Reversible
Reactions
Those chemical reactions which take
place in both the directions and never proceed to completion are called
Reversible reaction.
For these type of reaction both the
forward and reverse reaction occur at the same time so these reaction are
generally represented as
Reactant □ Product
The double arrow □ indicates that the
reaction is reversible and that both the
forward and reverse reaction can
occur simultaneously.
Some examples of reversible reactions
are given below
1. 2Hl □ H2 + l2
2. N2 + 2 H2 □ 2 NH3
Irreversible
Reactions
Those reactions in which reactants
are completely converted into product are called Irreversible reaction.
These reaction proceed only in one
direction. Examples of such type of
reaction are given below
1. NaCl + AgNO3 —-> AgCl + NaNO3
2. Cu + H2SO4 —-> CuSO4 + H2
Equilibrium
State
The state at which the rate of
forward reaction becomes equal to the rate of reverse reaction is called
Equilibrium state.
Explanation
Consider the following reaction
A + B □ C + D
It is a reversible reaction. In this
reaction both the changes (i.e. forward &
backward) occur simultaneously. At
initial stage reactant A & B are separated from each other therefore the
concentration of C and D is zero.
When the reaction is started and the molecules of A and B react with each other the concentration of reactant is decreased while the concentration of product is increased. With the formation of product, the rate of forward reaction decreased with time but the rate of reverse reaction is increased with the formation of product C & D.
When the reaction is started and the molecules of A and B react with each other the concentration of reactant is decreased while the concentration of product is increased. With the formation of product, the rate of forward reaction decreased with time but the rate of reverse reaction is increased with the formation of product C & D.
Ultimately a stage reaches when the
number of reacting molecules in the forward reaction equalizes the number of
reacting molecules in the reverse direction, so this state at which the rate of
forward reaction becomes equal to the rate of reverse reaction is called
equilibrium state.
Law of Mass
Action
Statement
The rate at which a substance reacts is proportional to its active mass and the rate of a chemical reaction is proportional to the product of the active masses of the reactant.
The rate at which a substance reacts is proportional to its active mass and the rate of a chemical reaction is proportional to the product of the active masses of the reactant.
The term “active mass” means the
concentration in terms of moles/dm3.
Derivation of Equilibrium Constant Expression
Consider in a reversible reaction “m”
mole of A and “n” moles of B reacts to give “x” moles of C and “y” moles of D
as shown in equation.
mA + nB □ xC + yD
In this process
The rate of forward reaction ∞ [A]m
[B]n
Or
The rate of forward reacting = Kf
[A]m [B]n&
The rate of reverse reaction ∞ [C]x
[D]y
Or
The rate of reverse reaction = Kf
[C]x [D]y
But at equilibrium state
Rate of forward reaction = Rate of
reverse reaction
Therefore,
Kf [A]m [B]n = Kf [C]x [D]y
Or
Kf / Kr = [C]x [D]y / [A]m [B]n
Or
Ke = [C]x [D]y / [A]m [B]n
This is the expression for
equilibrium constant which is denoted by Ke and defined as
The ratio of multiplication of active
masses of the products to the product of active masses of reactant is called
equilibrium constant.
Equilibrium
Constant for a Gaseous System
Consider in a reversible process, the
reactants and product are gases as
Shown
A(g) + B(g) □ C(g) + D(g)
When the reactants and products are in
gaseous state, their partial
pressures are used instead of their
concentration, so according to law of
mass action.
Determination of Equilibrium Constant
The value of equilibrium constant
K(C) does not depend upon the initial concentration of reactants. In order to
find out the value of K(C) we have to find out the equilibrium concentration of
reactant and product.
1. Ethyl
Acetate Equilibrium
Acetic acid reacts with ethyl alcohol
to form ethyl acetate and water as shown
CH3COOH + C2H5OH □ CH3COOC2H5 + H2O
CH3COOH + C2H5OH □ CH3COOC2H5 + H2O
Suppose ‘a’ moles of acetic acid and
‘b’ moles of alcohol are mixed in this reaction. After some time when the state
of equilibrium is established suppose ‘x’ moles of H2O and ‘x’ moles of ethyl
acetate are formed while the number of moles of acetic acid and alcohol are a-x
and b-x respectively at equilibrium.
According to law of mass action
K(C) = [CH3COOC2H5] [H2O] / [CH3COOH]
[C2H5OH]
K(C) = [x/V] [x/V] / [a-x/V] [b-x/V]
K(C) = (x) (x) / (a-x) (b-x)
K(C) = x2 / (a-x) (b-x)
2. Hydrogen
Iodide Equilibrium
For the reaction between hydrogen and
iodine suppose a mole of hydrogen and ‘b’ moles of iodine are mixed in a scaled
bulb at 444ºC in the boiling sulphur for some time. The equilibrium mixture is
then cooled and the bulbs are opened in the solution of NaOH. Let the amount of
hydrogen consumed at equilibrium be ‘x’ moles which means that the amount of hydrogen
left at equilibrium is a-x moles. Since 1 mole of hydrogen reacts with 1 mole
of iodine ‘o’ form two moles of hydrogen iodide hence the amount of iodine used
is also x moles so its moles at equilibrium are b-x and the moles of hydrogen
iodide at equilibrium are 2x.
According to law of mass action
K(C) = [Hl]2 / [H2] [l2]
K(C) = [2x/V]2 / [a-x/V] [b-x/V]
K(C) = 4×2 / (a-x) (b-x)
Applications of Law of Mass Action
There are two important applications
of equilibrium constant.
1. It is used to predict the
direction of reaction.
2. K(C) is also used to predict the
extent of reaction.
To Predict the Direction of Reaction
The value of equilibrium constant
K(C) is used to predict the direction of
reaction. For a reversible process.
Reactant □ Product
With respect to the ratio of initial
concentration of the reagent.
There are three possibilities for the
value of K
1. It is greater than K(C)
2. It is less than K(C)
3. It is equal to K(C)
Case I
If [Reactant]initial /
[Product]initial > K(C) the reaction will shift towards the reverse
direction.
Case II
If [Reactant]initial /
[Product]initial > K(C) the reaction will shift towards the forward
direction.
Case III
If [Reactant]initial /
[Product]initial > K(C) this is equilibrium state for the reaction.
To Predict the Extent of Reaction
From the value of K(C) we can predict
the extent of the reaction.
If the value of K(C) is very large e.g.
If the value of K(C) is very large e.g.
For 2 O3 □ 3 O2 ……….. K(C) = 10(55)
From this large value of K(C) it is
predicted that the forward reaction is almost complete.
When the value of K(C) is very low
e.g.,
2 HF □ H2 + F2 ……….. K(C) = 10(-13)
From this value it is predicted that
the forward reaction proceeds with
negligible speed.
But if the value of K(C) is moderate,
the reaction occurs in both the direction and equilibrium will be attained
after certain period of time e.g., K(C) for
N2 + 3 H2 □ 2 NH3 …………. is 10
N2 + 3 H2 □ 2 NH3 …………. is 10
So the reaction occurs in both the
direction.
Le Chatelier’s
Principle
Statement
When a stress is applied to a system at equilibrium the equilibrium position changes so as to minimize the effect of applied stress.
The equilibrium state of a chemical reaction is altered by changing concentration pressure or temperature. The effect of these changes is explained by Le Chatelier.
When a stress is applied to a system at equilibrium the equilibrium position changes so as to minimize the effect of applied stress.
The equilibrium state of a chemical reaction is altered by changing concentration pressure or temperature. The effect of these changes is explained by Le Chatelier.
Effect of Concentration
By changing the concentration of any
substance present in the equilibrium mixture, the balance of chemical
equilibrium is disturbed. For the reaction,
A + B □ C + D
A + B □ C + D
K(C) = [C][D] / [A][B]
If the concentration of a reactant A
or B is increased the equilibrium state shifts tc right and yield of products
increases.
But if the concentration of C or D is
increased then the reaction proceed in the backward direction with a greater
rate and more A & B are formed.
Effect of Temperature
The effect of temperature is
different for different type of reaction.
For an exothermic reaction the value of K(C) decreased with the increase of
For an exothermic reaction the value of K(C) decreased with the increase of
temperature so the concentration of
products decreases.
For a endothermic reaction heat is
absorbed for the conversion of reactant into product so if temperature during
the reaction is increased then the reaction will proceed with a greater rate in
forward direction.
ENDOTHERMIC
REACTION
Temperature increase —-> More
products are formed
Temperature decrease —-> More
reactants are formed
EXOTHERMIC REACTION
Temperature increase —-> More
reactants are formed
Temperature decrease —-> More
products are formed
Effect of Pressure
The state of equilibrium of gaseous
reaction is distributed by the change of pressure. There are three types of reactions
which show the effect of pressure change.
1. When the Number of Moles of Product are Greater
In a reaction such as
PCl5 <—-> PCl3 + Cl2
The increase of pressure shifts the
equilibrium towards reactant side.
2. When the Number of Moles of Reactant are Greater
In a reaction such as
N2 + 3H2 <—-> 2NH3
The increase of pressure shifts the
equilibrium towards product side because the no. of moles of product are less
than the no. of moles of reactant.
3. When Number of Moles of Reactants and Products are Equal
In these reactions where the number
of moles of reactant are equal to the number of moles of product the change of
pressure does not change the equilibrium state e.g.,
H2 + l2 □ 2 Hl
Since the number of moles of
reactants and products are equal in this reaction so the increase of pressure
does not affect the yield of Hl.
Important Industrial Application of Le
Chatelier’s Principle
Haber’s Process
This process is used for the
production of NH3 by the reaction of nitrogen and hydrogen. In this process 1
volume of nitrogen is mixed with three volumes of hydrogen at 500ºC and 200 to
1000 atm pressure in presence of a catalyst
N2 + 3 H2 □ 2 NH3 …………… ΔH = -46.2
kJ/mole
1. Effect of Concentration
The value of K(C) for this reaction
is
K(C) = [NH3]2 / [N2] [H2]3
Increase in concentration of
reactants which are nitrogen and hydrogen the equilibrium of the process shifts
towards the right so as to keep the value of K(C) constant. Hence the formation
of NH3 increases with the increase of the concentration of N2 or hydrogen.
2. Effect of Temperature
It is an exothermic process, so heat
is liberated with the formation of product. Therefore, according to Le
Chatelier’s principle at low temperature the equilibrium shifts towards right
to balance the equilibrium state so low temperature favours the formation of
NH3
3. Effect of Pressure
The formation of NH3 proceeds with
the decrease in volume, therefore, the reaction is carried out under high
pressure or in other words high pressure is favourable for the production of
NH3.
Contact
Process
The process is used to manufacture
H2SO4 on large scale. In this process the most important step is the oxidation
of SO2 to SO3 in presence of a catalyst vanadium pentoxide.
2 SO2 + O2 □ 2 SO3 ………………. ΔH = – 395
kJ/mole
1. Effect of Concentration
The value of K(C) for this reaction
is
K(C) = [SO3]2 / [SO2]2 [O2]
Increase in concentration of SO2 or
O2 shifts the equilibrium towards the right and more SO3 is formed.
2. Effect of Temperature
Since the process is exothermic, so
low temperature will favour the formation of SO3. The optimum temperature for
this reaction is 400 to 450ºC.
3. Effect of Pressure
In this reaction decrease in volume
takes place so high pressure is favourable for the formation of SO3.
Common Ion
Effect
Statement
The process in which precipitation of an electrolyte is caused by lowering the degree of ionization of a weak electrolyte when a common ion is added is known as common ion effect.
The process in which precipitation of an electrolyte is caused by lowering the degree of ionization of a weak electrolyte when a common ion is added is known as common ion effect.
Explanation
In the solution of an electrolyte in water, there exist an equilibrium between the ions and the un dissociated molecules to which the law of mass action can be applied.
In the solution of an electrolyte in water, there exist an equilibrium between the ions and the un dissociated molecules to which the law of mass action can be applied.
Considering the dissociation of an
electrolyte AB we have
AB □ A+ + B-
And
[A+][B-] / [AB] = K (dissociation
constant)
If now another electrolyte yielding
A+ or B- ions be added to the above solution, it will result in the increase of
concentration of the ions A+ or B- and in order that K may remain the same, the
concentration AB must evidently increase. In other words the degree of
dissociation of an electrolyte is suppressed by the addition of another
electrolyte containing a common ion. This phenomenon is known as common ion
effect.
Application of Common Ion Effect in Salt
Analysis
An electrolyte is precipitated from
its solution only when the concentration of its ions exceed from the solubility
product. The precipitates are obtained when the concentration of any one ion is
increased. Thus by adding the common ion, the solubility product can be
exceeded.
In this solution Ou(OH)2 is a weak base while H2SO3 is a strong acid so the pH of the solution is changed towards acidic medium.
In this solution Ou(OH)2 is a weak base while H2SO3 is a strong acid so the pH of the solution is changed towards acidic medium.
When Na2CO3 is dissolved in water, it
reacts with water such as
Na2CO3 + 2 H2O □ 2 NaOH + H2CO3
In this solution H2CO3 which is weak
acid an NaOH which is a strong base are formed. Due to presence of strong base
the medium is changed towards basic nature.
Solubility
Product
When a slightly soluble ionic solid
such as silver chloride is dissolved in water, it decompose into its ions
AgCl □ Ag+ + Cl-
These Ag+ and Cl- ions from solid
phase pass into solution till the solution becomes saturated. Now there exists
an equilibrium between the ions present in the saturated solution and the ions
present in the solid phase, thus
AgCl □ Ag+ + Cl-
AgCl □ Ag+ + Cl-
Applying the law of mass action
K(C) = [Ag+][Cl-] / [AgCl]
Since the concentration of solid AgCl
in the solid phase is fixed, no matter how much solid is present in contact
with solution, so we can write.
K(C) = [Ag+][Cl-] / K
K(C) = [Ag+][Cl-] / K
Or
K(C) x K = [Ag+][Cl-]
Or
K(S.P) = [Ag+][Cl-]
Where K(S.P) is known as solubility
product and defined as
The product of the concentration of
ions in the saturated solution of a sparingly soluble salt is called solubility
product.
the value of solubility product is
constant for a given temperature.
Calculation of Solubility Product From
Solubility
The mass of a solute present in a
saturated solution with a fixed volume of solvent is called solubility, which
is generally represented in the unit of gm/dm3. With the help of solubility we
can calculate the solubility product of a substance e.g., the solubility of
Mg(OH)2 at 25ºC is 0.00764 gm/dm3. To calculate the K(S.P) of Mg(OH)2, first of
all we will calculate the
concentration of Mg(OH)2 present in
the solution.
Mass of Mg(OH)2 = 0.00764 gm/dm3
Moles of Mg(OH)2 = 0.00764 / 58 moles
/ dm3
= 1.31 x 10(-4) moles/dm3
The ionization of Mg(OH)2 in the
solution is as follows.
Mg(OH)2 □ Mg(+2) + 2 OH-
And the solubility product for
Mg(OH)2 may be written as,
K(S.P) = [Mg(+2)] [OH-]2
Since in one mole of Mg(OH2) solution
one mole of Mg++ ions are present
while two moles of OH- ions are present,
therefore in 1.31 x 10(-4)
mole/dm3 solution of Mg(OH)2, the
concentration of Mg(+2) is 1.31 x 10(-4) moles/dm3 while the concentration of
OH- is 2. 62 x 10(-8) moles/dm3. By substituting these values
K(S.P) = [Mg(+2)][OH-]2
= [1.31 x 10(-4)] [2.62 x 10(-4)]2
= 9.0 x 10(-12) mole3 / dm9
So in this way the solubility product
of a substance may be calculated with
the help of solubility.
Calculation of Solubility from Solubility
Product
If we know the value of solubility
product, we can calculate the solubility of the salt.
For example, the solubility of
PbCrO4at 25ºC is 2.8 x 10(-13) moles/dm3.
m = n2 / w1 in kg
m = (w2 / m2) / (w1 / 1000)
m = w2 / m2) x (1000 / w1)
Hydration
Addition of water or association of
water molecules with a substance without dissociation is called Hydration.
Water is a good solvent and its polar
nature plays very important part in dissolving substances. It dissolves ionic
compounds readily.
When an ionic compound is dissolved
in water, the partial negatively charged oxygen of water molecule is attracted
towards the cation ion similarly the partial positively charged hydrogen of
water molecule is attracted towards the anions so hydrated ions are formed.
Diagram Coming Soon
Diagram Coming Soon
In solution, the number of water
molecules which surround the ions is indefinite, but when an aqueous solution
of a salt is evaporated the salt crystallizes with a definite number of water
molecules which is called as water of crystallization E.g., when CuSO4
recrystallized from its solution the crystallized salt has the composition
CuSO4. 5H2O. Similarly when magnesium chloride is recrystallized from the
solution, it has the composition MgCl2.6H2O. This composition indicates that
each magnesium ion in the crystal is surrounded by six molecules. This type of salts
is called hydrated salts.
It is observed experimentally that
the oxygen atom of water molecule is attached with the cation of salt through
co-ordinate covalent bond so it is more better to write the molecular formulas
of the hydrated salts as given below.
[Cu(H2O)5]SO4 …………….. [Mg(H2O)6]Cl2
[Cu(H2O)5]SO4 …………….. [Mg(H2O)6]Cl2
It is also observed that these
compound exist with a definite geometrical structure e.g., the structure of
[Mg(H2O)6]Cl2 is octahedral and [Cu(H2O)4]+2 is a square planar.
Diagram Coming Soon
Factors for Hydration
The ability of hydration of an ion
depend upon its charge density.
For example the charge density of Na+ is greater than K+ because of its smaller size, so the ability of hydration for Na+ is greater than K+ ion. Similarly small positive ions with multiple charges such as Cu(+2), Al(+3), Cr(+3) posses great attraction for water molecules.
For example the charge density of Na+ is greater than K+ because of its smaller size, so the ability of hydration for Na+ is greater than K+ ion. Similarly small positive ions with multiple charges such as Cu(+2), Al(+3), Cr(+3) posses great attraction for water molecules.
Hydrolysis
Addition of water with a substance
with dissociation into ions is called Hydrolysis.
OR
The reaction of cation or anion with water so as to change its pH is known as Hydrolysis.
Theoretically it is expected that the solution of salts like CuSO4 or Na2CO3 are neutral because these solutions contain neither H+ ion nor OH-, but it is experimentally observed that the solution of CuSO4 is acidic while the solution of Na2CO3 is basic. This acidic or basic nature of solution indicate but H+ ions or OH- ions are present in their solutions which can be produced only by the dissociation of water molecules.
OR
The reaction of cation or anion with water so as to change its pH is known as Hydrolysis.
Theoretically it is expected that the solution of salts like CuSO4 or Na2CO3 are neutral because these solutions contain neither H+ ion nor OH-, but it is experimentally observed that the solution of CuSO4 is acidic while the solution of Na2CO3 is basic. This acidic or basic nature of solution indicate but H+ ions or OH- ions are present in their solutions which can be produced only by the dissociation of water molecules.
Theory of Ionization
1n 1880, a Swedish chemist Svante
August Arrhenius put forward a theory known as theory of ionization, in order
to account for the conductivity of electrolytes, electrolysis and certain
properties of electrolytic solutions. According to this theory.
1. Acids, Bases and Salts when
dissolved in water yield two kinds of ions, one carry positive charge and the
other carry negative charge. The positively charged ions are called cations
which are derived from metals or it may be H+ ion but the negatively charged
ions which are known as anions are derived from non-metals
NaCl —-> Na+ + Cl-
H2SO4 —-> 2 H+ + SO4(-2)
KOH —-> K+ + OH-
2. Ions in the solution also
recombine with each other to form neutral molecules and this process continues
till an equilibrium state between an ionized and unionized solid is attained.
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