-unit) the pH of the blood serum at 20 C.
40
Review Exercises
8.15 Fuel-Cell
Liquid hydrogen reacts explosively with oxygen and for this reason is used as rocket
fuel. A fuel cell also employs the oxidation of hydrogen, but does so in a much more
controlled manner, see the figure below.
H2(g) and O2(g) are supplied from external tanks to the cell in which the controlled
combustion takes place. The electrodes are hollow cylinders of porous carbon and
immersed
in
an
aqueous
KOH-solution.
a) Specify the half-reactions and the cell reaction.
b) Which electrode in the figure is negative, and which positive?
c) Would the cell also work without KOH being present?
8.16 The Henry constant for oxygen in water at 25 C is Kz= 4.4 · 104 bar. The total air
pressure is ptot= 1.0 bar. The mole fraction of oxygen in the air is equal to xz= 0.2.
a) What is the partial pressure pz of the oxygen in the open air?
b) Calculate how many grams of oxygen will dissolve in 1 m 3 water (density ρ= 1 kg
L-1), that is in contact with the open air.
c) What is the equilibrium pressure of a solution?
-What is Raoult’s law?
-What concentration measure is used in Raoult’s law for the solvent, and why?
d) Is the evaporation of a liquid always exothermic or endothermic? Explain your
answer.
41
Chapter 9 The Chemical Potential
Theoretical Exercises
9.1
The equilibrium condition for the reaction 3NO → N2O + NO2 is:
A μ (N2O) + μ (NO2) = 3 μ (NO)
B μ (N2O) – μ (NO2) = 3 μ (NO)
C μ (N2O) + μ (NO2) + 3 μ (NO) = 0
D μ (N2O) – μ (NO2) – 3 μ (NO) = 0
9.2
Given
is
the
following
reaction,
which
takes
place
in
cells
of
a
flower
C6 H 12O6 + 2NAD+ + 2ADP + 2Pi ® 2C3H4O3 + 2NADH + 2H+ + 2ATP + 2H2O
a) Give an expression for the equilibrium condition for this reaction in terms of the
chemical potential.
b) Does it make any difference for the expression by a) if H 2O is in a liquid phase or
a gas phase?
Math-related Exercises
9.3
æ ¶V ö æ ¶V ö for an ideal gas. Also show that for an ideal gas
,
è ¶T ÷ø p çè ¶p ÷ø T
Derive ç
¶ æ ¶V ö
¶ æ ¶V ö
=
ç
÷
ç
÷
¶T è ¶p ø T p ¶p è ¶T ø p T
is valid. Assume n is kept constant.
9.4
a) Formulate the total differential for the Gibbs energy G as function of p,T and moles
nA, nB of substances A and B.
b) Derive the chemical potential for an ideal gas at pressure p. Hint: start from the
fact that the chemical potential is the molar Gibbs free energy.
9.5
Derive the chemical potential for an incompressible substance at pressure p and
temperature T. Do you understand the meaning of all the symbols and their units?
9.6
a) Write down the equations for the chemical potential of an ideal solution with
solvent mole fraction x, and for an ideal gas with pressure p.
b) Use the results from a) to formulate the equilibrium condition for an ideal solution
(L) in equilibrium with its vapor (G).
42
c) Show from the result in b) that the equilibrium constant K for the vaporization of
solvent from a solution equals K = p / x. Hint: make use of the general result
ln K =
9.7
-m0
; Dm 0 = DG 0 per mole .
RT
Calculate the mole fractions of NaCl and H2O in an aqueous solution of
a) 1 g NaCl per liter and
b) 100 g NaCl per liter.
c) Calculate the corresponding osmotic pressures at T=298 K.
9.8
Find out how to calculate the Taylor expansion of a function f(x). Calculate the first
two terms in the Taylor expansion around x = 0 for the following functions:
a) f(x) = exp(x)
b) f(x) = 1/(1–x)
c) f(x) = ln(1+x)
d) f(x) = ln(1–x)
e) f(x) = sin(x)
Which of these expansions is used in the derivation of Van ’t Hoff’s osmotic pressure
law? What is the justification for this use?
9.11 a) Formulate the equilibrium criterion for the gas-phase reaction 2 I ⇄ I2 in terms of
chemical potentials.
b) Show how this criterion leads to the equilibrium constant:
æ Dm 0 ö
K = exp ç ; Dm 0 = DG 0 per mole
÷
è RT ø
9.12 Show that the expression for the chemical potential of an ideal gas is equivalent to
the ideal gas law (i.e. do the derivation backwards).
9.13 For a certain chemical reaction A → B it turns out that ΔG=–100 kJ mol-1.
a) How much non-volume work can be extracted from the conversion of one mole of
A to B?
b) How much non-volume work is minimally needed to convert one mole of B to A?
c) To which maximal height h the reaction A→B could lift a mass M of 1000 kg?
43
Review Exercises
9.15 In a fuel cell, hydrogen gas is oxidized to water. The electrodes are in a solution of
potassium hydroxide (KOH) in water. The fuel cell works under standard conditions.
a) What reaction takes place at the cathode, and at the anode? What is the net cell
reaction?
b) The (reversible) standard cell voltage appears to be equal to E0= +1.23 V.
Calculate the corresponding value for ΔG0. What is the maximum amount of electrical
work (per mole H2) that the cell can produce under standard conditions?
c) Does the cell reaction occur spontaneously or not? Why?
d) The enthalpy change for the cell reaction is ΔH0= –286 kJ mol-1. How much heat
q does the cell produce during the reversible power supply?
44
Chapter 10 Electrolyte Solutions
Theoretical Exercises
10.1 For the reaction between an acid Z and a base B the following can be written:
Z  B + H+
Formulate the Henderson-Hasselbalch equation for this reaction.
10.2 A solution of 0.1 mol NaOOC-COOH in pure water with a total volume of 1 dm 3 is
made. Formulate the conservation law and the electron neutrality condition for the
individual concentrations in the solution.
Math-Related Exercises
10.3 The dissociation constant of water is pK w = 14. How many water molecules in a
collection of 1 billion water molecules will be ionized (split into OH- and H+) at a
neutral pH of 7?
10.4 Calculate the pH of the solution that one obtains by dissolving 2.30 gram of the
monovalent acid HCOOH (formic acid) in water with a volume of 1.00 dm 3. The
temperature is 298 K and the dissociation constant of the acid is pK a = 3.75.
10.5 Tartaric acid (a divalent organic acid of which the undissociated form is abbreviated
as T) dissociates in watery solutions according to the following reactions:
pK a1
pK a2
T Û T - Û T 2Given are the pKa1 = 3.0 and pKa2 = 4.4. Calculate the amount of W, W- and W2- in a
solution of 0.085 mole tartaric acid of which the pH is brought to 3.8.
10.6 (Warning: this is an extensive exercise.)
The ionization-equilibrium of glycine in water is given by:
pK a1
pK a2
G+  G  G−
=2.34 =9.60
0.010 mol of glycine is dissolved in pure water with a total volume of 1.0 dm3. The
pH of this solution is 6.05.
a) Calculate (accurate to 10%) the concentration of G + and G- in the solution.
b) What is the pH of the solution after adding 0.0020 mol HCl?
45
Review Exercises
10.7 In oxygenated water (pH 4.5), an iron nail rusts over time, while aluminium remains
intact in the same water. Which answer is correct, and explain why the others are
not.
A The oxidation reaction of iron is thermodynamically favourable, whereas that with
aluminium is not.
B The oxidation reaction with iron and aluminium are both thermodynamically
favourable, but the oxidation of aluminium stops almost immediately because the
metal is covered with an oxide layer.
C Aluminium is already in the highest oxidation state, whereas iron can still be
oxidized.
D Aluminium will spontaneously take up electrons from the water, which protects
itself from oxidation, whereas iron loses electrons to water, causing the iron to
dissolve.
10.8 An ideal gas consisting of 1 mole of molecules goes through a circular process. The
3 state changes (1 → 2, 2 → 3, 3 →1) are indicated in the p,T diagram below (on the
y-axis is p/105 Pa).
a) What can you say about ΔU for this circular process? Explain your answer.
b) What is the volume of the gas in state 3?
c) Indicate the state changes in a p,V diagram.
d) Calculate all the changes in energy, warmth, work, and enthalpy in kJ.
Fill in the following table:
step
Type of process
∆U / kJ
q / kJ
1→2
w / kJ
+ 1,57
2→3
Isobaric
3→1
Isochoric
1→2→3→1
Circular process
+ 5,67
− 3,39
46
∆H / kJ
10.9 5The movements of atoms within molecules are considered to be harmonic
vibrations with a certain frequency ν. Because of this, molecules are also
considered to be harmonic oscillators. The thermodynamic energy U of a set of N
molecules is given by:
U=
Nhn
hn
e kT -1
where h is Planck’s constant, ν the frequency of the oscillator, k the Boltzmann’s
constant and T the absolute temperature.
The heat capacity of this system is given by:
C=
dU
dT
a) Show that the heat capacity is given by:
hn
e kT
æ hn ö
C = Nk ç ÷
è kT ø æ hn ö 2
kT
çè e -1÷ø
2
b) What happens with the value of the heat capacity when T goes to 0 K? Explain
your answer using a calculation or explanation.
[Hint: Calculate the limit of the heat capacity with T → 0 K. Use the expression from
a), even if you were not able to derive it.]
5 Adapted from: Mathematics for Physical chemistry, Chapter 6, Problem 32
47
Chapter 11 Diffusion in Liquids and Gases
Theoretical Exercises
11.1 It is often claimed that Brown observed the Brownian motion of pollen grains. Pollen
grains of Clarkia Pulchella have diameters in the range 50-100 microns. Suppose you
would observe the grains in water under a microscope for one hour: calculate the
diffusive displacement of a 50-micron grain in that hour. Conclusion?
11.2 Explain why Brownian motion does not stop due to viscous friction between colloid
and surrounding solvent.
11.3 One could argue that a colloid receives heat from its environment and converts it to
work of motion, and vice versa. However, according to the Second Law it is impossible
for heat to be fully converted to work in a closed cycle. Does Brownian motion
contradict the Second Law?
11.4 Estimate the time it would take oxygen molecules to diffuse in water
(D
= 18·10-6 cm2 s-1) at room temperature a distance equal to 1) the typical thickness of
a bacteria – about 1 µm; 2) the typical thickness of a human being. Verify that
diffusive transport of oxygen from the environment to the lungs is not an alternative
to oxygen transport by red blood cells. Do you expect that an oxygen molecule in air
diffuses much slower or much faster than in water? See for example S. Vogel “Life’s
Devices; the physical world of animals and plants” (Princeton University Press, 1988).
11.5 Explain that for diffusion of particles the average displacement < 𝑥 > = 0 𝑚.
11.6 Explain in your own words the random walk.
Math-Related Exercises
11.8 Remember the hot cup of tea from Figure 1.4. Suppose we place a sugar cube at the
bottom of the cup, how long does it take for sugar to reach the top of the cube without
stirring? Assume the diffusion only happens in 1D (along the height of the cup.)
Data: height of cup = 10 cm, Dtea = 10-9 m2 s-1.
11.9 Calculate for the Particle Quartet in Table 11.1 diffusion coefficients in water at T=298
K.
11.10 How far does each reference particle travel by diffusion in a) 1 hour; b) one year?
48
11.11 Calculate the molar volumes for the Particle Quartet, assuming each member forms
a random sphere packing with a sphere volume fraction of φ=0.64.
11.12 A marble with diameter 0.5 μm is put in water for two weeks. Calculate the rmsvalue of the displacement r for this marble. Assume T=298 K and the viscosity of
water is 0.89 mPa·s.
11.13 The diffusion coefficient for carbon in iron is 2.9 · 10 -8 cm2 s-1 at 500 °C. Assume
1D diffusion. How far does a carbon atom diffuse in 1 year (in the x-direction)?
11.14 Calculate the diffusion coefficient of a spherical protein (radius 0.1 m) in water at
T=298 K.
11.15 Show that the random walk gives the result that the quadratic displacement is
proportional to time, i.e. <x2> ~ t. What is the order of this equation?
11.16 a) A bug is flying through the air travels by a speed of 2.1 km h -1 for 50 seconds.
Calculate its diffusion coefficient, assuming its path is in a straight line.
b) It’s raining and the bug falls in a puddle of water. He is still moving and ‘swims’
with a speed of 5 m h-1. After 50 seconds, he gives up and drowns. Calculate its
diffusion coefficient in water when he was still alive.
11.17 Gravity accumulates colloids at the earth’s surface whereas Brownian motion tends
to homogenously disperse them. In equilibrium a constant concentration profile is
present, also known as the barometric height distribution.
a) This distribution is a Boltzmann distribution, so the distribution function has the
form P(h) = C exp(-Energy/kT) with C=constant. Here P(h) is the distribution function
for particles at height h; k is the Boltzmann constant. Which energy term should we
substitute here for otherwise ideal (=except for the gravity) particles?
b) What is the meaning of P(h) dh?
c) Calculate the constant C via the normalization: ∫P(h)dh = 1
d) Calculate the average height <h> of the particles above the Earth's surface located
at h=0.
e) Evaluate <h> for nitrogen particles and for colloids in the form of small water
droplets with a diameter of 1 µm and mass density 1g mL-1 in an atmosphere at T =
298 K. Discuss the two outcomes.
f) Derive a formula for the root-mean-square height of particles in a barometric
distribution.
49
11.18 We are going to use our knowledge of diffusion to understand a bit more about the
health of freshwater fish.
a)
In a tube with length L, an ideal gas starts out released at the left side. After some
time, it will be spread over the entire tube homogeneously.
Assuming stationary diffusion, rewrite Fick’s first law to an equation in terms of
concentrations Cleft, Cright, diffusion coefficient D and tube length L.
b)
In a water pool with a depth of 10.0 meters, perpendicular walls and a flat horizontal
bottom, dissolved oxygen diffuses from top to bottom at a constant temperature of
25 C. Estimate the speed of diffusion in mol m-2 h-1 with (your answer from a and)
the following rough model:
The top layer of water (thickness 0.5 m) is in equilibrium with the oxygen in the
atmosphere, and according to Henry’s law has an oxygen concentration of 2.7 · 10-4
mol dm-3. In the bottom layer of water (thickness 0.5 m) the oxygen concentration
is zero because there all oxygen is consumed. You may assume undisturbed
stationary diffusion in the intermediate layer.
c)
At the bottom of the pool is a fish that (for the purpose of this exercise) does not rise
higher than 0.5 m above the bottom and therefore lives on a certain calculated
oxygen supply. Assume that the fish absorbs all of the supplied oxygen. What is the
maximum weight the fish could be?
The oxygen consumption of freshwater fish is about 0.1 gram O 2 per hour per kg of
body weight. The water pool has a surface of 145 m2.
Review Exercises
11.19 A loose piece of platinum in 0.1 M hydrochloric acid looks completely inert. If a piece
of zinc is placed in contact with the platinum, bubbles are formed on the platinum.
What is the best explanation for this?
A The reaction of platinum with zinc is exothermal; the heat that is released causes
steam bubbles to form.
B When platinum and zinc are brought into contact with each other, the electrical
circuit is closed and water is electrolytically split into hydrogen gas and oxygen gas.
C The oxidation of zinc produces electron with which platinum can reduce hydrated
protons.
D Cathodic hydrogen development on the platinum is thermodynamically already
possible, before the platinum is in contact with zinc, but it is very slow. Zinc operates
as a catalyst.
50
Chapter 12 Kinetic Theory
Theoretical Exercises
12.1 a) Calculate the average free volume < 𝑣𝑓 > (in Å3) per water molecule in liquid water,
assuming a mass density of ρ = 1 g cm-3 at T=298 K.
b) In a) we have not specified the pressure p that is exerted on the liquid water.
Discuss whether this is a serious omission.
c) Calculate <vf> (again in Å3) for water particles in a water vapor at
T=298
K and a pressure of p=1 bar. Assume the vapor obeys the ideal gas law.
d) Estimate the average distance between the particles in the vapor phase.
12.2 Given are 2 cars with speed u = 80 km h-1, 4 cars at u = 100 km h-1 and
10
cars at u = 140 km h-1.
a) Calculate <u>.
b) Calculate the RMS-speed of the cars.
c) Comment on the difference between the answers in a) and b).
12.4 In an isolated vessel, He atoms all start at time t=0 with the same speed of 2.74 ·
104 cm s-1. Next the atoms collide with each other as a result of which the MaxwellBoltzmann distribution is established. Assume that the vessel does not adsorb any
heat.
a) What is the temperature of the gas at equilibrium?
b) What happens to the entropy of the isolated system starting at t=0?
12.5 Marshlands and sewage generate an inflammable gas. A volume of this gas effuses
through an orifice in 12.6 minutes. At the same pressure and temperature the same
volume of oxygen takes 17.8 min to effuse through the same orifice. What is the
molar mass of the gas? What might this gas be?
12.6 An equimolar mixture of hydrogen and oxygen gas in a vessel. A effuses at
25
C through an orifice to vessel B. After a short while, the connection is closed.
°
Evaluate the composition of the gas mixture (in mole fractions) in vessel B.
Math-related Exercises
12.7 A cold, extremely dilute hydrogen gas with a temperature of T=3 K contains a number
density of hydrogen molecules of 1 cm-3.
Data: MH = 1.008 g mol-1 , mH = 1.008 a.m.u.
a) Calculate the hydrogen pressure.
51
b) Estimate the mean free path for hydrogen atoms in the gas. Assume the size of
the molecules is d = 10-10 m.
c) Estimate their collision frequency; how often does a hydrogen atom collide in one
week and one year?
12.8 Estimate the collision frequency on one particle in a gas with a volume fraction φ =
0.5 · 10-4. Assume that particles (with diameter 0.1 nm) move on average at the
speed of sound.
12.9 Calculate the mean free path in a gas for the Particle Quartet from Table 11.1, for a
particle volume fraction of one per cent.
12.10 Calculate the average molar translational kinetic energy at 300 K.
12.11 From the distribution function for particle speeds derived in Chapter 12, show that
the most probable speed is given by:
æ 2kT ö
umax = ç
è m ÷ø
1/2
12.12 Derive a formula for the average square of the velocity x-component <vx2> using
the appropriate Boltzmann distribution. How large is <vx> and why?
12.13 Verify that:
æ m ö
C=ç
è 2p kT ÷ø
1/2
for the distribution in equation (12.53) of Chapter 12 is correct.
12.14 Solve integral (12.69) of Chapter 12.
12.15 For N2 molecules at 500 K, calculate:
a) their root-mean-square speed.
b) their most probable speed.
c) their mean speed of N2 molecules at 500 K.
Assume the N2 gas is in a container at a pressure of 110 Pa. Assume a collision area
of 0.43 nm2.
d) Evaluate the collision frequency for these N2 molecules.
e) What is the average time between two collisions?
52
f) And what is the distance a molecule traverses in this time? Compare this distance
to the diameter of the nitrogen molecule.
12.18 Calculate the ratios umax : <u> : urms.
12.19 At which altitude above the earth is the average kinetic energy of a N 2 molecule
equal to its gravitational potential energy? Do the same calculation for a He atom.
12.20 At which temperature will urms of He atoms be equal to urms of N2 molecules at 25
C?
°
Review Exercises
12.21 a) A volume of gas of 0.280 L (at 1 bar and 273.15 K) weighs 0.400 grams. Calculate
the molar mass of the gas.
b) Baking powder (NaHCO3) releases CO2 gas when heated. Calculate the maximum
amount CO2 can be released in litres, by 5.0 g baking powder at 180 C and 1.3
bar.
c) A solution of glucose in water contains 1 weight percent of glucose. Calculate the
average kinetic energy per mole, at 293 K, of both the glucose and water molecules.
d) Calculate the ratio between the average speed of a glucose molecule and that of
a water molecule, at 293 K.
12.22 Indicate whether the following statements are true or false, without further
explanation. In a sugar solution, sugar molecules start to spontaneously to
associate with each other. Because of this irreversible association:
a) the vapour pressure of the solution decreases.
b) the osmotic pressure of the solution increases.
c) the boiling point of the solution decreases.
d) the weight concentration of the sugar remains the same.
e) the Maxwell-Boltzmann velocity distribution of the sugar molecules does not
change.
53
54
Answers
55
Chapter 1 A First Encounter with the First and Second Law
Theoretical Exercises
1.1
a) A reaction that proceeds by itself, without any external assistance.
b) Exothermic reactions produce heat; endothermic reactions take up heat from the
surroundings.
1.2
a) Mass m quantifies the amount of matter in a body; it also quantifies the inertia of
the body, a resistance to a change dv/dt of its speed, according to Newton’s second
law F=mdv/dt. b) Weight is the force F that the earth exerts on a body: F=mg where
g is the gravitational acceleration.
1.3
a) CH4 + 2O2 ®CO2 + 2H2O
-1
E 604.5 ×10 3 J mol -1
-12 J mol
-12
b) m = 2 =
kg mol -1
8
-1 2 = 6.72 ×10
2 -2 = 6.72 ×10
c
(3.00 ×10 m s )
m s
NB: such a change in mass is too small to be measured but it does imply that mass
is not conserved in this chemical reaction.
1.4
A PM of the first kind creates energy out of nothing: a PM of the second kind is the
cyclic conversion of heat into work (which does not violate the First Law).
When the boat comes to a halt at the quay in Figure 1.6, it donates all its energy of
motion to the quay where this energy is dissipated as heat into disordered molecular
motions. It is extremely unlikely that all molecules involved simultaneously return
their randomly spread kinetic energies to the boat such that it can make another
cycle.
1.5
a)
4 11 H ®
4
2
He + 2 0+1e
4
Here 2 He refers to a helium nucleus with atomic number 2 (its two protons) and a
mass number of 4 (two protons + two neutrons). In addition to one helium nucleus,
also two positrons are formed; their charge (2+) compensates for the loss of two
positive protons.
b) mass loss Δm= 0.0286 g
Þ E = D m×c 2 = (0.0286 ×10 -3 kg)(3.00 ×10 8 m s -1 ) 2 = +2.57 ×1012 J
c) No violation: mass is proportional to energy
1.6
a) Mw(H2O) = 18.016 g mol-1
b) 1.0 dm3= 103 g, which corresponds to
concentration is 55.5 mol dm-3.
1.7
r=
m
0.040 kg
=
= 8.0 ×10 2 kg m - 3
-3
3
V 0.050 ×10 m
56
10 3 g
= 55.5 mol , hence the
18.016 g mol -1
1.8
The molar weight depends on the gravitational acceleration, which is different for the
moon and on earth. Hence, the molar weight is not the same. Though: the molar
mass is the same, which has units of g mol-1, whereas the molar weight has units of
N mol-1.
Math-related Exercises
1.9
a) 1. ½
2. ln(2)
b) 1.ln(AB)
4. (1-e-1)/a
3. ln(2)
2. ln(A/B)
3. ln(B/A)
4. 1
5. 2c
6. p∆V.
5. -2
6. ea+b 7. e
8. e2
1.10 a) ᵅ' (ᵆ) = 4 ᵆ3
b) ᵅ' (ᵆ) = ᵅ ᵆᵅ − 1
c) ᵅ' (ᵆ) = ᵅᵆ
d) ᵅ' (ᵆ) = ᵅ' (ᵆ) ᵅᵅ (ᵆ)
4
e) ᵅ' (ᵆ) = (4 ᵆ3 + 1) ᵅᵆ + ᵆ + ᵆ
f) ᵅ (ᵆ) = 1
ᵆ
g) ᵅ (ᵆ) =− 12
ᵆ
( ) =ᵆ
i) ( ) = 2ᵆ ᵆ
j) ( ) =−
h)
∂ᵅ
∂ᵆ ᵆ
∂ᵅ
∂ᵆ ᵆ
∂ᵅ
∂ᵆ ᵆ
2
ᵆ
ᵆ2
1.11 a) ᵃ (ᵆ) =
1
ᵆᵅ + 1 + ᵃ
ᵅ+1
with C is a constant
b) ᵃ (ᵆ) = ᵆ4 + ᵃ
c) ᵃ (ᵆ) = ᵅᵆ + ᵃ
d) ᵃ (ᵆ) = ᵅᵅ (ᵆ) + ᵃ
3
e) ᵃ (ᵆ) = ᵅᵆ + ᵄ + ᵃ
f) ᵃ (ᵆ) = ln |ᵆ| + ᵃ
g)
∫4ᵆ · ᵆ dᵆ = 4ᵆ · ᵆ − ∫ ᵆ · 4dᵆ = ᵆ − ∫ ᵆ dᵆ = ᵆ − ᵆ + ᵃ = ᵆ + ᵃ
1
3
2
3
1
3
4
3
3
4
4
3
3
4
3
4
1
3
4
4
1.12 Avogadro’s number times the Boltzmann constant equals the gas constant
R = 8.31451 J mol-1 K-1
1.13 Not allowed: a, c, d, f, h, k
1.14 b, g and i have the unit Nm=J
1.15 1
Pa
=
Nm-2
1
5
,
1atm
=
1.01325·105 Pa
12
1 bar = 10 Pa, 1 light year = 9.46 · 10
1.16 a)
ò
V2
V1
,
1
atm=
1.01325
km , 1 nano-ton = 1 mg.
V2 dV
æV ö
æV ö
æV ö
nRT
dV = nRT ò
= nRT ln ç 2 ÷ = pV ln ç 2 ÷ = C ln ç 2 ÷
V1
V
1
V
V
è V1 ø
è V1 ø
è V1 ø
pdV = ò
V2
b) pV in (N m-2) · m3 = J so C has the dimension of energy.
c) pV = nRT; for n=1 mol and T=298 K: C = 2.48 kJ
57
bar
,
1.17 The quadratic formula gives as solutions
− ᵄ + ᵄ2 − 4ᵄᵅ
− ᵄ − ᵄ2 − 4ᵄᵅ
ᵆ=
Vᵆ=
2ᵄ
2ᵄ
1.18 Total number of heart beats per day = 60 · 60 min h-1 · 24 h day-1 =
8.64 · 104 day-1.
A total energy of 1 J · 8.64 · 104 day-1 = 8.64 · 104 J day-1 is needed, which
corresponds to
8.64 ×10 4 J day - 1
= 2.79 mol of ATP .
31×10 3 J mol - 1
1.19 a) p =
F Fw mg 65 kg × 9.81 m s - 2
=
=
=
= 2.55 ×10 4 Pa = 0.252 atm
A A
A
0.025 m 2
b) p =
65 kg × 9.81 m s- 1
= 3.19 ×10 6 N m - 2 = 3.19 ×10 6 Pa = 31.5 atm
2.0 ×10 - 4 m 2
1.20
p=
F mg
=
A A
Mass m = volume V · mass density ρ
volume column V = area A · height h
Hence, p =
V r g Ahr g
=
= r gh
A
A
a) ᵅ = 1 . 0 · 103 kg m-3 · 9 . 81 m s-2 · 10 m=9.8· 104 Pa=0.97 atm
b) ᵅ = 13 . 5 · 103 kg m-3 · 9 . 81 m s-2 · 10 m=1.3· 106 Pa=13.1 atm
58
Chapter 2 Thermodynamic States and Ideal Molecules
Theoretical Exercises
2.1
The air can be approximated quite well as an ideal gas. It is relatively diute and the
molecules don’t have strong interactions. Strictly speaking the answer would be ‘no’
as it is not fully ideal, but therefore ‘yes’ is a reasonable answer.
2.2
V RT 8.314 J mol - 1 K - 1 × 298 K
=
=
= 2.48 ×10 - 2 m 3 mol -1 = 24.8 L mol -1
5
n
p
1 ×10 Pa
2.3
a) Vtot = 1.0 L
-1
= 0.936 g
-1
= 0.258
= 1.207
g g
80 vol% N2, hence VN2 = 0.80 L , so mN2 = 0.80 L ×1.17 g L
20 vol% O2, hence VO 2 = 0.20 L , so mO 2 = 0.20 L ×1.29 g L
So, the total mass is 0.936 g + 1.207 g = 1.194 g.
b) One cannot use the ideal gas law, since the vol% given do not correspond to the
same fractions of partial pressures (i.e. not 0.8 bar and 0.2 bar for respectively N2
and O2.) One can only make the calculation
ptot = pN 2 + pO 2 by knowing the mole
fractions of N2 and O2, which are still unknown in this question.
For clarity, by giving the following calculation, you can see a significant difference in
the mass of the gas mixture, which is incorrect:
pt o t = pN2 + pO 2 = 0.8 bar + 0.2 bar = 1.0 bar
pN 2 =
nN2
RT Þ n N2 = 32.3 mmol; n O 2 = 8.1 mmol
V
So according to this, the mass of one liter of air is:
m = (32.3×10 - 3 mol ×28 g mol -1 ) + (8.07 ×10 -3 mol ×32 g mol -1 ) = 1.16 g
V RT
=
= 0.27 L mol - 1
n
p
2.4
V=
2.5
pV (1.01325×10 5 N m - 2 )(22.414 ×10 -3 m -3 )
R=
=
= 8.314 J mol -1 K -1
nT
(1 mol)(273.15 K)
2.6
10 g glucose L-1 ;
2.7
P =
n 10
n
=
mol L-1 Þ P = RT = 1.36 atm
V 180
V
n
c (= weight)
RT =
RT
V
V ×M
59
Þ M=
2.8
(20 kg m - 3 )(8.314 J mol - 1 K - 1 )(298 K)
= 65 kg m ol - 1 = 65 kDa
763 N m - 2
The final equilibrium pressure is the pressure that would result from removing the
piston. The equal gas volumes mix spontaneously to a final pressure of
p = (1 + 2) / 2 = 1.5 bar .
2.9
a) The volume fraction φ for a vessel volume Vvessel is defined as:
j =
N
Vvessel
×Vmarble
, where N is the number of marbles, each with a volume
Vmarble = (p / 6)d 3 . The vessel volume that accommodates one mol of marbles is
therefore:
Vvessel (per mole) = V =
=
b) L
3
N av × Vmarble
j
6.023×10 23 mol -1 × (p / 6) ×10 -6 m 3
= 4.93 ×10 8 km 3 mol -1
0.64
= 10 -12 mol ×V Þ L = 79.0 m
Math-Related Exercises
2.10
pV
pV p V
= nR = constant , so 1 1 = 2 2 , so
T
T1
T2
V2 =
2.11
p1V1T2 80 ×10 3 Pa ×100 cm 3 × 273 K
=
= 70.5 cm 3
p2 T1
10 5 Pa × 310 K
pH 2 = xH 2 × ptot
2g
nH
2.0 g mol - 1
xH 2 = 2 =
= 0.5
2g
2g
16 g
n to t
+
+
2.0 g mol - 1 4.0 g mol - 1 32.0 g mol - 1
so pH 2 = 0.5 ×12.0 kPa = 6.0 kPa
2.12 molar volume of the gases is the same for each gas:
V RT 8.314 J mol - 1 K - 1 × 298 K
=
=
= 24.78 L mol -1
n
p
10 5 Pa
60
mN 2 mO 2 + mC H 4
M N 2 xM O 2 + (1 - x)M C H 4
=
Û
=
V
V
V
V
Þ M N2 = xO 2 M O 2 + (1 - xO 2 )M C H 4
r N 2 = r O 2 +C H 4 Û
28.02 = x(32.0) + (1 - x)(16.04)
solve for x to obtain xO 2 = 0.75
pO 2 = xO 2 × pt o t = 0.75 ×1 bar = 0.75 bar
2.13 Reaction equation: C2H5OH + 3 O2 à 2 CO2 + 3 H2O,
mole ratio C2H5OH : O2 = 1 : 3
1.5g
1.5 g C2H5OH =
46.0 g mol-1
= 0 . 033 mol, times 3 = 0.098 mol O2
ᵄO2 = 0 . 21 · 1 dm3 = 0 . 21 dm3
ᵅᵄ2 ᵄᵄ 0 . 098 mol·8.314 J mol-1 K-1 · 303 K
ᵅᵄᵅᵅ = ᵅᵄ2 =
=
= 1 . 2 · 106 Pa=12 bar
ᵄ ᵄ2
0.21· 10-3 m3
2.14 ᵅᵄ2 , ᵆᵅᵆ = ᵅᵄ2 , ᵅᵅᵅᵆᵅᵄᵅ + ᵅᵄ2 , ᵅᵅᵄᵅᵆᵅᵅᵅ
Volume gaseous mixture = 2.000 dm3 – 0.200 dm3 = 1.800 dm3
ᵅᵄ
101000 Pa·1 . 800 · 10− 3 m3
=
= 0 . 072 mol
ᵄᵄ
8 . 314 J mol-1 K-1 · 303 K
0 . 100 mol
ᵅᵄ2 , ᵅᵅᵄᵅᵆᵅᵅᵅ =
= 0 . 050 mol
2
ᵅᵄ2 , ᵅᵅᵅᵆᵅᵄᵅ =
ᵅᵄ2 , ᵆᵅᵆ = 0 . 072 mol+0.050 mol=1.12 mol
ᵅᵄ2 , ᵅᵅᵅᵄᵅ =
ᵅᵄᵄ
1 . 12 mol·8.314 J mol-1 K-1 · 303 K
=
= 1 . 71 · 106 Pa=171 kPa
ᵄ
1.8· 10-3
Review Exercises
2.15 a) 4565 mol H2
b) Mass H2=9.13 kg; mass air=137.9 kg. So the balloon can lift 129 kg.
c) Helium is heavier: mass He = 4565 mol · 4.00 g mol-1 = 18.26 kg. So the balloon
can lift 119.6 kg.
d) n =
pV
= fixed when p, V, T are fixed; n is fixed by an equation of state, here the
RT
ideal gas law.
61
Chapter 3 The First Law: the Total Energy Cannot Change
Theoretical Exercises
3.1
A
3.2
C (Not D! The reaction being spontaneous does not inform you about the rate.)
3.3
B
3.4
D
3.5
Processes that are slow enough such that they can be reversed at any moment. Or:
quasi-static processes that are a sequence of equilibrium states.
Or: any process that does not change the total entropy.
Some chemical reaction are almost reversible, such as redox reactions in a galvanic
cell where the electrical current is very small, see Chapter 8.4 on the Nernst equation.
Reversible processes are essential to calculate changes in state functions such as
energy and entropy – see, for example, Lecture Notes sections 3.4 and 4.4.
3.6
DU = q + w
 DU: internal energy of the system
 q: heat (positive for transport of heat into the system)
 w: work (positive for work performed on the system)
3.7
Heat is the transport of energy (via disordered molecular motions) due to a difference
in temperature.
3.8
Here sometimes more than one answer is correct, depending on how you define the
system, but normally we would say:
a) isolated
3.9
b) open
c) open
d) closed
Since heat depends on the path followed to exert a process.
3.10 Trick question: you can agree or disagree but have to provide context!
If one were to drive reversibly, then all of the energy would be converted into motion,
so indeed the car would attain maximum efficiency. However, in reality, friction and
heat dissipation will of course always occur. Notice also, that for true reversible
driving, exhaust gases would have to be converted into gasoline spontaneously upon
driving backward (in the same way as we encountered for the perpetuum mobile of
the second kind). So while this statement is true for an idealized world or in a useful
mathematical exercise, it is false in the real world in the sense that is impossible /
violates the Second Law.
3.11 Virtually no NO present
3.12 a) Evaporation of a solid, or any liquid
b) Burning of coal; the thermite reaction
c) The reverse of any reaction in b)
d) Condensation of water molecules in an unsaturated vapor. NB: the molecules
attract each other and heat is liberated when they condense. However, condensation
62
of gas implies a high entropy penalty which makes condensation a non-spontaneous
process.
e) Spreading of ideal gas molecules in vacuum, or ideal solutes in a solvent.
f) Compression of ideal molecules in gases or solutions.
3.13 By oxidation of Zn in a galvanic cell (see chapter 8.1); then the oxidation delivers
electrical work.
3.14 No, because dimensions are wrong.
3.15 Here sometimes more than one answer is correct, depending on how you define the
system, but normally one would say:
a) closed, w negative as electrical work is done on the resistance, q may be negative
if the battery also loses heat in this process.
b) closed, q negative as heat is lost in the form of friction.
c) open, as matter (air) is able to be exchanged.
3.16 a) See Lecture Notes section 3.4
b) -nRTln(2) J = -1.0000 mol · 8.3145 J mol-1 K-1 · 298.15 K · ln(2)=-1.7183 kJ ;
negative because the gas delivers work thereby losing energy.
c) q = +RTln(2). For an ideal gas at constant T, DU = 0.
3.18 a) 3600 L
b) Step 1: steam from 500 °C to 100 °C:
à q1 = 2 kJ kg-1 K-1 · 400.0 K · 1.000 kg = 800.0 kJ.
Step 2: steam of 100 °C to liquid water of 100 °C:
à q2 = 2258 kJ.
Step 3: water of 100 °C to water of 25 °C:
à q3 = 4.180 kJ kg-1 K-1 · 75.00 K · 1.000 kg = 313.5 kJ.
qtot = 800.0 kJ + 2258 kJ + 313.5 kJ = 3372 kJ.
Comment: if you burn your fingers with boiling water, you only suffer from the heat
of condensation in step 3; if steam of 500 °C is cooled to room temperature the heat
release is almost ten times as large.
3.19 Since pV = nRT = constant, and since the energy of an ideal gas does not change in
an isothermal process we find from the definition of enthalpy
H=
U-pV: DH = DU + D(pV) = 0. Conclusion: the enthalpy of an ideal gas does not change
if you compress or expand the gas. The same holds for the concentration and dilution
of ideal solute molecules in a solution.
3.20 a) The enthalpy change for 125 g of iron oxide =
125 g
159 . 67 g mol− 1
· 851 kJ mol− 1 =− 666 kJ
Assuming that the reaction only delivers volume-work qp = DH such that the heat
production (a positive quantity) is qp = +666 kJ.
63
b) To heat up 1 kg water with 75 K you need: 4.18 kJ kg
Thus,
starting
from
1
kg
of
iron
-1
K -1 ×75 K = 313.5 kJ kg -1 .
oxide
we
can
heat
up
5328 kJ
= 17 kg = 17 L of water .
313.5 kJ kg - 1
3.21 If T increases, the equilibrium will shift to the endothermic side, so less [H+], so the
pH will increase.
3.22 a) For one mole ideal gas: CV = (3/2)R = 12.47 J K-1 (temperature is not needed
here).
b) Argon is a mono-atomic gas that obeys the ideal gas law very well; the higher
heat capacity of oxygen manifests additional degrees of freedom to store energy,
here the rotational motions of the di-atomic molecule O2. NB: two rotations
contribute such that O2 has in total 5 degrees of freedom (including 3 translations)
such that CV = (5/2)R = 20.78 J K-1.
3.23 a) pV = nRT Þ
nR 1 mol ×8.314 J mol - 1 K - 1
æ ¶V ö
=
=
= 8.314 ×10 - 5 m 3 K - 1
çè ÷ø
5
¶T p p
10 Pa
For p = 2 and 3 bar: (/)
V
Tp
= 4.16 · 10-5 m3 K-1 and 2.77 · 10-5 m3 K-1,
respectively.
b) This partial derivative is the volume of an ideal gas that increases linearly with
temperature when the pressure is constant.
Math-related Exercises
3.24 W = Fgh = mgh = 75 kg · 9.81 m s-2 · 10 m = 7.4 kJ.
3.25 Ideal molecules in either gases or solutions have only kinetic energy. Thus U is
proportional to temperature T implying that (V/ T)  0
p
.
Comment: this result expresses that ideal molecules do not interact; when the
average distance between molecules changes because they are brought into a smaller
or large volume, this has no effect on the internal energy of gas or solution. In other
words: the energy or enthalpy (see also exercise 3.19) of dilution of ideal molecules
is zero.
3.26 From the definition of enthalpy H = U + pV it follows that DH = DU + D(pV). Since for
an ideal gas at constant temperature D(pV) = D(nRT) = 0 and also DU = D(3nRT/2)
= 0, it follows that DH = 0.
3.27 a)
- (1 - 4x 2 ) sin(x) - 8x cos( x)
e)
1 - ln(x)
x2
64
b)
c)
6 + 18x - 3x2 - 4 x3 - 3x 4
(3 + x3 ) 2
x
f)
( 5 + 3x) e x
2 + x2
g)
4x - 3
2x 2 - 3x + 1
(
d) 1 + ln( x)
h) 2 x 1 + 2 x
2
)e
2 x2 + 3
3.28 First, rewrite each expression to either V as function of p, or p as a function of V and
calculate dp/dV for the latter case, since dV/dp=(dp/dV)-1, one can determine dV/dp.
a) V =
nRT
Þ
p
dV
nRT
=- 2
dp
p
æ 1 nB ö
+
Þ
è V V 2 ÷ø
dp
nRT
æ 1 2nB ö
= nRT ç - 2 - 3 ÷ = - 3 ( V + 2nB )
è V
dV
V ø
V
b) p = nRT ç
Þ
dV
V3
=dp
nRT ( V + 2nB )
c) p =
Þ
nRT
dp
nRT
p
Þ
=2 =V - nb
dV
V - nb
(V - nb )
dV
V - nb
=dp
p
d) p =
nRT
n2 a
- 2 Þ
V - nb V
dp
nRT
2n 2 a
=+
2
dV
(V - nb ) V 3
dV æ 2n 2 a
nRT ö
Þ
=ç 3 dp è V
(V - nb ) 2 ÷ø
-1
Review Exercises
3.29 a) dᵅ =- ᵄᵅᵅ è
dᵆ
dᵅ
ᵄᵅ
=−
dᵆ è
ᵅ
ᵄᵄ
ᵄᵄ
()
ᵅ
ln ᵅ =−
0
ᵄᵅ è
ᵅ = ᵅ0 exp
ᵆ
ᵄᵄ
(− ) = ᵅ exp (− )
ᵄᵅᵆ
ᵄᵄ
ᵄ
ᵄᵄ
0
b) ᵄ = ᵄᵅᵆ
()
3.30 a) ᵆᵅᵅᵆ =− ᵅᵄᵄ ln ᵄ2 =− ᵄᵄ ln⁡(2)
ᵄ1
Sign is negative, which means that the system delivers work.
b) ᵆ
ᵅᵅᵆ =−
=− ᵄᵄ ln
(
∫
ᵄᵄ
ᵄ
−ᵄ
ᵅ
dᵄ =− ᵄᵄ
∫
ᵅ
dᵄ =− ᵅᵄᵄ
ᵄ − ᵄᵅ
) =− ᵄᵄ ln⁡(2 . 11)
4 . 0 ᵅᵅ3 − 0 . 2 ᵅᵅ3
2 . 0 ᵅᵅ3 − 0 . 2 ᵅᵅ3
65
∫
(
ᵄ − ᵄᵅ
1
dᵄ =− ᵅᵄᵄ ln ᵄ2 − ᵄᵅ
ᵄ − ᵄᵅ
1
)
3.31 a) 9.81 kJ
b) 9.81 kJ
c) total molecular kinetic energy = (3/2) nRT = 20.6 · 106 J.
d) 20.9 km
66
Chapter 4 The Second Law: the Total Entropy Cannot Decrease
4A) Theoretical Exercises
4.1
C
4.2
A
4.3
D
4.4
C
4.5
D
4.6
A
4.7
a) False
b) False
c) True
d) True
e) False
f) False
g) False
h) True (depends on how you argue; Avogadro’s constant has the unit mole-1, but
very often the number and constant are considered identical).
i) True
4A) Math-related Exercises
4.8
D U = qrev + w rev = 0 and volume work is
a) For a reversible isothermal expansion:
given by: wv,rev =
V2
ò - pdV = - nRT ln V .
1
V
Þ qrev = nRT ln 2
V1
Thus entropy is given by: D S sys =
qrev
T
Þ
V
D S sys = nR ln 2
V1
b) V2 > V1, hence ln⁡ ᵄ2 > 0, so ∆S > 0; molecules are divided over a bigger space,
ᵄ1
hence more microstates / disorder.
c) p1V1 = nRT; p2V2 = nRT; fill in in working book equation (4.1) gives
D S = nR ln
p1
p2
which is equal to D S = - nR ln
p2
.
p1
d) dS = dqrev/T is correct. The conclusion dS = 0 is wrong because working book
equation (4.3) is only valid for reversible processes. qrev can be calculated by
67
performing the expansion reversibly: to do this you have to break the isolation.
Hence, heat exchange will occur (see Lecture Notes sections 3.4 and 4.4).
æ ¶U ö
, see Lecture Notes page 52, the energy of n moles
è ¶T ÷ø V
4.9 a) For derivation of C V = ç
of ideal gas is U = (3/2)nRT such that Cv = (3/2)nR.
æ ¶H ö
æ ¶U ö
æ ¶V ö
=ç
+ pç ÷ .
÷
÷
è ¶T ø p è ¶T ø p
è ¶T ø p
b) Cp is given by: C p = ç
æ ¶U ö
æ ¶U ö
çè
÷ø = çè
÷ , because U only depends on T, so it is the same with p fixed as
¶T p
¶T ø V
with V.
nR
æ ¶V ö
çè ÷ø =
¶T p
p
; V=
nRT
from ideal gas law
p
æ ¶V ö
Þ p ç ÷ = nR
è ¶T ø p
Þ C p = C V + nR
c) Part of the heat that is added to the constant pressure gas is returned to the
environment as volume-work, whereas the heat added to the constant volume gas is
all used to increase the gas temperature, which is why the heat capacity CV is smaller
than Cp.
4.10 See Lecture Notes Figure 4.12. A heat source with temperature Th delivers heat qh, which
is partly converted to work w. The remaining heat qc enters the cold sink with temperature
Tc.Assuming the device works reversibly:
D U to t = q + w = q h ,rev + qc, rev + wrev = 0
Þ - wr ev = qc ,rev + qh ,r ev
The efficiency is defined as e rev =
energy output as work -w rev
=
energy input as heat
qh,rev
Since the heat machine delivers work wrev<0, to make the efficiency a positive number we
define it as:
e rev = 1 +
qc,rev
q h,rev
; qc,rev < 0 and qh,rev > 0
68
Since for individual entropy changes D S =
qre v
, we have:
T
q h,rev qc,rev
+
=0
Th
Tc
q
q
Þ h,rev = - c,rev
Th
Tc
T q
Þ - c = c,rev
Th qh,rev
D S tot =
Þ e rev = 1 -
Tc
Th
nRT 1.000 mol × 8.314 J mol - 1 K - 1 × 283.15 K
=
= 0.0232 m 3 = 23.2 L
5
p
10 Pa
4.11
V=
4.12
dS = dqrev/T , which gives by integration: DS = qrev/T since T is taken constant.
DS = (100.0 J) / (293.15 K) = 0.341 J K-1
4.13
V2
V2
V1
V1
w = ò - pex dV = pex ò dV = - pe x (V2 - V1 )
w = - (1.000 atm ×101325 pa atm - 1 ) × (3.000 ×10 -3 m 3 - 1.000 ×10 -3 m 3 )
= -202.7 Pa m 3 (= kg m -1 s -1 × m 3 = kg m 2 s -2 ) = -202.7 J
The absolute value of the volume work is: |w| = |-202.7 J| = 202.7 J
4.14
dS = dqrev / T ; dqrev = C dT
æT ö
C
283.15 K
dT = C ln ç 2 ÷ = 12.0 J K -1 × ln
= -0.416 J K -1
T1 T
è T1 ø
293.15 K
DS = ò
T2
4B) Theoretical Exercises
4.15 Because the energy U of an ideal gas only comprises molecular kinetic energy, which
is proportional to temperature. Whether the expansion is reversible or not does not
make any difference here.
4.16 C: The gas does not perform work, w=0, and since ΔU=0, it follows that also q=0.
For a reversible isothermal expansion, work is non-zero, and q=-w such that ΔU=0.
4.17 C: H = U + pV

DH = DU + D(pV). At constant temperature ΔU=0 for an ideal gas.
Further;pV = nRT = constant, so DH=0.
69
4.18 A
4.19 B: in the isothermal reversible expansion of gas, the added heat is completely converted to
work. And when heat is added, boiling water remains at 100 °C.
4.20 a) No; b) Yes; c) Yes; d) No; e) Yes
4.21
1.000 J ×
kg m 2 s -2
g
cm 2
erg
×10 3 ×10 4 2 ×
= 10 7 erg
2 -2
J
kg
m g cm s
70
4B) Math-related Exercises
4.23 a) w = -
ò
V2
V1
pex dV = - ò
V2
V1
nRT
V
V
dV = - nRT ln 2 = - p1V1 ln 2
V
V1
V1
= -101325 Pa ×1.000 ×10 -3 m 3 ln
0.500 L
= +70.2 J
1.000 L
b) DUT,n = 0 for an ideal gas such that q = -w.
DS = qrev/T because T is constant.
Together, this gives DS = -wrev/T = –(70.2 J) / (293.15 K) = -0.2395 J K-1
c) H º U + pV
DH = DU + D(pV)
DUT,n = 0 for ideal gas.
D(pV) = 0 for ideal gas at constant T and n because pV = nRT.
DH = DU + D(pV) = 0 + 0 = 0
4.24 a) D S =
qrev
= +92 J K -1 and +67 J K -1 for, respectively, 0 °C and 100°C.
T
b) For copper idem.
c) We calculate in a) and b) the entropy change DSsys of a system, here a block of
metal, not the total entropy to which the lecturer refers. When heat is transferred
reversibly to the system from the surroundings at (almost) the same temperature,
the entropy change DSsurr in the surroundings compensatesDSsys such that: DStot =
DSsurr + DSsys = 0 in agreement with the Second Law.
NB: reversible heat transfer to a block of metal can be approached by immersing the
block in a large thermostat filled with water at a constant temperature T + DD
 that is slightly above that of the block temperature T, i.e. DT << T.
4.25 a) It is an isolated system, hence no change in total energy, so ΔU = 0 J, the total
amount of heat is constant, so q = 0 J, hence w = 0 J since ΔU = q + w. ΔH = 0 J
since pressure is kept constant and there is no electrical work, hence ΔH = qp = 0 J.
dS = d ᵅᵅᵅᵆ, since T is not constant, one can rewrite dqrev to dqrev = CdT, i.e. dᵄ= ᵃdᵄ,
ᵄ
ᵄ
hence Δᵄ=ᵃ ln
()
ᵄ2
ᵄ1
Since the system is isolated, ΔStot = ΔSsys (ΔSsur = 0 J K-1) and there are two systems,
namely two copper blocks:
Δ ᵄᵆᵅᵆ = Δ ᵄᵄᵅᵅᵅᵅ100 + Δ ᵄᵄᵅᵅᵅᵅ0 = ᵃ ln
( ) + ᵃ ln ( ) = ᵃ (ln ( ) + ln ( ))
ᵄᵅ , 100
ᵄᵅ , 100
ᵄᵅ , 0
ᵄᵅ , 0
ᵄᵅ , 100
ᵄᵅ , 100
ᵄᵅ , 0
ᵄᵅ , 0
Tf,100 = Tf,0 = 50 °C = 323 K, Ti,100 = 100 °C = 373 K, Ti,0 = 0 °C = 273 K, m = 104 g,
so:
( ( )
323 K
( )) =+93.4 J K
323 ᵃ
Δᵄ ᵆᵅᵆ = 0 . 385 J K-1 g-1 · 104 g ln 373 K + ln 273 ᵃ
71
-1
b) The sign of ΔS is positive, since this equals ΔStot this is an irreversible process.
4.26 a) The expression for the reversible volume is given by
ᵆᵆ , ᵅᵅᵆ =−
∫ᵅ ᵅᵄ, in case p
ex
ᵅᵆ
equals p, we can substitute the ideal gas law for p, inserts
ᵅᵄ = ᵅᵄᵄ ⟹ᵅ =
ᵅᵄᵄ
ᵄ
Substitution yields
ᵄ2
ᵆᵆ , ᵅᵅᵆ =−
∫
ᵄ1
ᵄ2
ᵅᵄᵄ
dᵄ =− ᵅᵄᵄ
ᵄ
∫ ᵄ1 dᵄ =− ᵅᵄᵄ [ln (ᵄ)] ᵄᵄ =− ᵅᵄᵄ ln (ᵄᵄ )
ᵄ1
2
2
1
1
Here, we can substitute V1 = 3V2, which yields:
ᵆᵆ , ᵅᵅᵆ =− ᵅᵄᵄ ln
( )
()
ᵄ2
1
=− ᵅᵄᵄ ln
=− ᵅᵄᵄ ln (3− 1) = ᵅᵄᵄ ln⁡(3)
3 ᵄ2
3
b) Again, the general expression
ᵆᵆ , ᵅᵅᵆ =−
∫ᵅ ᵅᵄ can be used, where p equals p.
ex
ᵅᵆ
However, there is no change in volume given, but a change in pressure, which means
that the integration variable dV has to be changed. For an ideal gas, we can use the
ideal gas law to substitute V as ᵄ = ᵅᵄᵄ, substitution yields:
ᵅ
ᵄ(ᵅ * )
∫
ᵆᵆ , ᵅᵅᵆ =−
ᵄ(ᵅ)
ᵅ*
ᵅ*
ᵅ*
∫
∫
ᵅᵄᵄ
1
ᵅdᵄ
=−
ᵅd ᵅ− 1 ,
ᵅd
=− ᵅᵄᵄ ᵅd =− ᵅᵄᵄ
ᵅ
ᵅ
ᵅ
ᵅ
ᵅ
∫
To evaluate this integral, we can use the given standard integral, which yields
(reading x as p):
ᵅ*
ᵆᵆ , ᵅᵅᵆ =− ᵅᵄᵄ
∫
ᵅ*
ᵅd ᵅ
−1
=− ᵅᵄᵄ
ᵅ
∫
ᵅ*
ᵅ (− ᵅ
) dᵅ = ᵅᵄᵄ ∫ ᵅ
−2
ᵅ
−1
dᵅ = ᵅᵄᵄ ln
ᵅ
()
ᵅ*
ᵅ
Note that in the first integral, the integration variable is V, so the boundaries of the
integral have to be volume quantities, which are represented by the volume by
pressure p, V(p), and the volume by pressure p*, V(p*). By making the substitution
for V by nRT/p, the boundaries have to be pressure quantities, p and p*.
c) If p* > p, then the value of ln(p*/p) is positive, (usually written as ln(p*/p) > 0),
such that ᵆᵆ , ᵅᵅᵆ is positive (because n, R and T are positive quantities too). A positive
value for work indicates that work should be performed on the system, so work should
be performed to transfer the n moles of gas from pressure p to p* when p* > p.
4.27 a) Derivation of (4.29) from (4.21): see Lecture Notes.
b) When VA = VB, you will get:
( (
)
(
))
ᵄᵃ
ᵄᵃ
1
1
ᵅᵆᵅᵆ
+ ᵅᵆᵅᵆ
2
ᵄᵃ + ᵄᵃ
2
ᵄᵃ + ᵄᵃ
1
1
=− ᵄ ᵅᵆᵅᵆ 0 . 5 ᵅᵅ
+ 0 . 5 ᵅᵅ
=+ ᵄ ᵅᵆᵅᵆ ln⁡(2)
2
2
So the mixing entropy per mole of gas is ᵮ ᵄᵅᵅᵆ = 0 . 7ᵄ =+ 5 . 76 ᵃ ᵅᵅᵅ− 1 ᵃ− 1
ᵮ ᵄᵅᵅᵆ =− ᵄ
(
()
( ))
ᵅᵆᵅᵆ
Since mole fractions xA and xB are positive, ΔSmix > 0, hence spontaneous.
c) When VA = 10VB, you will get:
( ( (( )
) ( )) (
))
ᵄᵃ
10ᵄᵃ
1
1
ᵮ ᵄᵅᵅᵆ =− ᵄ ᵅᵆᵅᵆ
+ ᵅᵆᵅᵆ
1
10
2
ᵄᵃ + 10ᵄᵃ
2
ᵄᵃ + 10ᵄᵃ 72
=− ᵄ ᵅᵆᵅᵆ 0 . 5 ᵅᵅ
+ 0 . 5 ᵅᵅ
=+ ᵄ ᵅᵆᵅᵆ (1.25)
11
11
So the mixing entropy per mole of gas is ᵮ ᵄᵅᵅᵆ = 1 . 25·ᵄ =+ 10 . 36 ᵃ ᵅᵅᵅ− 1 ᵃ− 1
ᵅᵆᵅᵆ
c) By increasing the ratio between the volumes, the change in entropy of mixing will
increase. This is due to more disorder in particle distribution. With respect to the
smallest volume, the higher degree of disorder will mainly affect ΔSmix.
4.28 By using three gases, the general expression for ΔSmix will be:
æ
æ
ö
æ
ö
æ
öö
VA
VB
VC
D S mix = - R ç xA ln ç
+
x
ln
+
x
ln
B
C
çè V + V + V ÷ø
çè V + V + V ÷ø ÷ø
è VA + VB + VC ÷ø
è
A
B
C
A
B
C
By filling in the values as given:
1
2
= 0.25 ; xB = = 0.5
4
4
æ
æ V ö
æ 2V ö
æ 3V ö ö
D S mix = - R ç 0.25ln ç ÷ + 0.5 ln ç ÷ + 0.25ln ç ÷ ÷
è 6V ø
è 6V ø
è 6V ø ø
è
xA = xC =
æ
æ 1ö
æ 1ö
æ 1 öö
= - R ç 0.25ln ç ÷ + 0.5 ln ç ÷ + 0.25 ln ç ÷ ÷ = +9.73 J mol -1 K - 1
è 6ø
è 3ø
è 2 øø
è
Review Exercises
4.29 ΔS consists of two contributions, namely the freezing of water at 0 °C, followed by
the change in temperature from 0 to –10 °C.
Note: 1 mole of water equals 18.0 g (Mw(H2O) = 18.0 g mol-1)
Freezing of water at 0 °C (=273 K): ΔG = 0 J, so ΔH – TΔS = 0, hence
Δᵄ =
Δᵃ
ᵄ
, ΔH = –332 J g-1 so Δᵄ= Δᵃ = − 332 J g-1 · 18 . 0 g = –21.9 J K-1. Note the minus sign of ΔH,
ᵄ
273 K
since freezing is the opposite of melting.
()
Change in temperature: Δᵄ=ᵃ ln ᵄ2 (see answer of 4.3a). Here, C refers to the
ᵄ1
heat capacity of the ice, since one has frozen the water in step 1). T1 = 0 °C = 273
( )
K and T2 = –10 °C = 263 K, so Δᵄ = 2 . 2 J K-1 g-1 · 18 . 0 g ln 263 ᵃ = –1.5 J K-1
273 ᵃ
So, ΔS for this irreversible process is the sum of these two contributions, so ΔS =
–21.9 J K-1 + (–1.5 J K-1) = –23.4 J K-1.
NB: This value of ΔS refers to the system, which is compensated by the change in
entropy of the surroundings, such that ΔStot is positive. Freezing at –10 °C occurs
spontaneously.
4.30 The gas constant R is given by ᵄ = ᵅB ᵄᵃᵆ, (see Lecture Notes equation 2.9),
substitution in working book equation 4.6 gives:
ᵅᵄ = ᵅ ᵅB ᵄᵃᵆ ᵄ
73
which is equal to (a bit reorganizing)
ᵅᵄ = ᵅ ᵄᵃᵆ ᵅB ᵄ
Here, ᵅ ᵄᵃᵆ represents the number of moles gas times Avogadro’s number, which is
equal to the total number of molecules N, so substitution yields working book
equation (4.7):
ᵅᵄ = ᵄ ᵅB ᵄ
You do not have to make any additional assumptions, since only the definition of the
gas constant and of the number of particles have been substituted. These two
equations are valid for all cases where ideal gases are involved.
74
Chapter 5 The Gibbs Energy
5A) Theoretical Exercises
5.1
A (same as in the other cases where we drop the ‘sys’ subscript; otherwise we add
the ‘tot’ subscript)
5.2
d = infinity small change, Δ is total change between initial to final state.
5.3
a) DH = -6 kJ mol-1 (Nb: formation enthalpy of oxygen is by definition zero).
b) per mole of the reaction as it is written.
c) for the reaction: 2SO 2 + O 2 ® 2SO 3
; D H = -12 kJ mol - 1
Equilibrium so D G = D H - T D S = 0 J mol
-1
Þ DS =
D H -6 kJ mol -1
=
= -20.1 J mol -1 K -1
T
298 K
d) Note that the entropy decrease in c) is that of the reaction (the system). In
equilibrium this decrease is compensated by the entropy increase in the surroundings
such that DStot = 0 J.
5.4
a) D G = 0 J
D S = +261.6 J K -1
Þ
b) Positive: liquid molecules migrate to a gas with higher entropy, which
compensates for the positive enthalpy change (evaporation is endothermic) such
that DG = 0 J.
c) An equilibrium (constant) does not depend on pressure. A change in pressure
will only change the boiling point of benzene.
5.5
a) A galvanic cell is an electrochemical cell that generates electrical energy from
redox reactions that take place spontaneously within the cell.
b) reversible = one can turn over the reaction taking place
isothermal = temperature is kept constant
isobaric = pressure is kept constant.
c) DU = –141.0 kJ, DH = –142.9 kJ, DS = –82.6 J K-1, DG = –118.3 kJ.
d) DG equals the delivered electrical work.
5.6
a) True, b) True, c) True, d) False, e) True, f) False
5A) Math-related Exercises
5.7
ΔU = 0 J ; ΔH = 0 J ; w = –5743 J
5.8
a) dG = Vdp
∆ ᵃ = ᵅᵄᵄᵅᵅ
; DS = +19.14 J K-1 ;
DG = –5743 J
( ) = 1 ᵅᵅᵅ ∙ 8 . 3145 J mol K ∙ 298 K ∙ ᵅᵅ ( ) = 1717 J
ᵅᵅ
ᵅᵅ
-1
-1
2 ᵄᵄᵅ
1 ᵄᵄᵅ
b) ΔG = 3435 J
c) Sign is positive, since the compression of a gas is a forced reaction.
5.9
ΔG is defined as G(p2) – G(p1) = ΔG = ΔH – TΔS at constant T.
75
æ V2 ö
æpö
= nR ln ç 1 ÷ .
÷
è V1 ø
è p2 ø
for an ideal gas at constant T we found that: D S = nR ln ç
ΔH = 0 J because T=constant and pV=nRT = constant. Thus filling in for ΔG gives:
æp ö
D G = nRT ln ç 2 ÷ .
è p1 ø
5B) Theoretical Exercises
5.10
a) T, b) F NB: ‘standard’ only relates to a pressure of 1 bar. c) F d) Very False e) F,
f) T.
5.11
a) T b) F c) F d) T e) T
5.12
a) ΔGo = ΔHo – TΔSo = –2864 kJ
b) 0.402 mmol fructose and 0.804 mmol glucose
c) the same since G is a state function,
d) by coupling to other reaction(s) with negative DG.
5.13
a)DG = +2066 J mol-1 (no)
b) DG = –7506 J mol-1 (yes)
c) K = 6.08 · 10-3
d) pH 2 = 12.8 bar
e) K = 164
∆G° = –21.21 kJ mol-1
;
f) endothermic
5.14
a) It is DG if all gas have the standard pressure po = 1 bar (T not specified).
b) K =
(p
/ p0
NH 3
)
0 3
2
(p / p ) (p / p )
H2
c) D G = D G
Þ ln K = -
0
N2
0
+ RT ln Q ; D G 0 = - RT ln K
-33.2 kJ mol - 1
= 13.4 Þ K = e 13.4 = 6.60 ×10 5
RT
K is very big, so H2 and N2 are almost completely converted into NH3.
d) Le Chatelier: the equilibrium moves into the direction of “less molecules” to
counteract the increase of pressure, into NH3 (K remains constant!)
2
2
æ p0 ö
æ p0 ö
DG 0
e) K = ç
Þ
ln
=
çè x ÷ø
è x ÷ø
RT
æ p0 ö
Þ ln ç ÷ = 6.7 Þ x = 1.2 ×10 -3 atm
è xø
76
77
5B) Math-Related Exercises
5.15
a) All gas pressures are at p0 =1 bar.
b) DH > 0 J so reaction is endothermic; entropy increases because the reaction
increases the number of gas molecules from two to three molecules.
c) DGo = =+30 kJ mol-1
d) T=541 K
5.16
5.17
a) D G
0
= 40.6 kJ mol -1 - 298 K ×108.71 J mol - 1 K - 1 = +8.2 kJ mol -1
b) D G
0
= 0 J mol -1 ; water boils at 100 °C, but not at 25 °C.
(1) Use dG = Vdp with V = constant, hence ΔG = V(p2 – p1) = 1.82 J
æ p2 ö
= 1717 J
è p1 ÷ø
(2) Use dG = Vdp = nRT ln ç
5.18
DHo = +80 kJ mol-1
5.19
DG = –28.2 kJ mol-1, so yes: the reaction will occur spontaneously.
5.20
D G = D G 0 + RT ln Q , spontaneously when ΔG < 0 J mol-1, for calculation one has to
assume ΔG = 0 J mol-1, hence:
æ
ö
æ DG 0 ö
63.7 ×10 3 J mol -1
Q = exp ç =
exp
= 1.8 ×10 -11
-1
-1
÷
ç
è RT ø
è 8.314 J mol K × 310 K ÷ø
In addition: Q =
[ C 2 H 4 O ][ NADH ] éëH + ùû = [ C 2 H 4 O ] éë H+ ùû × [ NADH ] = 1.8 ×10 - 1 1
[ C2 H 5 OH ] éëNAD + ùû
[ C2 H 5 OH ] éëNAD+ ùû
1.8 ×10 -11 × [ C2 H 5 OH ]
1.8×10 -11
=
[ C 2 H 4 O] éëH + ùû
[ C 2 H 4 O] éëH + ùû
[ C 2 H 5 OH ]
Þ
[ NADH ] =
=
1.8 ×10 -11 ×1.0 ×10 -5 mol dm -3
mol dm -3 = 0.185
-4
-3
-7
-3
1.0 ×10 mol dm ×10 mol dm
éëNAD+ ùû
5.21 On substitution of (A) in (B) and solving for pA using the quadratic formula we find:
pA =
-1 ± 1 + 4K ptot
2K
Disregard the negative root, pressures must be positive:
pB = ptot - pA = ptot -
-1 ± 1 + 4K ptot
2K
78
5.22 a) ln K = –DGo/RT
;
D G 0 = D G 0f ,H 2 O(l)
;
DGo for H2 and O2 is 0 J.
æ -237.129 kJ mol -1
ö
Þ K = exp ç
= 3.5 ×10 41 ; in equilibrium there is almost no
-1
-1
è 8.314 J mol K × 298 K ÷ø
H2 and O2.
b) DG = DGo + RT ln Q
1
Q=
=
pO 2 × pH 2
1
1
=
1× 2 2
Þ D G = -237.129 kJ mol - 1 + RT ln
1
= -238.85 kJ mol - 1
2
c) This makes sense, since DG < 0 J.
d) DGo = DHo – TDSo
D H 0 - D G 0 -285.83 kJ mol - 1 - 237.129 kJ mol -1
=

T
298 K
= -0.16 kJ mol -1 K -1
Þ D S0 =
DS° < 0 J, because molecules go from big (gas) to small (water) volume.
e) < E >=
Þ
3
R
kT ; T = 298 K ; k =
= 1.38 ×10 -23 J K -1
2
N av
< E >= 6.17 ×10 -21 J
f) DfHo < 0 J. Assuming this is not pressure-dependent, reaction (1) is exotherm with
other pressures p ¹ 1 atm. So if T increases, it moves to the endothermic side: more
H2 and O2.
5.23 a) ΔG° = ΔH° - TΔS°, so
Δᵄ° =
(
− 2816 kJ mol-1 − − 2879 kJ mol-1
Δᵃ° − Δᵃ°
=
ᵄ
298 . 15 K
) = +0.211 kJ mol-1 K-1
b) Since ΔS° increases, the number of microstates increases too.
c) This is the exact opposite reaction of (1), so ΔG° = +2879 kJ mol-1
d) Additional energy has been added, for example via sunlight.
Review Exercises
5.24 a) Isothermal process = temperature remains constant during the process.
b) By using equation (4.11), dqrev = nCdT, substitution into equation (4.10) gives:
S2
T2 dT
dq rev
T
= nC ò
= nC ln 2
q1
T1 T
T
T1
D S = ò dS = ò
S1
q2
79
5.25 a) See Lecture Notes par. 5.1. Performing the integration of (5.8) gives
0
D S surr
=-
0
D H sys
.
T
b) For an equilibrium process under standard pressure, ∆G°= 0 J, hence
∆H° – T∆S° = 0 J, where all terms refer to the system. Rewriting gives
0
D S sys
=+
0
D H sys
. The signs are different because in equilibrium, ∆Stot = 0 J, i.e. ∆Ssys
T
= –∆Ssurr, so the change in entropy for the surroundings has the opposite sign of the
change in entropy of the system.
80
Chapter 6 Liquid-Gas Equilibria
Theoretical Exercises
6.1
A
6.2
C
6.3
B
6.4
C
6.5
C The thermodynamic constant must be dimensionless, so a) and d) are anyhow
wrong: K = (x=1)/ (p/p0) = 1/0.5 = 2.
6.6
a) Water, since a KD > 1 means a higher solubility in that liquid phase.
b) No, since you cannot have negative concentrations, KD has values higher or equal
to 0. A value between 0 and 1 means a higher solubility in the second liquid phase
and a value above 1 means a higher solubility in the first liquid phase.
c) KD < 1, so it will dissolve more in water. You can first immerse the mixture in
water, the solute will dissolve mostly in water. This can be extracted by a separatory
funnel. Afterwards, the water can be evaporated to yield the pure solute.
NB: You can also evaporate the hexane directly, since it has a relatively low boiling
point.
6.7
*
a) pi = xi pi
*
; for each component i with mole fraction x ; p i = vapor pressure of the
pure liquid.
pB
xB pB*
xB pB*
=
=
b) y B =
pB + pT xB pB* + xT p*T p*T + xB ( p*B - pT* )
c) y B =
(Last step: xT=1-xB)
0.5× 0.513 atm
- 0.73
0.185 atm + 0.5 × ( 0.513 atm - 0.185 atm )
pT = 0.5 ×0.185 atm = 0.0925 atm
6.8
The HCl fountain
a)
V=
nRT 1 mol ×8.314 J mol -1 K -1 × 298 K
=
= 24.8 dm 3
5
p
10 Pa
Þ z = 3 24.8 dm 3 = 29.2 cm
b) HCl dissolves spontaneously (and also very fast) in water. The pressure in the
vessel will drop such that the external pressure pushes water upwards through the
capillary; more water enters the vessel in which more HCl can dissolve.
c) D H
0
= +92.3 kJ mol -1 - 167.2 kJ mol -1 = -74.9 kJ mol -1
D U 0 = D H 0 + D ( pV ) , but since no expansion-work is involved:
D U 0 = D H 0 = -74.9 kJ mol - 1
81
D S 0 = 56.5 J mol - 1 K - 1 - 186.9 J mol - 1 K - 1 = 130.4 J mol -1 K -1 ;
negative:
HCl
molecules loose entropy going from the gas phase to a smaller solution volume.
D G 0 = D H 0 - T D S 0 = -74.9 kJ mol - 1 - (298.15 K)( - 130.4 J mol - 1 K - 1 )
= -36.0 kJ mol - 1
negative; dissolution is spontaneous: the negative enthalpy change ‘wins’ from the
entropy loss.
d) Since at constant pressure:
D H = q p + we,rev
and at constant temperature:
D G = we,rev = D H - T D S , it follows that for constant p, T:
D S = ( q p / T ) Þ q p = -38.9 kJ mol -1
(dissolution is exothermic).
e) Then the energy decrease is fully delivered as heat: DU0. More heat is delivered
than in the set-up in Figure 5.5, where part of the energy decrease is delivered as
non-volume work.
0
f) D S sur =
- D H 0 +74.9 kJ mol -1
=
= +251.2 J mol -1 K -1
T
298.15 K
0
0
-1
Þ D S tot0 = D S sur
+ D S sys(HCl
K - 1 - 130 J mol -1 K -1
dissolution) = 251.2 J mol
= +120.8 J mol - 1 K - 1
0
Þ T D S tot
= 298.15 K ×120.8 J mol -1 K -1 = +36.0 kJ mol -1
Apparently: D G
0
= -T D S tot0 . This is generally true: the Gibbs energy decrease of a
spontaneous reaction equals the total entropy increase.
é-DG 0 ù
é
ù 14.5
+36 kJ mol -1
=
exp
= e = 2.0 ×10 6
ú
ê
-1
-1
ú
ë RT û
ë 8.314 J mol K × 298.15 K û
g) K = exp ê
In equilibrium virtually all HCl molecules will be in solution.
h) The work needed to lift a mass m to height h equals mgh (g = gravitational
acceleration). The maximal work HCl dissolution can deliver is ΔG0 = we,rev. Hence
the
m=
maximal
DG 0
gh
=
water
mass,
per
mole
HCl,
that
can
be
lifted
10
cm
is
-1
36.0 ×10 3 J mol -1
3 N mol
=
36.7
×10
= 36.7 ×10 3 kg (per mole HCl)
-2
-2
9.81 m s × 0.1 m
ms
That is a lot of water, lifted by one mole of HCl! Two comments here: 1) The work
actually delivered by a spontaneous, irreversible process will be much smaller than
the maximal reversible work, and 2) gravitational potential energies are modest in
comparison to molecular energies. For example, the height H at which the
82
gravitational potential of one mole of oxygen equals the thermal energy of one mole
is about:
H =
RT
2.48 kJ mol -1
=
= 7.9 km
Mg 32 ×10 -3 kg mol -1 × 9.81 m s -2
6.9 a) According to Raoul´s law and the Clausius-Clapeyron equation the vapor pressure
of a solution is:
p = xH 2 O p*ref e - y
; y =
DH 0 æ 1 1 ö
R çè T Tref ÷ø
*
where pref = 1 bar is the vapor pressure of pure water at Tref=373 K (=100 °C). We wish
the salt solution to boil on top of Mount Everest (where p= 0.28 bar) at T=Tref. Hence
= 0 and
xH 2 O = p / p*ref = 0.28 bar / 1 bar = 0.28 . The number of moles, ns, of solute
particles follows from:
xH 2 O =
nH 2 O
nH 2O + ns
= 0.28 Þ ns = 2.57 × n H 2 O
Assuming that NaCl molecules fully dissociate, each NaCl molecule produces two solute
particles (here two ions): n NaCl = ns / 2 = 1.29 ×n H 2 O . So the prediction is that to each
mole of water, 1.26 moles of NaCl should be added.
Discussion: This outcome is clearly not realistic: it corresponds to adding 71.3 moles
of NaCl (4.2 kg of salt) to 55.5 moles (one liter) of water. This is more than 10 times
the maximal salt solubility in water: at room temperature one liter of water can dissolve
a maximal amount of about 360 gram NaCl – the salt concentration in the Dead Sea,
see question 6.8b. Raoul’s law only applies to dilute solutions of ideal solutes and
clearly not to super-saturated salt solutions.
b) 360 g/L corresponds to 6.2 moles NaCl L-1, which yields 12.4 moles of ions per liter.
Thus x H 2 O =
55.6 mol
= 0.8 .
12.4 mol + 55.6 mol
p = xH 2 O p*ref e - y
; y =
DH 0 æ 1 1 ö
R çè T Tref ÷ø
Substitute: atmospheric pressure = 1 bar, ΔH0= + 40 kJ mol-1, pref = 1 bar, Tref= 373
K. The boiling point of Dead Sea water turns out to be about
T » 380 K =
107 °C. Comment: vapor pressure lowering is a small effect: even the maximal salt
concentration of 6M, raises the boiling point only about seven degrees.
83
c) Since the vapor pressure above salt water is lower than above pure water, water
vapor from the pure-water compartment will spontaneously transport towards the
vapor above the salt solution, where it will condense onto the salt solution. So as a net
effect, the salt solution is spontaneously diluted – in a process that is also referred to
as ‘isothermal distillation’.
Math-related Exercises
6.10 a) + 41.2 kJ mol-1
b) at 100 °C: ∆S = +110.4 J mol-1 K-1, at 75°C: ∆S = +118.4 J mol-1 K-1
– positive sign as the entropy increases by the evaporation,
– the entropy change is lower at 100 °C since molecules had a higher thermal motion
(as liquid) to begin with.
6.11 a) D G
0
= G I02 - 2G I0 = -121.17 kJ mol -1
é-DG 0 ù
é
ù 48.9
121.17 kJ mol - 1
21
=
exp
ú
ê
-1
-1
ú = e = 1.7 ×10
RT
8.314
J
mol
K
×
298
K
ë
û
ë
û
b) K = exp ê
0
c) D S m = -100.89 J mol
d) Calculate ΔH0: D H
0
-1
K -1
= -151.24 kJ mol -1
Þ D G 0 = D H 0 - T D S 0 = -151.2 kJ mol - 1 - (298 K)( - 100.89 J mol -1 K -1 )
= - 121.2 kJ mol -1
Note that the entropy change of the reaction is negative (two gas molecules merge
into one gas molecule) so the entropy change is, via -TDS°, a positive contribution to
DG. It is the strongly negative enthalpy change that makes the reaction (under
standard condition) a spontaneous one.
6.12 At 273.16 K, liquid water and ice are in equilibrium hence DG = DH - TDS = 0. Since
DH = qp it follows that at 273.16 K: DS = +21.97 J mol-1 K-1. Note the plus-sign:
liquid water has a higher entropy than crystalline ice.
6.13 For the integration: see footnote in the Lecture Notes on page 109.
The physical meaning of C is a dimensionless number, which is the change of pressure
compared to the reference point, see for example Figure 6.2.
6.14 The expression for the equilibrium constant is:
æ pH2 I ö
çè p2 ÷ø
p2HI
0
K =
=
pH 2 pI 2 pH × pI
2
2
×
p0 p0
84
Let x be the change in partial pressures of one of the gases, say H2, then we will end
up with this table:
H2 (g)
I2 (g)
HI
Initial pressure [atm]
1.980
1.710
0
Change in pressure [atm]
-x
-x
+2x
Equilibrium pressure [atm]
1.980-x
1.710-x
2x
Filling in all equilibrium partial pressures and K = 92.6 into the expression for the
equilibrium constant gives
92.6 =
(+2x)2
(1.980 - x)(1.710 - x)
This expression leads to a quadratic equation:
88.6x 2 - 341.694 x + 313.525 = 0
Applying the quadratic formula (see exercise 1.18) gives:
341.694 ± (341.694)2 - 4(88.6)(313.525)
2(88.6)
x=
The two solutions are:
x=1.5044 atm
;
x=2.3522 atm
So x = 1.5044 atm, which gives that at equilibrium the partial pressures are:
pH 2 = 0.4756 atm ; pI2 = 0.2056 atm ; pHI = 3.0088 atm
6.15 C
K =
c benzene
cH O
2
[X] = 0.10 mol dm-3
100 cm3 solution è 10 mmol X
c benzeneVbenzene + c H OVH O = 10 mmol
2
2
if cbenzeneVbenzene = 7.5 mmol
c b en zen eVb en ze ne
V
7.5
= 5 b en z ene =
=3
c H OVH O
VH O
2.5
2
So
2
2
Vb en zen e
= 0.6 Þ Vbe nz en e = 60 cm 3
VH O
2
Review Exercises
6.16 a) ΔU = 0 J since the temperature has been kept constant for the ideal gas.
b) Trajectory B (V = constant): dU = dq + dw = CdT – pdV, dw = 0 J since the
volume has been kept constant, so dU = CdT, hence ΔU = CV(T2 – T1)
Trajectory C (p = constant): dU = CdT – pdV = Cp(T1 – T2) – p(V1 – V2)
c) ᵃ = ᵅᵅ , so find an expression for qp:
ᵅ
Δᵄ
85
Δᵃ = Δᵄ + ᵅΔᵄ = ᵅᵅ + ᵆᵅ + ᵅΔᵄ = ᵅᵅ − ᵅΔᵄ + ᵅΔᵄ = ᵅᵅ
( ) +ᵅ( )
So, ᵃᵅ = ᵅᵅ = Δᵃ = Δᵄ + ᵅΔᵄ = Δᵄ + ᵅ Δᵄ = ∂ ᵄ
Δᵄ
Δᵄ
Δᵄ
Δᵄ
( ) = ᵅᵄ = ᵃ
ᵅᵄ = ᵅᵄᵄ, so ᵄ =
, hence ( ) =
3
2
ᵄ = ᵅᵄᵄ, hence
∂ᵄ
∂ᵄ ᵅ
3
2
ᵅᵄᵄ
ᵅ
∂ᵄ
∂ᵄ ᵅ
∂ᵄ ᵅ
Δᵄ
ᵆ
∂ᵄ
∂ᵄ ᵅ
ᵅᵄ
ᵅ
Combining these expressions yields ᵃᵅ = ᵃᵆ + ᵅᵄ
d) From ᵅ = ᵃΔᵄ, one can conclude that Δᵄ = ᵅ. Since q is the same, gives that the
ᵃ
system with the lowest heat capacity gives the highest rise in temperature. Since ᵃᵅ > ᵃᵆ
, the temperature will increase the most when the volume has kept constant.
V RT 8.314 J mol - 1 K - 1 × 298 K
6.17 a)
=
=
= 0.0247 m 3 mol -1 ; the molar volume is
5
n
p
10 Pa
equal for each gas.
0
0
0
0
b) D H reaction = D H C2 H 5OH(g) - (D H C 2 H 4 (g) + D H H 2 O(g) )
0
D H reac
= -235.10 kJ mol -1 - (52.26 kJ mol -1 - 241.82 kJ mol -1 ) = -54.54 kJ mol -1
c) ΔG = 0 J mol-1 in equilibrium
DG = DH - TDS = 0
D S = +282.70 J mol - 1 K - 1 - (219.56 J mol - 1 K - 1 + 188.83 J mol - 1 K - 1 )
Þ D S = -125.7 J mol - 1 K - 1
D G = -54.54 ×10 3 J mol - 1 - T ( -125.7 J mol - 1 K - 1 ) = 0
Þ T=
54.54 ×10 3 J mol - 1
= 433.9 K
125.7 J mol - 1 K - 1
d) When the volume V is increased isothermally, the pressure will increase
according to the ideal gas law (pV=nRT). This means that the amount of ethanol
will increase (the total amount of molecules will decrease).
86
Chapter 7 Osmosis and Osmotic Pressure
Theoretical Exercises
7.1
Your blood cells will start to swell up, as they will start taking up water through
osmosis, as the concentration of ions in your blood will be much lower than in your
blood cells.
7.2
a) Disagree, the dissolution of NaCl into Na+ and Cl- increases the entropy of the
system, and will therefore occur spontaneously. The strong Coulomb attraction that
the ions feel toward each other is shielded by the presence of water, as water will
interact with the ions.
b) By measuring the electrical conductivity of a solution of NaCl compared to that of
a solution of sucrose, where the same amount of molecules as NaCl are added. If
NaCl dissolutes into two ions, the electrical conductivity should be higher for the NaCl
solution.
7.3
a) Vapor pressure: pressure in a closed vessel exerted by vaporized molecules.
Vapor pressure can be measured by means of a manometer for a pure vapor in vessel,
hold at constant temperature.
b) pO 2 = 0.21×1 bar = 0.21 bar
c) x O 2 =
d)
pO 2
0.21
=
= 3.98 ×10 - 6
3
K
52.8 ×10
nO 2 xO 2 × ntot xO 2 × n H 2 O
=
»
as [O2] << 1
V
V
V
1 L water =
n
1000
1000
mol L- 1 Þ O 2 = 3.98 ×10 - 6 ×
mol L- 1 = 2.21×10 -4 mol L-1
18
V
18
(Luckily, we have some red blood cells, because this is way too slow…)
e) π = ctotRT; ctot=1 mol L-1; T=310 K ; π = 2577340 N m-2 (»25,8 bar.)
7.4
a) There is no interaction between the glucose molecules; the size of the glucose
molecules may be regarded as negligible; the molecules have kinetic energy but no
potential energy (vibrational,rotational energy); and they behave according to the
ideal gas law (i.e. no alternative equation of state defined).
b) <ᵃ ≥ 3 ᵄᵄ (per mole)
2
→ <ᵃ> =
c) P =
3
∙ 8 . 3145 J mol-1 K-1 ∙ 298 K= 3716 J/mol
2
n
1 mol ×8.314 J mol -1 K -1 × 298 K
RT =
= 0.136 bar
V
180 ×10 -3 m 3
87
d) According to Raoult’s law: pi = xipi*, so the vapor pressure is higher above water
than for the glucose solution. Hence, water vapor will move from the right side to the
left side and condense on the glucose solution. Because pi is always lower than pi*
(pure water), this transfer will not stop.
NB: the difference in pressure in Figure 7.1 is
D p = p*i - pi = (1 - xH 2 O ) p*i = xgluc p*i > 0
7.5
The
First
Law
could
not
care
less,
because
energy
is
conserved.
The Second law dictates that liquid water will spontaneously migrate (via the
membrane) from pure solvent into the solution: the Gibbs energy of a solution is
always lower than that of pure solvent, see also Lecture Notes Chapter 5, page 9697. Thus liquid water transport from solution to pure solvent is non-spontaneous and
would imply here a decrease in total entropy.
At the left the liquid level rises until the pressure of the extra liquid equals the osmotic
pressure of the solution. This extra liquid pressure also increases the vapor pressure
of the solution to the value of pure water. In equilibrium all spontaneous net transport
of water has stopped; no work can be obtained from the osmotic equilibrium sketched
in the figure below.
Math-related Exercises
7.6
a) Mw, NaCl = 58.44 g mol-1
1 gram NaCl = 1.71 · 10-2 mol NaCl = 3.42 · 10-2 mol Na++Cl- ions.
nRT
V
-2
3.42 ×10 mol ×8.314 J mol -1 K -1 × 298 K
P =
= 8.48 ×10 4 Pa
-3
3
10 m
P V = nRT Û
P =
b) h = 8.65 m
c) 1 gram protein = 1.67 · 10-5 mol
P =
nRT 1.67 ×10 -5 mol ×8.314 J m ol -1 K -1 × 298 K
=
= 41.3 Pa
V
10 -3 m 3
88
1 gram of table salt dissolves into many more ions than 1 gram protein, and thus
the solution in a) has a much higher osmotic pressure.
7.7
ᵅ
ᵄ
ᵰᵄ = ᵅᵄᵄ = > =
ᵰ
7 . 2 ∙ 105 Pa
=
= 279 ᵅᵅᵅ ᵅ− 3
ᵄᵄ
8 . 3145 J mol-1 K-1 ∙ 310 K
This is number of ions, NaCl releases two ions per salt molecule, hence the
concentration of NaCl is half of this value, i.e.
ᵅᵄᵄᵃᵅ =
279 ᵅᵅᵅ ᵅ− 3
2
=
0.140 ᵅᵅᵅ L− 1 =
0.140 M
7.8
Assuming that the sea water is a solution of ideal solutes, we can apply Van ’t Hoff’s
osmotic pressure law to find that the osmotic pressure of sea water is about:
P = MRT = (1.13 mol L-1 ) ( R = 0.0821 L bar mol -1 K -1 ) ( 298 K ) = 27.6 bar
7.9
From pV= nRT the number of moles follows as:
n=
0.0100 bar ×1 L
= 4.09 ×10 -4 mol
-1
-1
0.0821 L bar mol K × 298 K
So the molecular weight is:
M=5.0 g/4.09 × 10-4 mol = 12200 g mol-1
7.10 0.33 L
A small hint to the solution: Calculate the pressures at the surface and 100 m deep
in the water and calculate the corresponding mole fractions of nitrogen in the blood.
This difference corresponds to the amount of nitrogen gas in the blood in mole per
liter. (Keep in mind: what information do you need to calculate the volume of the gas
→ Ideal gas law, so ᵄ
ᵅᵄᵄ, all parameters are known, except for n, how do you find
N2 , gas = ᵅ
N2
this value? This is related to the difference in nitrogen amounts between the surface
and in the water.)
7.11 a) The osmotic pressure drops with a factor of two (from 30 bar to 15 bar).
b) For an ideal gas we had (see Lecture Notes Chapter 4):
V
p
D S = nR ln 2 = nR ln 1
V1
p2
; pV = nRT
For ideal solutes, the gas pressure p has to be replaced by the osmotic pressure,
hence:
D S = nR ln
P1
P2
; P V = nRT
The Gibbs free energy change for the mixing of 1 m3 fresh water and 1 m3 of salty
sea water is then:
D G = D H - T D S = -T D S = nRT ln
P1
= - nRT ln(2)
P2
89
(Here we used from previous exercises, that for an isothermal change of ideal solutes,
ΔU and ΔH are zero.) 1 m3 of salty seawater contains about 599 moles of salt, so the
reversible work (the maximal work) is about:
∆ ᵃ = ᵆᵅ , ᵅᵅᵆ =− 1029 kJ
NB: Please notice that one has to use n as amount of salt instead of the number of
mole of ions, since each cation combines with one anion.
That is a substantial amount of work! Dutch rivers discharge about
3
m s
-1
3000
in the North Sea, implying vast amounts of useful work just lost!
Review Exercises
7.12 a) ΔU = 0 J, because an ideal gas only has kinetic energy, which is only dependent
on the temperature.
b) wrev = 0 J, because the gas does not work when expanding against a vacuum.
c) wrev = -
dV
V2
ò pdV = - nRT ò V = - nRT ln V = -nRT ln(2)
1
The sign is negative because the gas loses energy (and absorbs energy as
reversible heat so that ΔU=0 J).
d) ΔH = ΔU + Δ(pV) = 0 J, because ΔU=0 J and Δ(pV)= Δ(nRT) = 0 J.
ᵅ
ᵮᵄ = rev =
− ᵆrev
ᵄ
ᵄ
=+ ᵄ ᵅᵅ ( 2)>0 J/K
because the amount of microstates of the molecules increase.
ᵮᵃ = ᵮᵃ − ᵄᵮᵄ = 0 − ᵄ( + ᵄ ᵅᵅ 2 ) =− ᵄᵄ ᵅᵅ 2 <0 J
because it’s a spontaneous process.
7.13 a) ᵅᵄ2 = ᵆᵄ2 ᵅ* = 0 , 2 atm
Henry: ᵅᵄ2 = ᵃᵆ ᵆoctantol
ᵄ2
octanol
è xO
=
2
pO
2
Kz
=
0.2 atm
= 2.0 · 10 -7
1.0 · 10 6 atm
ᵆoctanol
=
ᵄ2
n oc tan o l =
o ct an o l
è nO
2
ᵅ ᵄ2
ᵅᵄ2 + ᵅoctanol
=
ᵅ ᵄ2
ᵅoctanol
weight 100 cm 3 · 0.83 kg dm - 3
=
= 0.638 mol
M
130 g mol - 1
= 0.638 · 2 · 10 - 7 = 1.28 · 10 - 7 mol
octanol
3
c octanol O 2 concentration in octanol nO 2 / 100 cm
=
= water
Nernst: K =
b)
c water
O 2 concentration in water
n O / 200 cm 3
2
è c w ate r =
nOw at er
2
200 cm
=
3
nOo ct an o l
2
K · 100 cm
=
3
1.28 · 10 - 7 mol
4.5 · 100 cm 3
90
c water = 0.284 · 10 - 9 mol cm -3
è
n Owater = 0.57 · 10 -7 mol
2
c)
ᵅᵄ2 = ᵃᵆ ᵆwater
ᵄ2
xOwater =
2
pO
2
Kz
=
0.2 atm
= 4.65 · 10 -6
3
43 · 10 atm
This is in clean water. In b) the mole fraction of O2 is:
x
water
O2
nOwater
0.57 · 10 -7 mol
=
=
= 5.1 · 10 -9
3
-3
-1
nH O 200 cm · 1 g cm / 18 g mol
2
2
This is a lot less, thus the concentration of O2 decreases.
d) No influence. The volumes do not matter, only the concentrations. Proof:
octanol ᵄ
ᵅ ᵄ2
ᵅᵄ
2
octanol
Henry: ᵅᵄ = ᵃᵆ ᵆᵄ = ᵃᵆ ᵅᵄ2 = ᵃᵆ
= ᵃᵆ
ᵰ
2
Nernst: ᵃ =
2
ᵅoctanol
ᵰoctanol
octanol
octanol
ᵄoctanol ᵄ
ᵄ2
ᵅoctanol
ᵄ2
ᵅwater
octanol
ᵅᵄ
Henry + Nernst = ᵅwater
= 2
ᵄ2
ᵃ
=
ᵰoctanol ᵅᵄ2
ᵃ ᵄoctanol ᵃᵄ
Herein nothing is mentioned on dimensions/the thickness of the oil layer, thus the
even a very thin layer of oil will cause an an O2 shortage for the fish.
91
Chapter 8 Redox Equilibria
Theoretical Exercises
8.1
C
8.2
a) cathode: Cu2+ + 2e- ® Cu
anode: Zn ® Zn2+ + eCell reaction: Zn + Cu2+ ® Zn2+ + Cu
b) Yes, because E
0
= E anode - E cathode = +0.3419 V - (-0.7618 V) = +1.097 V
2+
RT éëZn ùû
D E = +1.097 V ln
2 F éëCu 2 + ùû
8.3
a) cathode:
anode:
CH3CHO + 2 H+ + 2 e- ® CH3CH2OH
NADH ® NAD+ + H+ + 2e-
b) Pt | NAD+, NADH, H+ || CH3CHO, H+, CH3CH2OH | Pt
8.4
a) Oxidation of Au; Ecel = Ek – Ea = (0.000 V) – (1.498 V) = –1.498 V; nonspontaneous
b) Oxidation of Na; Ecel = Ek – Ea = (0.00 V) – (–2.71 V) = +2.71 V; spontaneous
c) Oxidation of H2; Ecel = Ek – Ea = (–2.71 V) – (0.00 V) = –2.71 V; non-spontaneous.
Compared to b) this reaction is reversed.
d) Ecel = Ek – Ea = +2.71 V as in b). Potentials and voltages are intensive parameters.
e) Oxidation of Na; Ecel = Ek – Ea = (–0.8277 V) – (–2.71 V) = 1.88 V ; spontaneous
The potential for which H2O is reduced depends on the pH, not for Na+.
Therefore, the cell potential depends similar on the pH as for the reduction of H2O.
f) Red. and ox. van water; Ecel = Ek – Ea = (0.000 V) – (1.229 V) = –1.229 V; nonspontaneous
g) Ecel = Ek – Ea = (–0.828 V) – (0.401 V) = –1.229 V
Same answer as for f), because both reactions have the same pH-dependence.
h) Oxidation of Fe2+; Ecel = Ek – Ea = (1.229 V) – (0.771 V) = 0.458 V ; spontaneous
i) Reduction of AgCl: AgCl + e- ® Ag + Cl- Oxidation of Ag: Ag ® Ag+ + eE°cel = Ek – Ea = (0.22233 V) – (0.7996 V) = –0.57727 V; non-spontaneous
E°cel = [RT/(nF)] ln(K)
gives K = 1.75 · 10-10 (equal to literature).
j) Oxidation of Zn; Ecel = Ek – Ea = (–0.3419 V) – (–0.7618 V) = 1.1037 V ;
spontaneous
This is Daniel cell.
DG° = –nFE°cel = – (2)(96485 C mol-1)(1.1037 V) = –213,0 kJ mol-1
Q=1000 gives: E = E° – [RT/(nF)] ln(Q) = 1.0150 V
DG = –195.6 kJ mol-1
8.5
a) The two reactions are: Pb + SO42- ® PbSO4 +2e-
(8.1)
PbO2 + SO42- +4H+ +2e- ® PbSO4 + 2H2O
(8.2)
92
Cell reaction: Pb + PbO2 + 2H2SO4 ® 2PbSO4 + 2H2O
(8.3)
b) Reaction (8.1) deposits electrons on the negative electrode.
Reaction (8.2) adsorbs electrons from the positive electrode.
Electrons flow (spontaneously) from the negative to the positive electrode via the
wire.
c) When all (or most) SO42- has been converted to lead sulfate, according to reaction
(8.3), the battery is empty.
The battery can be ‘charged’ by passing a current through it in backward direction.
This
reverses
reactions
(8.1)
and
(8.2).
(ΔG>0J).
NB: The charging process does not lead to extra storage of charge: the number of
protons and electrons in the battery remains constant. What changes is the
concentration of chemical compounds such that ΔG>0. The battery is charged with
‘chemical potential energy’ that has the capacity to deliver electrical work.
Math-related Exercises
8.6
The battery performs 1.5 J of work on each Coulomb that passes through. (This
follows from the definition of ‘Voltage = Joule per Coulomb’.) For a current of 1 A,
the total charge in 1 h is 1A · 3600 sec = 3600 C. The total work is
1.5 J C-1 ·3600 C = +5.4 kJ.
8.7
a) I =
D V 1.56 V
=
= 0.78 A
R
2.0 W
b) One ampere is one coulomb of charge per second. Thus:
0.78 C s-1
· 250 s = 195 C
c) The Faraday number is 96485 coulombs per mole of electrons, so 2.0 millimoles
of electrons move in 250 seconds.
d)I2Rt = (0.7 C s-1)2.(2.0 Ω).(250 s) = 304 J
e) The spontaneous electron flow produces in 250 sec 304 J of irreversible heat.
According to Clausius the entropy increase is:
DS >
8.8
qrev 304 J
=
= +1.02 J K -1
T
298 K
ᵅᵃ
37
ln ᵃ = ᵄᵄ ᵃᵅ ; n=2 => K = 1.6 · 10
at 298.15 K
Thus Zn2+ replaces virtually all Cu2+.
8.9
E= [RT/(nF)] ln(Q)
= ± [(8.314 J mol-1 K-1)(293,15 K) / (96485 C mol-1)] ln[(10-1)/(10-2)]
= ± 0.0582 V
93
At 25.00°C E = ±0.0592 V (±, depending on the way the electrodes are connected).
For both cases the signal is about 60 mV per decade (power of 10 in [H+]).
Furthermore, temperature affects the measurement of the pH electrode.
8.10 a) 2H+ + 2e- ® H2
Na+ + e- ® Na
E0 = 0 V
;
E0 = –2.71 V
;
Na + H+ ® Na+ + (1/2)H2
;
E0 = Ecathode – Eanode = +2.71 V
E > 0 V, so sodium will dissolve spontaneously.
b) Au+ + H+ ® Au
E0 = +1.692 V
;
+
0
Au + H+ ® Au + (1/2)H2 ; E = –1.692 V
E < 0 V, so gold will not dissolve spontaneously.
E0 = -1.40 V, gold will not spontaneously dissolve.
0
8.11 a) D E = D E -
RT
RT 10 -3 ×10 -4
ln Q = (1.44 V - (-0.771 V))ln
nF
F 10 -3 ×10 - 3
D E = 2.211 V -
RT
ln10 = 2.211 V - 0.06 V = +2.151 V
F
b) cell reaction: Fe2+ + Ce4+ ® Fe3+ + Ce3+
D G = nF D E = -96485 C mol -1 × 2.151 V = -208 kJ mol - 1
D G 0 = -nF D E 0 = -96485 C mol -1 × 2.211 V = -213 kJ mol -1
c) D G
0
é DG 0 ù
= - RT ln K Û K = exp ê ú
ë RT û
é
ù
-213 ×10 3 J mol - 1
Þ K = exp ê= 2.17 ×10 37
-1
-1
ú
ë 8.314 J mol K × 298 K û
8.12 Fe2+ + 2e- ® Fe
;
E0 = -0.447 V
O2 + 4H+ + 4e- ® 2H2O
;
E0 = 1.229 V
D E 0 = 1.229 V + 0.447 V = 1.676 V
D G 0 = -nF D E 0 = - 2 × 96485 C mol -1 ×1.676 V = -323.5 kJ mol - 1
-1
0
K = e - DG /RT = e323.4 kJ mol /(8.314 J mol
-1
K - 1×298 K)
= 4.9 ×10 56
So yes, the equilibrium constant is in favour of the formation of Fe2+(aq).
8.13 a) 2Cu+ ® 2Cu2+ + 2eCu2+ + 2e- ® Cu
;
;
E0 = -0.153 V
E0 = 0.3419 V
Total reaction: 2Cu+ ® Cu + Cu2+
;
ΔE0 = 0.1889 V
94
D G 0 = -nF D G 0 = -2 × 96485 C mol - 1 × 0.1889 V = -36.5 kJ mol -1
0
é
36.5 ×10 3 J m ol -1
ù
K = e - D G /RT = exp ê
= 2.45 ×10 6
-1
-1
ú
ë8.314 J m ol K × 298 K û
No, Cu+ will not be a common ion in water.
b) Pt | Cu2+(aq), Cu+(aq) || Cu+(aq) | Cu
c) DG · 2 = -7.3 kJ mol-1, E will not change.
8.14 Nernst: E = E
0
RT
ln Q and EMK= E – E0
nF
-
⇒ᵃ − ᵃ0 =−
[
[ ]
ᵄᵄ
ᵃ0 − ᵃ
ᵃ0 − ᵃ
ᵅᵅ ᵄ ⇔ ᵅᵅ ᵄ =
⇒ ᵄ = ᵅᵆᵅ
ᵄᵄ
ᵅᵃ
ᵄᵄ
ᵅᵃ
ᵅᵃ
]
0 . 0250 V
= ᵅᵆᵅ [9 . 90 ⋅ 1 0− 1] = 2 . 69
8 . 314 J mo l− 1 K− 1 ⋅ 293 . 15 K
96485 C mo l− 1
+
[ᵃ ]ᵃ
ᵄ = + = 2 . 69 ; [ᵃ+]ᵃ = 1 0− 6 . 88 = 1 . 38 ⋅ 1 0− 7 mol L− 1
[ᵃ ]ᵃ
⇒ [ᵃ+] = 5 . 13 ⋅ 1 0− 8 mol L− 1 ⇒pH=7.30 for the bloodserum
ᵄ = ᵅᵆᵅ
ᵃ
Review Exercises
8.15 a,b) Hydrogen is pumped into the negative electrode and reacts to donate electrons
to it (oxidation, as always, at the anode):
2H2 + 4OH- ® 4H2O + 4eOxygen is pumped into the positive electrode:
O2 + 2H2O + 4e- ® 4OHc) Conductivity of pure water is much too low to achieve electrolysis at a reasonable
rate: you must add electrolyte. KOH is a good choice: K+ does not reduce, and OHtakes part in the oxidation of hydrogen.
8.16 a) pZ = x Z ptot = 0.2 ×1.0 bar = 0.2 bar
b) Apply Henry’s law:
xZ =
nZ
n
» Z
nw + nZ nw
xZ =
pZ
0.2 bar
=
= 4.5 ×10 -6
K Z 4.4 ×10 4 bar
; because nZ << nw
1m3 water has nw= 55.56 · 103 mol water.
Þ n Z = 4.5 ×10 -6 ×55.56 ×10 3 = 0.25 mol O2 Þ 8.0 g oxygen in 1 m 3 water
c) The vapour pressure in an isolated vessel at constant temperature.
95
*
- Raoult: p = x H 2 O p
- Mole fraction: it concerns the number of solute and solvent molecules.
d) It is always endothermic: liquids need to absorb energy to move molecules from
their attracting neighbours to the much more dilute vapour phase.
96
Chapter 9 The Chemical Potential
Theoretical Exercises
9.1
A
9.2
a)
m (C 6 H1 2 O6 ) + 2 m (NAD + ) + 2 m (ADP) + 2m (Pi ) = 2 m (C3 H 4 O3 ) + 2m (NADH) +
2 m (H + ) + 2 m (ATP) + 2 m (H 2 O)
b) No, for the chemical potential, all components should be included. μ is phaseindependent.
Math-related Exercises
9.3
nR
- nRT
;
p
p2
9.4
a) G = G (n1 , n2 , p, T )
; both second derivatives equal
- nR
p2
æ ¶G ö
æ ¶G ö
æ ¶G ö
æ ¶G ö
dG = ç
dn1 + ç
dn 2 + ç
dp + ç
dT
÷
÷
÷
è ¶T ÷ø n1 , n 2 , p
è ¶p ø n1 , n2 ,T
è ¶n1 ø n 2 , p,T
è ¶n 2 ø n1 , p,T
æ ¶G ö
æ ¶G ö
dG = m 1dn1 + m 2 dn 2 + ç
dp + ç
dT
÷
è ¶T ÷ø n1 ,n 2 , p
è ¶p ø n1 , n 2 ,T
b) First realize that the chemical potential is the molar Gibbs energy for a certain
pressure/temperature point in state space:
Thus, we need to find G(p) for an ideal gas, which we can do using integration of the
ideal gas law:
dG = Vdp =
nRT
dp ; T = constant
p
97
Now that we have a formula for G as a function of p, we can take the derivative with
respect to n to arrive at the chemical potential:
9.5
dG = Vdp
G ( p) - G ( p0 ) = V ( p = p0 )
G ( p) = G ( p0 ) + V ( p - p0 ) ; p0 = 1 bar
G ( p) = G ( p0 ) + V ( p - 1)
¶V
so, m = m 0 +
( p - 1)
¶n
m = m 0 + V ( p - 1)
9.6
0
æ xö
; x0 = 1
÷
0
èx ø
a) m (x) = m (x ) + RT ln ç
æ pö
m ( p) = m ( p0 ) + RT ln ç 0 ÷ ; p0 = 1 bar
èp ø
b) Equilibrium between solution and vapor implies that the chemical potential of
solution and vapor are equal:
æ pö
æ xö
m 1 = m 2 Þ m (x) = m (x 0 ) + RT ln ç 0 ÷ = m ( p0 ) + RT ln ç 0 ÷
èx ø
èp ø
æ pö
Þ RT ln ç ÷ = - ( m ( x0 ) - m ( p0 ) ) = - D m 0
è xø
Here
D m 0 is the standard chemical potential difference for water molecules going
from solution to the vapor phase. Comparison with ln K = - D m
(per mole) shows that the equilibrium constant is K = p/x
9.7
a) 17.11 mmol NaCl L-1
Þ xNaCl =
nNaC l
17.11×10 - 3 mol
»
3.08 ×10 - 4 for nNaCl << nH 2O
n Na Cl + n H 2 O
55.56 mol
b) x NaCl =
1.71
= 0.030
1.71 + 55.56
98
0
/ RT ; Dm 0 = DG 0
Comment: we ignore here the volume occupied by NaCl in the solution.
ntot 2 ×17.11×10 -3 mol
n
=
= 34.22 mol m - 3 Þ P = tot RT = 0.85 bar
c)
-3
3
V
10 m
V
For 100 g NaCl L-1: P » 85 bar
Comment: This is an estimate: for Van ’t Hoff’s osmotic pressure law to be accurate,
ions should behave ideal which is unlikely to be the case at such high salt
concentrations.
9.8. The Taylor expansion of a function f(x) around the value x=a is:
f (x) = f (a) + (x - a)
a) e
b)
x
f '(a)
f ''(a)
f '''(a)
+ ( x - a) 2
+ (x - a) 3
+ ...
1!
2!
3!
1
= 1 + x + x 2 + ...
2
1
= 1 + x + x2 + ...
1- x
c) ln(1 + x) = x -
1 2
x + ...
2
d) ln(1 - x) = - x -
e) sin( x) = x -
1 2
x + ...
2
x3
+ ...
3!
Note that in the expansion of sin(x) the quadratic term is zero. For a more detailed
explanation, see Steiner: The Chemistry Math book (2nd edition, pages 208 – 214).
=> The expansion ln(1–xs) = –xs is used in Van ‘t Hoff’s law, which is justified/valid
for small values of the solute mole fraction xs.
9.11 a) 2 m I = m I 2
0
b) m I = m I + RT ln( pI )
; m I2 = m I02 + RT ln( pI 2 )
2 m I = m I2
2 m 0I + 2 RT ln( pI ) = m I02 + RT ln( pI 2 )
-( m I02 - 2 m I0 ) = RT ln( pI2 ) - 2RT ln( pI )
æ pI ö
- D m 0 = RT ln ç 22 ÷
è pI ø
- D m 0 = RT ln K
99
é Dm 0 ù
; D m 0 = D G 0 per mole ,
ú
ë RT û
Hence, K = exp ê -
æ DG 0 ö
è RT ÷ø
which is equal to K = exp ç -
9.12
æ pö
m ( p) = m ( p0 ) + RT ln ç 0 ÷ ; p0 = 1 bar
èp ø
Þ dm =
Since
RT
dp ; T = constant
p
æ ¶m i ö
çè ¶p ÷ø = V i
T ,ni
V=
, it follows that
RT
Þ pV = nRT
p
9.13 a) Reversible work: –100 kJ; any irreversibility will decrease the delivered work.
b) If the conversion were reversible, +100 kJ would be needed; any irreversibility
will increase the invested work).
Comment: a ΔG of -100 kJ is only delivered fully as non-volume work in a reversible
process. To raise the Gibbs energy to its initial value via the non-spontaneous
reaction B → A, more work than 100 kJ must be invested: part of the investment is
lost due to entropy production.
c) Put potential energy Mgh equal to –ΔG = 100 kJ mol-1 ;
maximally h = 10m, per mole converted A.
Review Exercises
9.15 a) anode: H2 ® 2H+ + 2e- but in KOH solution becomes H2 + 2OH- ® 2H2O + 2ecathode: (1/2)O2 + 2H+ + 2e- ® H2O becomes (1/2)O2 + H2O + 2e- ® 2OHnet: H2 + (1/2)O2 ® H2O
b) D G
0
= -nF D E 0 = -2 ×96485 C mol -1 ×1.23 V = -237 kJ mol -1
This is also equal to the maximal electrical work: ΔG0= w
= -237 kJ mol-1
rev
c) spontaneous, because ΔG0< 0 J mol-1.
d) q p ¹ D H = -286 kJ mol
-1
, because there is also non-volume work .
For a reversible process: q rev = T D S and D G = D H - T D S
Thus: q rev = D H
0
- D G 0 = -286 kJ mol -1 - -237 kJ mol -1 = -49 kJ mol - 1
100
101
Chapter 10 Electrolyte Solutions
Theoretical Exercises
10.1 pH=ᵅ ᵃacid + ᵅᵅᵅ [ᵃᵄ]
[ ]
10.2 0.1 mol dm-3
= [NaOOC-COOH] + [-OOC-COOH] + [-OOC-OOC-] + [HOOC-COOH]
[Na+]+[H+]=[-OOC-COOH]+[OH-]+2[-OOC-COO-]
Math-related Exercises
10.3 1.8 average
10.4 Z ® B + H+, the molar weight of the acid is MW= 46.03 g mol-1, so there is
2 . 30 g
46 . 03 g mo l− 1
= 4.997 · 10-2 mol L-1 HCOOH
ᵃ=
[ ᵃ ] [ ᵃ+]
ᵅ ᵃᵄ =− ᵅᵅᵅ ᵃᵄ ⇒ ᵃᵄ = 1−04− ᵅ ᵃᵄ = 1 . 778−⋅21 0− 4 − 1
+
+
ᵃ
ᵃ
[ ] [ ] = ᵃ ⋅ [ᵄ] = 1 . 778 ⋅ 1 0 ⋅ 4 . 997 ⋅ 1 0 mol L = 8 . 8856 ⋅ 1 0− 6 mol L− 1 ⇒ [ᵃ ] =
[ᵄ ]
10.5 0.010 mol, 0.060 mol, 0.015 mol
10.6 a) 1.95 · 10-3 mmol dm-3 en 2.82 · 10-3 mmol dm-3
b) 3.16
Review Exercises
10.7 A is incorrect because both oxidation reactions are thermodynamically favourable:
Fe + 2H+ ® Fe2+ + H2
E0= +0.44 V
Al + 3H+ ® Al3+ + (3/2)H2
E0= +1.66 V
B is correct
C is incorrect, because aluminium metal (oxidation state = 0) can indeed be oxidized,
to oxidation state +3 (in Al3+ or Al2O3).
D is incorrect, because aluminium tends to donate electrons to water rather than the
other way around. d) is basically another way of wording for a).
10.8 a) ΔU is a state variable. When returning to the same state, ΔU = 0 J.
b) Using the ideal gas law (read values from graph):
V=
nRT 1 mol ×8.314 J mol -1 K -1 × 546 K
=
= 0.0227 m 3 = 22.7 L
p
2 ×10 5 Pa
c)
102
p [105 Pa]
2
3
2
1
1
0
0
10
20
30
V [dm3]
The curvature is caused by the fact that pV = c is a hyperbolic relationship.
d)
step
Type of process
∆U / kJ
q / kJ
w / kJ
∆H / kJ
1®2
Isothermal
0
- 1,57
+ 1,57
0
2®3
Isobaric
+ 3,39
+ 5,67
- 2,28
+ 5,67
3®1
Isochoric
- 3,39
- 3,39
0
- 5,67
1®2®3®1
Circular
0
+ 0,72
- 0,72
0
process
After filling in a couple of fields, the rest can be derived (ΔU = q + w, isobaric
process ΔH = q; isobaric process w = 0 J, the values for the circular process are the
sum of the values above.
10.9
a) Differentiation of the given expression for U to T gives:
U =
Nhn
hn
e kT - 1
æ hnkT ö
= Nhn ç e - 1÷
è
ø
-1
hn
-2
hn
2 2
æ hn
ö
dU
e kT
æ hn ö Nh n
C =
= - Nhn ç e kT - 1÷ × e kT × ç - 2 ÷ =
×
è kT ø
dT
è
ø
kT 2 æ hn ö 2
kT
çè e - 1÷ø
2
hn
e kT
æ hn ö
= Nkç ÷
è kT ø æ hn ö 2
kT
çè e - 1÷ø
The final step is done, by realizing that 1/k is equal to k/k2.
2
b)
hn
kT
e
e¥
1
æ hn ö
lim C = lim Nk ç ÷
=
¥
×
= ¥ × ¥ = 0 , because e-∞=0, so if T
¥ 2
T®0 K
T® 0 K
è kT ø æ hnkT ö
e
(e )
çè e - 1÷ø
goes to 0 K, then the heat capacity of a system also goes to 0 (J K-1).
103
Chapter 11 Diffusion in Liquids and Gases
Theoretical Exercises
11.1 For a grain radius of 50 micron the displacement in one hour is: <x2>1/2 = 15 μm.
So you don’t see grains moving when you peer through a microscope; what Brown
observed was not motion of grains, but tiny particles released by them, see Section
11.1.
11.2 The colloid indeed dissipates motional energy to the solvent but – in equilibrium
–receives on average the same kinetic energy in return.
11.3 The phenomenological Second law of thermodynamics states that no process is
possible with the only result that heat withdrawn from a reservoir is fully converted
to work. A bouncing ball comes to a stop when all its kinetic energy has been
dissipated as heat to the surroundings; the Second Law forbids that afterwards the
ball spontaneously jumps upwards by conversion of (part of) that dissipated heat to
its kinetic energy. A Brownian particle, however, loses kinetic energy to its
surroundings and receives on average the same quantity such that its average speed
remains constant. Brownian motion, therefore, shows that the Second Law is not an
absolute one: a soccer ball spontaneously jumping up from the table is not an
impossible event: its occurrence is simply extremely unlikely. The smaller the ball,
however, the more likely this event becomes: a colloidal ball randomly jumps over
the table.
11.4 Diffusion coefficients of oxygen in water and air (p=1 bar, T=20 °C) are, respectively,
2.1 × 10-9 m2 s-1 and 2.0 × 10-6 m2 s-1.
11.5 <x> = 0 m since diffusion is completely random, so the probability that a particle
moves to the left is equal to the probability that it will move to the right. Hence, <x>
= 0 m.
11.6 Random walk is the random motion of particles, in which its movement towards a
certain direction in the positive direction and the negative direction has an equal
probability, such that its motion is completely random.
Math-related Exercises
11.8 <x2> = 2Dt à t =
< x2 >
0.10 2 m 2
=
= 5 ×10 6 s = 1388 h = 58 days
-9
2 -1
2D
2 ×10 m s
Luckily, spoons are available to stir the tea.
11.9 M: 2.5 x 10-5 cm2 s-1, N: 4.9 x 10-11 m2 s-1,
C: 2.5 x 10-12 m2 s-1, G: 2.5 x 10-16 m2 s-1.
104
11.10 RMS displacement = <r2>1/2= (6Dt)1/2. For an M-sphere: 0.73 cm in one hour and
68.1 cm in one year. For an N-sphere: 0.10 cm, respectively 9.6 cm. For C: 0.23
mm, respectively, 2.2 cm. For G: 2.3 μm, respectively, 0.22 mm.
11.11 For Z spheres with volume Vp in a vessel volume the sphere volume fraction is
f = ZVp / V . So the molar volume is (NAV is Avogadro’s constant):
VM =
N av (4 / 3)p R 3
0.64
M: Vm = 3.9 cm3 mol-1
;
N: 0.49 m3 mol-1
;
C: 3.94 · 103 m3 mol-1
;
G: 3.94 · 106 km3 mol-1
11.12 Combination of (11.1) and (11.9) gives:
1
2
1
2
1
2
æ kT ö
æ 6kTt ö
æ kTt ö
rrms = ( 6 D t ) = ç 6 × × t ÷ = ç
=ç
÷
è
f ø
è 6ph R ø
è ph R ÷ø
1
2
1
æ 1.381×10 -23 J K - 1 × 298 K × (14 × 24 × 3600) s ö 2
=ç
= 2.7 ×10 -3 m = 2.7 mm
-4
-6
÷
è
p × 8.90 ×10 Pa s × 0.25 ×10 m
ø
11.13 <x2>=2Dt = 2 · 2.9 · 10-8 · 3.156 · 107 cm2 = 1.73 cm2
<x2>1/2 = 1.35 cm
11.14 ᵃ = ᵅᵄ = ᵅᵄ =
6ᵰᵰᵄ
ᵅ
1 . 381 ∙ 10− 23 J K− 1 ∙ 298 K
= 2.45 · 10-12 m2 s-1
6 ᵰ 0 . 89 ∙ 10− 3 Pa s 0 . 1 ∙ 10− 6 m
11.15 For the derivation, see lecture notes pages 190 and 191.
The quadratic displacement is first-order in time i.e. linear.
2
æ 2.1 km h -1
ö
2
ç
-1
-1 × 50 s÷
ø
< x > è 3.6 km m s h
=
= 8.5 m 2 s -1
11.16 a) D =
2t
2 × 50 s
2
æ 5 m h- 1
ö
2
ç
- 1 × 50 s ÷
ø
< x > è 3600 s h
=
= 4.8 ×10 - 5 m 2 s - 1
b) D =
2t
2 × 50 s
[
11.17 a) ᵄ(ℎ) = ᵃ ᵅᵆᵅ − ᵅᵅℎ
ᵅᵄ
]
b) Probability to find a particle in the interval h, h+dh;
c) ᵃ = ᵅᵅ
ᵅᵄ
d)
By
using
¥
< h >= C ò he
0
integration
¥
-C h
dh = - ò h de
0
by
¥
- Ch
parts,
¥
= - ò d ( he
- Ch
0
105
) + òe
0
-C h
dh =
one
1 kT
=
C mg
will
find:
¥
Alternatively: < h >= -C
d
d æ 1 ö 1 kT
e - C h dh = - C
ç ÷= =
ò
dC 0
dC è C ø C mg
e) 9.0 km and 0.8 µm. Obviously the heavy droplets stay very much closer to the
surface of the earth.
¥
f) < h
2
¥
¥
0
0
0
¥
from d) we can infer that
òe
-Ch
h dh =
0
< h 2 >=
Þ
b)
¥
0
0
1
C2
2
2
kT
hrms =< h 2 >1/2 =
= 2
2 Þ
C
C
mg
So the rms-height is a factor
11.18 a)
¥
>= C ò h 2 e - C h dh = - ò h 2 de - C h = - ò d ( h 2 e - C h ) + ò e - C h dh 2 = 0 + 2 ò e - C h h dh
jd = - D
2
larger than the average height from d).
C - C left D ( C left - C right )
dr
Dr
= -D
= - D right
=
dx
Dx
L -0
L
1 du D (C A - C B ) 2.0 ×10 -9 m 2 s -1 (2.7 ×10 -1 mol dm - 3 - 0 mol dm -3 )
=
=
A dt
l
9.0 m
= 6.0 ×10 -11 mol m -2 s - 1
per hour: 6.0 · 10-11 mol m-2 s-1 · 3600 s h-1 = 2.2 · 10-7 mol m-2 h-1.
c)
A=145 m2
O2 transport per hour: 2.2 · 10-7 mol m-2 h-1· 145 m2
= 3.13 · 10-5 mol h-1 =1.00 · 10-3 g h-1 O2
So x =
1.00 ×10 - 3 g O 2 h - 1
0.1 g O 2 h - 1 kg - 1
= 0.010 kg , the fish weighs 10 g max.
Review Exercises
11.19 A is incorrect, because there is no solid reaction between platinum and zinc (what
would that be, a nuclear reaction?). An electrochemical reaction is possible.
B is incorrect, because the splitting of water into hydrogen gas and oxygen gas
cannot take place spontaneously (E0= –1.23 V). There is no oxygen gas
development at the anode, but instead zinc dissolves (E0= +0.76 V).
C is correct.
106
D is incorrect, because nothing can happen on platinum as long as the electrical
circuit is not closed. In addition, zinc does not act as catalyst as it is used up during
the reaction.
107
Chapter 12 Kinetic Theory
Theoretical Exercises
12.1 a) Notice that you need the number density, whilst we are given the mass density. If
we call the mass density ᵰᵅ and the number density ᵰᵅ, then they are related as: ᵰᵅ
= (ᵰᵅ/M)·NAV (check that you agree/understand). So:
< ᵆᵅ > =
ᵄ ᵃ2 ᵄ
1
=
≈ 30 ⋅ 1 0− 30 m3 = 30
ᵰᵅ
ᵰᵅ ⋅ ᵄᵄᵆ
ᵅ 3
()
ᵃ
b) It is not: water is hardly compressible….
c) From the ideal gas law:
< ᵆᵅ > =
ᵅᵄ
4 . 12 ⋅ 1 0− 21 J
=
= 4 . 12 ⋅ 1 04
ᵅ
1 05 Pa
d) Particles in liquid water are at an average distance of about
.
For the vapor:
ᵅ 3
()
ᵃ
ᵅ 1/3
( )
< ᵆᵅ >1 / 3 = 30 ᵃ
ᵅ
=3.1 ᵃ
ᵅ.
< ᵆᵅ >1 / 3 = 34 . 5 ᵃ
12.2 a) < u >= 122.5 km h
-1
; b) < u
2
>1/2 = 124.7 km h -1 c) The RMS speed is always higher
than or equal to the average speed.
2
12.4 a) The initial energy of N helium atoms at t=0 is kinetic energy: NE kin = N (1 / 2)mv .
At equilibrium the average kinetic energy of N atoms is N < E kin >= (3 / 2)NkT . Since
in the isolated vessel energy is conserved we have: N (1 / 2)mv
at equilibrium the gas temperature is: T =
2
= (3 / 2)NkT . Hence,
mv 2 Mv 2
=
= 12.0 K ; M = 4.0 g mol -1
3k
3R
b) The entropy will increase since all particles cannot have the same speed, as that
is a non-equilibrium situation as explained in Lecture Notes section 12.1. The entropy
will increase spontaneously until the equilibrium distribution has been reached.
12.5 Apply Graham’s law: (please note: <ᵆ> = ᵰ, so <u> ~ t-1 )
ᵆ
æ Mgas ö
çM ÷
è O2 ø
1/2
=
Eff. time (gas) 12.6
=
Þ M gas = 16.0 g mol -1
Eff time (O2 ) 17.8
The gas very likely is methane (CH4).
12.6 Apply Graham’s law:
MO2
Eff. rate H 2
=
= 4 . So four hydrogen molecules enter B for every oxygen
Eff. rate O 2
MH2
molecule: x H 2 = 0.8 and xO 2 = 0.2
108
Math-related Exercises
12.7 a) p = 4.1 · 10-22 bar
b) For d ≈10-10 m : λ ≈ 23 · 109 km
c) <u> = 251 m s-1; z = 1.1 · 10-11 s-1. The time elapsed between two collisions is of
order 2.8 · 103 years.
12.8 z = 1.5 · 109 s-1
l
1
=
= 11.8
d
2 6f
12.9
M-spheres: λ=2.4 nm ; N: 117.9 nm ; C: 2.4 μm ; G: 2.4 cm
12.10
3
3
< E kin > (per mole) = kT × N av = RT = 3.7 kJ mol - 1 at T = 300 K
2
2
12.11 Determine the maximum of the Maxwell distribution by putting dP(u)/du=0.
æ m ö
12.12 < v >= ç
è 2p kT ÷ø
1/2
2
x
¥
é - mvx2 ù
kT
× ò v x2 × exp ê
dv x =
ú
-¥
m
ë 2kT û
;
< v x >= 0
12.13 See Lecture Notes Chapter 12.
12.14 See Lecture Notes Chapter 12.
æ 3RT ö
12.15 a) urms = ç
è M ÷ø
æ 2R T ö
b) umax = ç
è M ÷ø
1/2
= 667 m s -1 ; M = 28 ×10 -3 kg mol - 1
1/2
æ 8RT ö
è p M ÷ø
= 545 m s - 1
1/2
= 615 m s -1
c) < u >= ç
d) ᵆ = 2 <ᵆ>ᵰ ᵅ2 ᵅ = 2 · 615 m s− 1 · 0 , 43 · 10− 18 m2 ·
ᵅᵄ
1
110 Pa
= 5.96 · 106 s-
1 , 381 · 10− 23 J K− 1 · 500K
So even at this low pressure, a nitrogen molecule experiences almost 6 million
collisions per second.
e) Time between collisions = 1/z = 0.17 μs.
f) Mean free path: ᵰ = <ᵆ> =
ᵆ
Since  d 22  0.43 nm
615m s− 1
= 0.103 mm
5 . 96 · 106 s− 1
the diameter of a nitrogen molecule is about
ᵅ = 0 . 43 / ᵰ nm=0.37 nm,
so ᵰ ≈ 1 . 0 · 10− 4 m = 0.3 · 106
ᵅ
0 . 37 · 10− 9 m
Thus, a nitrogen molecule travels unhindered over a straight distance (the mean free
path) that is a few hundred thousand times its own diameter.
12.18 umax : <u> : urms = (2)1/2 : (8/π)1/2 : (3)1/2 = 1.41 : 1.60 : 1.73
109
or umax : <u> : urms = 1.00 : 1.13 : 1.22
Note that these speed ratios are the same for any gas; they are independent of gas
temperature and gas molar mass because the term RT/M cancels.
12.19
3
3
ᵅᵄ = ᵅᵅℎ or ᵄᵄ = ᵄᵅℎ à For N2: h = 1.4 · 104 m
2
2
;
for He: h = 9.5 · 104 m
12.20 42.6 K
Review Exercises
12.21 a) From the ideal gas law we can determine the amount of moles:
n=
pV
0.400 g
= 12.3 mmol Þ M =
= 32.4 g mol -1
-3
RT
12.3 ×10 mol
b) 1 mol baking powder gives 1 mol of carbon dioxide gas, so 5.0 gram of baking
powder gives:
n=
5.0 g
= 59.5 mmol CO2
84 g mol -1
59.5 ×10 -3 mol × 8.314 J mol -1 K -1 × 453.2 K
Þ V=
= 1.73×10 - 3 m 3 = 1.73 L
5
1.3 ×10 Pa
c) The glucose concentration is not relevant. The average kinetic energy for water
and glucose molecules are equal:
3
3
< E kin >= RT = × 293 K × 8.314 J mol -1 K -1 = 3.65 kJ mol -1
2
2
d)
< u > gl u c
=
< u >w
Mw
18 g mol - 1
=
= 0.316
Mgluc
180 g mol - 1
12.22 a) Incorrect, the concentration of dissolved particles decreases, the vapour pressure
drops less compared to the pure solvent (according to Raoult’s law).
b) Incorrect, there are less particles dissolved per volume, so the osmotic pressure
decreases.
c) Correct, because the vapour pressure decreases due to the association of the
sugar molecules, the solution will start to boil at a lower T.
d) Correct.
e) Incorrect, the molar mass increases thus the distribution shifts to lower speeds.
110