Chemical Equilibrium, Law of Mass Action & Applications | Physical Chemistry PDF Download

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.

Irreversible and Reversible Reactions

Irreversible Reactions

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

For example:

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 move in forward direction].

Reversible Reactions

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

But when Fe_(s) is heated and water vapor is passed over it in an open vessel, it is converted to Fe₃O_4(s) along with the evolution of hydrogen gas.

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

But, if the reaction is carried out in closed vessel, this reaction becomes reversible.

Reversible reactions:

•  Never goes for completion
•  Irrespective of the fact whether the reaction has started by reactants or products.

If we remove C (which is product), the 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.

State of Equilibrium

Equilibrium is a state in which the rate of the forward reaction equals the rate of the backward reaction. Although some chemical reactions reach completion, some other reactions do not completely occur.

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

• 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 reactions, for example. 
• When PCl_5(g) is heated in a closed continue, its dissociation starts with the formation of PCl_3(g) and Cl_2(g) are formed due to dissociation of PCl_5(g). After a certain interval of time, the concentration of PCl_5(g), PCl_3(g), and Cl2(g) each become constant. 
• It does not mean that at this point of time, dissociation of PCl_5(g) and its formation from m PCl_3(g) and Cl_2(g) has been stopped. Actually, the rate of dissociation of PCl₅ and the rate of formation of PCl_5(g) becomes equal. 
• This state is called the state of chemical equilibrium. So, the state of chemical equilibrium is dynamic.

This can be shown graphically.

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”.

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 a constant temperature, 

Applying law of mass action, 

Rate of forward reaction, 

Where k_f = rate constant of forward reaction 

Similarly, 

Rate of reverse reaction

Where k_r = rate constant of reverse reaction.

At equilibrium, 

Rate of forward reaction = Rate of reverse reaction

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

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

Illustration:

The rate of disappearance of A is given as:

  Calculate the value of the equilibrium constant.

Solution: 

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

Equilibrium Constant (Kp) in terms of Partial Pressure: 

Consider the same general reaction taking place at a constant temperature,

From the law of mass action, 

From ideal gas equation 

PV = nRT

At constant temperature,

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

P ∝[A]

Similarly, P ∝ [B]

So, equation (1) can be rewritten as

All Equilibrium constants (Keq)-

• Kc 
• Kp
• Kx
• Kn

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

Relationship Between K_p and K_c

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

 And                                            

From the ideal gas equation,

PV = nRT

So, P_B = [B] RT; P_D = [D]RT

Similarly, P_A = [A]RT; P_C = [C]RT

Substituting the values o f P_A, P_B, P_C, and P_D in equation (2), we get 

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.

Illustration: The value of K_p for the Reaction

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 K_p.

Application of Law of Mass Action: 

Reversible Reactions:

• Homogeneous 
• Heterogeneous 

Equilibrium Constant for

Homogeneous Reversible Reactions: 

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

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

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 P_T also remains constant and we will get the curve mention below.

Case II: Homogeneous:Δng > 0

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 P_T also increases and becomes constant at equilibrium as shown below.

Case III: Homogeneous and Δng < 0

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

Synthesis of Hydrogen Iodide:

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

Initial concentration (mol L^–1)

Equilibrium concentration (mol L^–1)

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

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

 P_HI = mole fraction of HI × Total pressure

Similarly, 

Substituting these values in equation (iii) we get

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

K_p = K_c 

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

Thermal Dissociation of Phosphorus Pentachloride

PCl_5(g) dissociates thermally according to the reaction,

Let us consider that 1 mole of PCl₅ has been taken in a container of volume V litre and at equilibrium x moles of PCl_5(g) dissociates. Thus 

Initial concentration (mo l L^–1) 1                        0              0                 0

Equilibrium concentration (mol L^–1)                  

According to law of mass action, at a constant temperature,

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

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

Mole fraction of PCl₃ = Mole fraction  

And mole fraction of PCl₅ 

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

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

Equilibrium Constant for Heterogeneous Equilibria

The equilibrium which involves reactants and products in different physical states. The law of mass action can also be applied on heterogenous equilibria as it was applied for homogenous equilibria (involving reactants and products in same physical states).

Thermal Dissociation of Solid Ammonium Chloride:

The thermal dissociat ion of NH₄Cl_(s) takes place in a closed container according to the equation

Let us consider 1 mole of NH₄Cl_(s) is kept in a closed container of volume ‘V’ litre at temperature TK and if x mole of NH₄Cl dissociates at equilibrium then

Initial moles                           1               0            0 

Moles at equilibrium             1- x          x             x

Applying law of mass action, K_c 

As NH₄Cl is a pure solid, so there is no appreciable change in its concentration. Thus, 

K_c = [NH₃] [HCl]

And K_p =P_NH3 * P_HCl

Thermal Dissociation of Ag₂CO₃:

A_g2CO_3(s) dissociates thermally according to the equation.

Applying law of mass action, at constant temperature, we get,

Now, let us consider that 1 mole of Ag₂CO_3(s) is heated in a closed container of volume V and x mo l of Ag₂CO_3(s) dissociates at equilibrium, then

Initial moles                                    1             0          0 

Equilibrium moles                          1 - x        x           x

Now, as Ag₂CO₃ and Ag₂O are solids, so their concentration can be assumed to be constants Thus

Significance of magnitude of Equilibrium Constant: 

• A very large value of K_c or K_p signifies that the forward reactions has gone to completion or very nearly so.
•  A very small value o f K_c or K_p signifies that the forward reaction has not occur to any significant extent.
•  If the numerical value of K_c or K_p is neither very large or very small, a reaction is most likely to reach a state of equillibrium in which both reactants and products are present

Predictions of Direction of the Reaction:

QC is called Reaction Quotient: 

QC is the ratio of the concentration of products and reactants each raised to their schoitiometric  coefficient as in balanced chemical equation. 

Hence, three cases arises: 

Note: While dealing wit h K_P and α, take initial moles of reactant as 1 because ultimately they will be cancelled out.

The Extent of Reaction:

Magnitude of equilibrium constant tells us the extent of reaction. It resembles the extent by which the reactant has been reacted or product has been formed at equilibrium.

Case I: 

If equilibrium constant (say K_C) is very-very small.

(Concentration of product is very low and concentration of reactant is very high) It clearly explains that numerator is very small and denominator is very-very high.

Hence, by this we can infer that reaction has a very low approach in forward direction and reactions has just started.

Case II: 

If equilibrium constant (i.e. K_C) is very-very high.

(Concentration of product is very high and concentration of reactant is very low).

Hence, it explains that reaction has almost completed.

Illustration:

Initial moles of reactant is 1 mole and volume of the container is 1L. Calculate the moles of B(g) at Equilibrium Kc = 10^–50

Solution: Since the value of KC is very low. 

x = 10^–25

Illustration: If initially 1 mole of all the species present one taken in 10 L closed vessel, find equilibrium concentration of each species. 

Solution:                

At any time concentration  

Since Q_C> K_C, so reaction will move in backward direct ion to attain equilibrium.

Equilibrium concentration  

Equilibrium concentration of A(g) = 0.16 

Equilibrium concentration of B(g) = 0.04 

Equilibrium concentration of C(g) = 0.04

Illustration: Calculate K_P for the reaction

When NO₂ is 30% dissociated and total pressure at equilibrium is 2 bar.

Solution:

Given, α = 0.3

We are dealing with KP and α. Hence initial mole of N₂O₄ = 1

Partial Pressure      

_{Put α = 0.3}

K_P = 0.79

Dealing with Degree of Dissociation (α)

In terms of moles

In terms of α