Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

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CHEMICAL EQUILIBRIUM

In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system. Usually, this state results when the forward reaction proceeds at the same rate as the reverse reaction. The reaction rates of the forward and backward reactions are generally not zero, but equal. Thus, there are no net changes in the concentrations of the reactant(s) and product(s). Such a state is known as dynamic equilibrium.

1. Irreversible and Reversible Reactions

Irreversible reactions

Those reactions which proceeds in forward direction and reaches almost to completion are called Irreversible reactions.

For example:

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev (Thermal decomposition)

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev (Strong acid-strong base neutralization reaction)

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev (Precipitation reaction)

Irreversible Reactions: 

· Always goes for completion.
· Reactants are completely consumed in these reactions 

Ex. Ppt. reaction, NaCl + AgNO3→AgCl(s) + NaNO3

case, if we remove AgCl (which is the product), reaction will mo ve in forward direction].

Reversible reactions

Those reactions which proceed in both forward and backward directions and never reaches completion are called Reversible reactions.

For example,

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev (Thermal dissociation)

CH3COOH + NaOH ⇌ CH3COONa + H2O (Naturalization reactions)

N2(g) + 3H2(g) ⇌ 2NH3(g) (synthesis reaction)

But when Fe(s) is heated and water vapour is passed over it in open vessel, it is converted to Fe3O4(s) along with the evolution of hydrogen gas.

3Fe(s) + 4H2O(g) → 3Fe3O4(s) + 4H2(g)

And when Fe3O4 is reduced with hydrogen gas, it gives Fe(s) and H2O

Fe3O4(s) + 4H2(g) → 3Fe(s) + 4H2O(g)

But, if the react ion is carried out in close vessel, this react ion beco mes reversible.

3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g)

Reversible reactions: 

· Never goes for completion 

· Irrespective of the fact whether the reaction has started by reactants or products. 

Ex: A + B ⇌ C + D

If we remove C (which is product), reaction will again start and moves in forward direction to attain equilibrium again. 

· These reactions can be initiated in any direction.
· Always takes place in a closed vessel.

2. State of Equilibrium

It has generally been observed that many changes (physical and chemical) do not proceed to complet ion when they are carried out in a closed container. Consider, for example, vaporisat ion of water,

Water ⇌ Vapour

At any temperature, vapourisation of water takes place, initially the concentration of water is much greater than the concentration of vapour, but with the progress of time, concentration of vapour increases whereas that of water remains constant and after a certain interval of time. There is no change in concentration of vapour, this state is known as state of physical equilibrium. In a similar way, this has also been found for chemical react ions, for example. When PCl5(g) is heated in a closed continue, its dissociat ion starts with the format ion of PCl3(g) and Cl2(g) are formed due to dissociation of PCl5(g). After a certain interval of time, the concentration o f PCl5(g), PCl3(g) and Cl2(g) each beco mes constant. It does not mean that at this point of time, dissociat ion of PCl5(g) and its format ion fro m PCl3(g) and Cl2(g) has been stopped. Actually t he rate of dissociation of PCl5 and the rate of formation of PCl5(g) beco mes equal. This state is called the state of chemical equilibrium. So, the state of chemical equilibrium is dynamic.

PCl5(g) ⇌ PCl3(g) + Cl2(g)

This can be shown graphically.

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

So, “state of chemical equilibrium can be defined as the state when the rate of forward reaction becomes equal to rate of reverse reaction and the concentration of all the species becomes constant”.

3. Quantitative Aspects of Equilibrium: Law of Mass Action

Guldberg and Waage in 1807 gave this law and according to this law, “At constant temperature, the rate at which a substance reacts is directly proportional to its active mass and the rate at which a chemical reaction proceeds is directly proportional to the product of active masses of the reacting species”.

The term active mass of a reacting species is the effective, concentration or its activity (a) which is related to the molar concentration (C) as

a = f C

where f = activity coefficient. F < 1 but f increases with dilution and as V → ∞ i.e. C → 0, f → 1 i.e., a → C Thus at very low concentration the active mass is essentially the same as the molar concentration. It is generally expressed by enclosing the formula of the reacting species in a square bracket.

To illustrate the law of mass action, consider the following general reaction at constant temperature, 

aA(g) + bB(g) ⇌ cC(g) + dD(g)

Applying law of mass action, 

Rate of forward reaction, 

Rf ∝ [A]a [B]b 

Or Rf = kf[A]a[B]b ………………………….(i)

Where kf = rate constant of forward reaction
Similarly,
Rate of reverse reaction

Rr ∝ [C]c [D]d 
Or Rr = kr[C]c [D]d …………………………(ii)

Where kr = rate constant of reverse reaction.

At equilibrium,

Rate of forward reaction = Rate of reverse reaction

i.e., Rf = Rr

So, from equation (i) and (ii), we get

kf[A]a [B]b = kr[C]c[D]d
or Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

or Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

where, Kc is the equilibrium constant in terms of molar concentration.

Illustration:

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

The rate of disappearance of A is given as: 

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev
Calculate the value of equilibrium constant.

Solution: We know, at equilibrium concentration of each species present in the given equilibrium. Become constant.

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Equilibrium Constant (Kp) in terms of Partial Pressure: Consider the same general reaction taking place at constant temperature,

aA(g) + bB(g) ⇌ cC(g) + dD(g)

From law of mass action, 

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev   .....................(1)

From ideal gas equation 

PV = nRT

Or Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

At constant temperature,

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Thus, we can say in a mixture of gases, Partial pressure of any component (say A)

P ∝ [A]

Similarly, P ∝ [B]

P ∝ [C]

P ∝ [D]

So, equation (1) can be rewritten as

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev ..............(2)

All Equilibrium constants (Keq)-

· Kc
· Kp
· Kx
· Kn

Kx and Kn are not considered as equilibrium constants, because these can be varied without change in temp. for gases.

4. Relationship Between Kp and Kc

From the above, for the same general reaction at constant temperature.

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev .....................(1)

And,  Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev ..............(2)

From the ideal gas equation,

PV = nRT

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

So, PB = [B] RT; PD = [D]RT
Similarly, PA = [A]RT; PC = [C]RT

Substituting the values of PA, PB, PC, and PD in equation (2), we get 

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev
Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Where, Δn = (c + d) - (a + b)

i.e;, Δn = sum of no. of moles of gaseous products-sum of no. of moles of gaseous reactants.

If Δn > 0, Kp > Kc

If Δn = 0, Kp = Kc 
And If Δn < 0, Kp < Kc

Illustration: The value of Kp for the Reaction

2H2O(g) + 2Cl2(g) ⇌4 HCl(g) + O2(g)

is 0.03 atm at 4270C, when the partial pressure is expressed in atm. What is the value of Kc.

Solution: Firstly, consider species that are present in gaseous phase only. So, we can write Kp.

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

· Kp and Kx: Kp = KxPTΔng 

5. Application of Law of Mass Action: Equilibrium Constant for Heterogeneous Equilibria

Reversible Reactions :
· Homogeneous
· Heterogeneous 

Homogeneous: Products and reactants are in same phase. Homogeneous gaseous equilibria are best classified into three categories i.e., in which no. of moles of product and reactant varies.
Case I: Homogeneous; Δng = 0

2A(g) ⇌B(g) + C(g)
Δng = 2 - 2 = 0

In this reaction, each and every species present is in same phase. That’s why this reaction is under the category of homogeneous equilibria. 

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev
Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev
Therefore, Kn = Kx = K= Kp.

PV = nRT (Ideal gas equation)
Here, V is constant. R is also constant.

P α n

Since no. of moles are constant i.e.  Initial = a
Final = a
Hence PT also remains constant and we will get the curve ment ion below.

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev
Case II: Homogeneous:Δng > 0

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

You will easily get these expressions in the same way we did in last.

Here, no. of moles increases at equilibrium i.e. from a to (a+x).
Hence PT also increases and becomes constant at equilibrium as shown below.

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Case III: Homogeneous and Δng < 0

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

This graph clearly resembles that first pressure is decreasing and then becomes constant at equilibrium.

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

1. Synthesis of Hydrogen Iodide:

Suppose ‘a’ moles of H2 and ‘b’ moles of I2 are heated at 444°C in a closed container of volume ‘V’ litre and at equilibrium, 2x moles of HI are formed

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev .................(i)

Substituting the equilibrium concentrations of H2, I2 and HI in equation (i), we get 

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev .................(iii)

Suppose the total pressure of the equilibrium mixture at 444°C is P, then 

PHI = mole fraction of HI × Total pressure

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Similarly, 

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Substituting these values in equation (iii) we get

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev ..............(iv)

So, one can see from equations (iii) and (iv), that 

Kp = Kc

This is so, because Δn = 0 for the synthesis of HI from H2 and I2.

2. Thermal Dissociation of Phosphorus Pentachloride

PCl5(g) dissociates thermally according to the reaction,

PCl5(g) ⇌ PCl3(g) + Cl2(g) 

Let us consider that 1 mole of PCl5 has been taken in a container of volume V litre and at equilibrium x moles of PCl5(g) dissociates. Thus 

PCl5(g) ⇌ PCl3(g) +  Cl2(g)

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

According to law of mass action, at constant temperature,

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev ..................(1)

And Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev ..................(2)

Substituting the values of equilibrium concentration, in equation (1), we have 

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev or Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Now, total number of moles at equilibrium = 1 – x + x + x = 1 + x

Mole fraction of PCl3 = Mole fraction Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

And mole fract ion of Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Suppose total pressure at equilibrium is P, then we have from equation (2),

Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev or  Introduction to Chemical Equilibrium - Chemical Equilibrium Chemistry Notes | EduRev

Similarly, we can apply law of mass action on any reaction at equilibrium.

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