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.
Those reactions which proceed in forward direction and reaches almost to completion are called Irreversible reactions.
For example:
Irreversible Reactions:
Ex.: Ppt. reaction, NaCl + AgNO_{3}→AgCl(s) + NaNO_{3}
case, if we remove AgCl (which is the product), reaction will move in forward direction].
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 Fe_{3}O_{4(s)} along with the evolution of hydrogen gas.
And when Fe_{3}O_{4} is reduced with hydrogen gas, it gives Fe(s) and H_{2}O
But, if the reaction is carried out in closed vessel, this reaction becomes reversible.
Reversible reactions:
If we remove C (which is product), the reaction will again start and moves in forward direction to attain equilibrium again.
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.
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”.
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.
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)
Kx and Kn are not considered equilibrium constants, because these can be varied without change in temp. for gases.
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 productssum of no. of moles of gaseous reactants.
Illustration: The value of K_{p} for the Reaction
is 0.03 atm at 427^{0}C, 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}.
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.
Suppose ‘a’ moles of H_{2} and ‘b’ moles o f I_{2} 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 H_{2} and I_{2}.
PCl_{5(g)} dissociates thermally according to the reaction,
Let us consider that 1 mole of PCl_{5} 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 PCl_{3} = Mole fraction _{ }
And mole fraction of PCl_{5}
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.
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).
The thermal dissociat ion of NH_{4}Cl_{(s)} takes place in a closed container according to the equation
Let us consider 1 mole of NH_{4}Cl_{(s) }is kept in a closed container of volume ‘V’ litre at temperature TK and if x mole of NH_{4}Cl dissociates at equilibrium then
Initial moles 1 0 0
Moles at equilibrium 1 x x x
Applying law of mass action, K_{c}
As NH_{4}Cl is a pure solid, so there is no appreciable change in its concentration. Thus,
K_{c }= [NH_{3}] [HCl]
And K_{p} =P_{NH}_{3 }* P_{HCl}
A_{g2}CO_{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 Ag_{2}CO_{3(s)} is heated in a closed container of volume V and x mo l of Ag_{2}CO_{3(s)} dissociates at equilibrium, then
Initial moles 1 0 0
Equilibrium moles 1  x x x
Now, as Ag_{2}CO_{3} and Ag_{2}O are solids, so their concentration can be assumed to be constants Thus
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.
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.
If equilibrium constant (say K_{C}) is veryvery 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 veryvery high.
Hence, by this we can infer that reaction has a very low approach in forward direction and reactions has just started.
If equilibrium constant (i.e. K_{C}) is veryvery 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 NO_{2} 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 N_{2}O_{4} = 1
Partial Pressure
_{Put α = 0.3}
K_{P} = 0.79
In terms of moles
In terms of α
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