[PDF]A Level Passbook Chemistry

. For

. 1042

Higher

Secondary . PASSBOOKS
Entrance A-Level


. Examinations . и
| Chemistry


ЈЕ. Chandler and R.J. Wilkinson


charge of iron ore
coke and limestona


cones
hopper
ҒеО, + ЗСО = 2Fe
refractory brick


slag formed


_ iron saturate


molten iron


48. A blast furnace


anc RS


‘A-Level Passhook
Chemistry


A-Level Passbook
Chemistry


J. E. Chandler, B.Sc.


and
R. J. Wilkinson, Ph.D.


Arnold-Heinemann


БОВЕ T. W 5, LIBRARY
Datc [0 Ы Ф 18


Ассп. Nø.. 1622.


First published 1976 by Intercontinental Book Productions


Published 1982 by Charles Letts & Co Ltd
t Charles Leits Books Lid


First Published in India 1984


Published by Gulab Vazirani for Arnold Heinemann
Publishers (India) Pvt. Ltd., AB/9, Safdarjang Enclave,
New Delhi-110029 and printed at Mayfair Press, 31,
Бари Park, New Delhi-110003.


Contents


Introduction, 7


Atomic and Molecular Mass Deter-
mination; Colloids, 9


Chemical and Physical Equilibria, 28


Thermochemistry and Reaction Kinetics,
49 4


Electrochemistry and Ionic Equilibria, 74


Electronic Structure and Bonding;
Radioactivity, 100


s and p Block Elements, 130


d Block and First Row Transition Ele-
ments, 173


Quantitative and Qualitative Analysis, 193
Organic Chemistry: Introduction, 213
Hydrocarbons, 227

Functional Groups, 243

Organic Syntheses and Analysis, 278
Macromolecules, 288


Industrial Aspects of Organic Chemistry,
' 300


Index, 315


Introduction


This book is intended for students taking the examinations for
the General Certificate of Education at Advanced Level (both
traditional and Nuffield type courses). In addition it will prove
useful to students taking Technical Education Council courses
in Chemistry to level III.


The contents take into account the majority of topics listed in
the various syllabuses of the regional examining boards at this
level, including aspects of qualitative and quantitative analysis.
S.I. (International System) units and nomenclature are used
throughout.


Qualifications in Chemistry are of considerable importance
both in their own right and in many fields including biology,
medical and veterinary sciences, engineering and allied subject
areas. An understanding of Chemistry to the level of this text
is important to enable a student to study to undergraduate
level in these fields.


Acknowledgements . 1
To Janet for her hard work in typing and checking the manu-
script and for her tolerance and encouragement during the
writing of this book.


Thanks also to Julia for her support and encouragement.


Chapter 1

Atomic and Molecular
Mass Determination;
Colloids


Relative atomic mass, A, (formerly known as atomic
weight) is defined as the average mass of the atoms of an
element on a scale on which one atom of "C isotope of carbon
has a mass of 12 units.


mass of one atom of element x 12
STAG AR EA
mass of one atom of "C


Relative atomic mass, А, =


Mass spectrometer


ion source photographic


vacuum


Figure 1. Mass spectrometer


The mass spectrometer has now superseded all previous
methods for measuring atomic mass. The sample is introduced
into the spectrometer and ionized either by electron
bombardment, electric discharge or by direct heating. The
ionized atoms are then accelerated by applying a high negative
potential to the plates P, and P;. The slit in P» produces a fine


9


e
beam of ions all moving with the same velocity. The beam
then enters a uniform magnetic field which deflects the ions in
a circular path. The radius of the path depends on the mass
and charge of the ion. At 180° to the original beam direction
the ions fall onto a photographic plate producing a series of
lines corresponding to the different charges on the ions. The
photographic plate is calibrated using a sample of "C and by
comparing the positions of the lines the atomic mass can be
found. Alternatively electronic detectors can be used. The
whole apparatus is kept under high vacuum.


Relative molecular mass, M,. (formerly known as molecular
weight) of an element or compound is defined as the average
mass of its molecules on a scale on which one atom of the "C
isotope of carbon has a mass of 12 units.


The relative molecular mass of a compound is also equal to the
sum of the relative atomic masses of the atoms in the mole-
cule. However relative molecular mass is normally used to
derive molecular formulae and not vice versa.


Molecular masses of gases and vapours


А mole is the amount of substance which contains the same
number of particles (i.e. atoms, molecules, ions etc.) as there
are atoms in 12 grams of "C. This means that the mass in
grams of a mole of compound is its relative molecular mass.
Avogadro's hypothesis states that equal volumes of gases at
the same temperature and pressure contain the same number
of molecules, from which it is found that one mole of all
gases at s.t:p. occupies 22-4 dm’.


Hence: relative molecular mass of a gas (or vapour) = mass in
grams of 22-4 dm' of the gas measured at s.t.p.


Molecular masses are more conveniently found from the rela-
tive or vapour density of a gas. Relative density of a gas is
defined as the ratio of the masses of equal volumes of the gas
and hydrogen measured under identical conditions.


mass of one volume of gas


Relative density
mass of one volume of hydrogen


т


mass of n molecules of gas
mass of n -molecules of hydrogen


(Avogadro's theory)


mass of 1 molecule of gas
mass of 1 molecule of hydrogen


(dividing by n)
_ mass of 1 molecule of gas
mass of 2 atoms of hydrogen


(since hydrogen is diatomic)
= į(relative molecular mass)


= M.
2-016


Relative density — Me ( more exactly)


Regnault’s method This method of finding relative molecu-
lar mass is by direct weighing of a gas in a bulb of about 1 dm?
volume. 1. The bulb is evacuated and weighed (M,). 2. The
bulb is filled with the gas at a known temperature and pressure
and reweighed (M;). 3. The bulb is filled with water and
weighed (Му. M; — Mi is the mass of water and hence the
volume of gas filling the bulb. This volume is reduced to s.t.p.
and since the mass of gas, (M; — M), is known then M, the
mass of 22-4 dm! of the gas can be found. Corrections for the
üpthrust of air on the bulb improve the accuracy.


Dumas' method This method is suitable for volatile liquids
and solids which vaporize on heating. A bulb with a fine
capillary neck is used.


seal tip
when ali

liquid has
vaporized


volatile -
liquid


Fieure 2. Dumas’ bulb


i. The bulb is weighed full of air, (My). 2. Some of the liquid is
introduced into the bulb. 3. The liquid is vaporized by immers-


11


ing the bulb in a bath at least 10°C above the boiling point o:
the liquid. The vapour fills the bulb completely, the excess
vapour which escapes expelling the remaining air. 4. The tip of
the bulb is sealed and when cool the bulb is weighed (M3). 5.
Тће tip of the bulb is broken under water, which rapidly fills
the bulb due to the vacuum caused by the condensed vapour.
The mass of the bulb and water is found (M3). The volume of
the bulb is found from (M; — M,). By multiplying this by the
density of air the mass of air in the bulb can be found. Then
mass of bulb = M, — mass of air in bulb = M4. The mass of
vapour in the bulb = M;— М,. The temperature of the heating
bath is measured and the volume of vapour (volume of bulb) is
corrected to s.t.p. From the mass and volume of vapour the
mass of 22-4 dm' (M,) can be found.


Example A bulb of volume 1:33 dm' and mass 107-08 g when
filled with air, has a mass of 112-1 р when filled with tetra-
chloromethane vapour. The temperature of the heating bath
was 373 K. Find the rel. mol. mass of ССІ, if the lab. temp. is
293 K, atmospheric pressure is 750 mm Hg and the density of
air at s.t.p. is 1-293 рат".


Р 50 х2 `
vol. of air in bulb at s.t.p. = 135 стал = 1:22 дт


mass of air = 1:22 х 1-293 = 1:58 р
True mass of bulb = 107-08 — 1-58 = 105-5 g
mass of vapour = 112-1 — 105-5 = 6:62


The volume of ; _ 1:33 x 750 х 273 _ 9.961 dm?
Gd se meet MS НИ
The mass of 0-961 dm? is 6-6 g
366x224
mass of 22:4 dm! = 3
ES Мој 1048
М, = 154


Diffusion or effusion method Graham's law of diffusion (р:


30) is
Ri_ Je
К pi


where Rj, R2, are the rates of diffusion and ри. p the densities
of two gases.


12


У . __ density of gas
RélaUveTd eti density of hydrogen


R [а
= = 1 2 where d, d. are relative densities
К; а,


To find the relative density of a gas its rate of effusion is
usually found. Effusion is the passage of a gas under pressure
through a small orifice; the same law applies as for diffusion.
The rates of effusion for two gases, one of known M, are
compared by finding the time taken for them to effuse through
the same apparatus under identical conditions. The rate of
effusion is inversely proportional to the times taken, t, t».


From this expression the unknown relative density, and hence
molecular mass, can be found.


Abnormal relative densities
of vapours


Association Occasionally substances have abnormally high
relative densities. Often the relative density is double the
expected value. This is due to molecules dimerizing in the


vapour state, e.g.
2FeCl; = Fe2Cla; 2AICl; = ALCl,


Thermal dissociation Vapours which show much lower
relative densities than expected are probably undergoing
thermal dissociation. Many compounds split up at high
temperatures and recombine on cooling


NH,Cl= NH; + НСІ
PCl; = PCI; + Cl;
N204= 2NO2
On complete dissociation the volume of vapour is doubled


while the mass remains the same and so the relative density is
halved. The value of the relative density enables the degree


of dissociation, с, to be determined.


13


For all the above systems, if а is the fraction of molecules
dissociating, the number of moles at equilibrium is given by, e.g.


N;0,22NO;


1-а 2a moles


The total number of moles present at equilibrium will be
l-a+2a=l+a


For a given mass of vapour the density is inversely propor-
tional to the volume or number of moles present.


density undissociated t
density dissociated 1


_ density undissociated -
~ density dissociated


-Molecular masses of solids:
Colligative properties


When a non-volatile solid is dissolved in a liquid


the vapour pressure of the liquid is lowered
the boiling point is raised

the freezing point is lowered

+ OSMOSIS can occur.


Bae tel ee


These colligative properties are used to determine molecular
weights of compounds. The laws relating to these properties
apply only to non-electrolytes in dilute solutions.


1. Lowering of vapour pressure It is found that the rela-
tive lowering of vapour pressure of a solution is equal to
the mole fraction of soiute. This is an alternative statement
of Raoult's law (see p. 33). The lowering of vapour pressure
depends on the number of particles of solute (dissolved


substance) and not on the type of molecule (provided there is
no ionization). à


Mathematical statement:


о
DD. qno mae
CP SAYE N (dilute solution)


14


where p? and p are the vapour pressures of solvent and
solution respectively and п and N are the number of moles of
solute and solvent respectively. An explanation on kinetic
theory grounds is that the dynamic equilibrium which exists
between the solvent and its vapour is upset by the presence of
involatile solute molecules. A dynamic equilibrium 15
achieved when the rate of molecules evaporating from the
liquid is equal to the rate of molecules condensing from the
vapour. These rates attain equilibrium at a constant tempera-
ture. The involatile solute can only occupy the liquid state and
hinders the evaporation rate. The condensation rate remains
constant and a new equilibrium is reached with less molecules
in the vapour and a lower vapour pressure.


By measuring the relative lowering of vapour pressure of a
solution the weight of one moie of non-volatile solute can be
determined. This is achieved by passing a stream of dry air .
through the solution and then the pure solvent and measuring
the loss in weight of each.


Example A solution contains 6:3g urea in 45g of water.
When air was bubbled successively through this solution and
then pure water the loss in weight of the solution was 1122
and of the water 0-0469 g. Find the molecular weight of urea.


Loss in weight of solution x p = 1-1 12g
Loss in weight of water ~ p*— p = 0:0469 р
Total loss in weight ~ p° = 1-112 + 0-0469 g


= 0-0405


Bence p^ 11589 N+n


If M, is the relative molecular mass of urea then n, number of
moles of solute


and N, number of moles of solvent


- 45
718


tias n | 63/М,
mole fraction, Ar = 35118 + 6-3] M,


М, = 60


= 0:0405


15


2. Elevation of boiling point


one atmosphere


pure solvent


vapour
pressure
pure solid


Figure 3


A liquid boiis when its vapour pressure is equal to the external
pressure. Figure 3 shows the lowering of vapour pressure
produced by adding an involatile solute to a solvent. !t can be
seen that the solutions must be raised to higher temperatures
(Tı, Ту) before the saturated vapour pressure is:equal to
atmospheric pressure (i.e. boiling point). For dilute solutions
the solution curves are parallel to the solvent curve and ABC


and ADE are similar triangles.
Thas AB zAC
AD AE


pond to the elevation of boiling point
pond to the lowering of vapour pres-


But AB and AD corres
and AC and AE corres
sure. Hence


elevation of b.p. for solution 1 _ lowering of s.v.p. for 1
elevation of b.p. for solution 2 lowering of s.v.p. for 2


but the lowering of v.
sol


Р. is proportional to the concentration of
ute. Therefore, the elevation of boiling point of a solvent
is proportional to the molar concentration of solute. This
can also be shown experimentally and is true provided there is
no dissociation or association of the solute. The elevation of
boiling point is independent of the type of solute, it
depends only on the number of particles present.


16


The elevation of boiling point produced by one mole of solute
contained in 100g (or 1000g) of a solvent is termed the
ebullioscopic (or elevation of boiling point) constant and has
the symbol K. The relative molecular mass (M,) of m grams of
solute in w grams of solvent can be calculated from the
following formula у


MSIE (1.1)


(100 is replaced by 1000 if values of K are quoted рег 1000 g of
solvent.)


The value of K can be found experimentally or from the


expression >
K=—— (1.2)


where T = boiling point, L = latent heat of vaporization of the
solvent and R = the molar gas constant.


Example Calculate the molecular weight of sulphur if 1-38 g
dissolved in 63 g of carbon disulphide (K = 23-7°C/100 g) raises
its boiling point by 0:203°С.


· From equation 1.1


M, = 138x 100 x 23-7
А 63 x 0-203
М, = 256 (= Sx)

3. Depression of freezing point The graph in fig. 3 shows
that the lowering of vapour pressure also lowers the fréezing
point (T; T, and Ts). It can be shown that the freezing
point depression is proportional to the molar con-
centration of solute, for dilute solutions of involatile sol-
utes which do not associate or dissociate. The depression of
freezing point produced by dissolving one mole of solute in
100g of solvent is termed the cryoscoplc (or depression of
freezing point) constant, symbol K (*C/100 g).


It may be found experimentally or from equation 1.2 where
T = melting point and L = latent heat of fusion of the solvent.


The depression of freezing point (t) is related to relative
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