FSc Notes Class 11 Chemistry Chemical Equilibrium
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Chemical Equilibrium
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.
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 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 reactin = 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
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.
- It is used to predict the direction of reaction.
- 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
- It is greater than K(C)
- It is less than K(C)
- 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.
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
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.
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
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 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.
Explanation
In the solution of an electrolyte in water, there exist an equilibrium between the ions and the disassociated 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.
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-
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
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.
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
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.
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.
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.
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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