THE HYDROGEN BONDING TO THE PHYSICAL PROPERTIES OF SUBSTANCES, INCULDING ICE AND WATER
-----------OF THE HYDROGENBONDING TO THE PHYSICAL PROPERTIES OF SUBSTANCES, INCULDING ICE AND WATER.
The electronegative elements like
fluorine, oxygen and nitrogen are responsible for hydrogen bonding. Physical
properties of such molecules are different from other molecules.
(1) In liquid state water molecules are linked with one another through
hydrogen bonding. This is why boiling point of hydrogen is more than covalent
hydrides of same group.
(2) The water molecules in ice become more regular in structure
through hydrogen bonding. The structure of ice is diamond like i.e; each oxygen
atom of water molecule have four tetrahedral corners. Two corners are occupied
by bonded pair and two by lone-pair of electrons, through lone pairs oxygen
atom form two hydrogen bonds with the neighboring water molecules. Due to wide
spacing of water molecules in ice, it become less denser thain liquid water and
floats over the surface of liquid water.
(3) It has been observed that the compounds showing hydrogen
bonding are mutually soluble. Thus, both water and ethanol have hydrogen
bonding, therefore they are soluble in all proportions.
(4) Cleaning action of soaps and detergents is also due to
hydrogen bonding.
(5) Nylon fibres have high tensile strength and highly elastic due
to hydrogen bonding between hydrogen of – NH group and oxygen of C=O group.
(6) Cotton shrinks due to hydrogen bonding.
(7) Biochemical polymers like DNA and protein have helical
structure which is also due to hydrogen bonding.
(8) The adhesive property of paints and dyes to hydrogen bonding.
(9) Substances like glue and honey are sticky due to hydrogen
bonding.
(10) Ammonia is a gas which liquefies easily under pressure due to
hydrogen bonding and similarly alcohol molecules do not exist in gas phase but
exist as liquid and solid due to hydrogen bonding.
PREDICTION OF TYPE OF STRUCTUREAND BONDING PRESENT IN A SUBSTANCE FROM QUOTED DATA:
It is possible to use knowledge of physical properties to predict
the nature of bonding in an unknown substance.
Consider the following data:
|
Substance |
Mr |
Melting
point (OC) |
Boiling
Point (OC) |
Electrical
Conductivity |
|
A |
60 |
1610 |
2205 |
None |
|
B |
74.5 |
772 |
1407 |
In liquid stats |
|
C |
86 |
-95 |
69 |
None |
Substance A and B both have high
melting points, suggesting giant lattice structures. Substance A does not
conduct at all, whereas substance B conducts when molten. Therefore; substance
A is a giant molecular compound. Substance B is an ionic compound. The very low
melting and boiling points of substance C shows that it must be simple covalent
molecule, which also agrees with its lack of electrical conductivity Thus;
Substance C is a simple covalent molecule.
QUESTION 1: Where 30 cm3 of a gaseous hydrocarbon was
completely burnt in an excess of oxygen, 120 cm3 of carbon dioxide
and 90 cm3 of water vapour were formed. (All volumes were measured
at the same temperature and pressure). Deduce the formula of the hydrocarbon.
SOLUTION:
According to Avogadro’s law v
v = kn or v/n
= k
CxH
Vol.
of hydrogen = Vol.
of carbon dioxide
No. of moles of hydrocarbon no. of moles of CO2
30 x = 120
x = 120/30 or x = 4
Vol.
of CxHy = Vol. of H2O
No. of moles of CxHy no. of moles of H2O
Y = 90/15 or y = 6
Hence, the formula of hydrocarbon
is; C4H6.
QUESTION 2: An
organic compound X contains 60% carbon, 13.3% hydrogen and 26.7% oxygen by mass
0.6 g of x occupies 0.336 dm3 at 137 0C and standard
pressure. Calculate the (a) empirical formula, (b) relative molecular mass and
(c) molecular formula of X.
SOLUTION:
(a) C H O
60 13.3 26.7
12 1 16
5 13.3 1.67
5_ 13.3 1.67
1.67 1.67 1.67
3 8 1
Empirical
formula of x is; C3H8O
(b) m = 0.6 g
T
= 137 0C + 273 = 410 K
P
= 1 atom or 101325 Nm-2
V
= 0.336 dm3 or 0.336 x 10-3 m-3
R
= 8.31 JK-1 mole-1
QUESTION:
A gaseous mixture which contains 4 g
of hydrogen and 96 g of oxygen has a total pressure of 200 KPa. Calculate the
partial pressures of the gases.
Solution:
Number of moles of Hydrogen,
N =
=
Number of moles of oxygen;
N =
=
Total no of moles = (2 + 3) = 5
moles.
Partial pressure of Hydrogen =
Partial pressure of oxygen =
QUESTION:
A small space
craft of capacity 20 m3 is connected to another of capacity 30 m3.
Before connection, the pressure inside the smaller craft is 100k Pa and that
inside the larger is 150k Pa. What is the total pressure on connection?
PV =
Mr =
= 0.6 x 8.31 x 410
101325 x 0.336 x 10-3
Mr = 60 g mole-1
Thus Mr of x is; 60
(c) Let the molecular formula of x be (C3
H8O)n (Cn H8O)n = 60 (36
+ 8 + 16)n = 60
60n =
6
n = 1
Therefore,
The molecular
formula of x is; C3H8O
QUESTION 3:
0.258 g of x, when vaporized at 182 0C and 740 mm Hg
pressure, occupied a volume of 115 cm3. Calculate the relative
molecular mass of x.
SOLUTION:
M = 0.258 grams…… R = 8.31 J mole-1 K-1.
T = 182 0C
+ 273 = 455 K
P = 740 mm or
= 0.973 x 101325
= 98658.55 Nm-2
V = 115 cm3
or
Mr =
=
= 85.98
By using Boyle’s
Law, the pressure caused by the gas in the smaller craft on connection;
P1 =
100 k Pa P2 = ?
V1 =
20 m3 V2
= (20 + 30) = 50 m3
P1V1 = P2V2
P2 =
=
= 40 K Pa
The pressure caused by the gas
in the larger craft on connection;
P1 = 150 K Pa P2 = ?
V1 = 30 m3 V2 = (20 + 30) = 50
m3
P1V1 = R2V2
P2 =
=
= 90 K Pa
By using Dalton’s lan of
partial pressure, Total pressure = 40 + 90
= 130 K Pa
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