Chemistry DYNAMIC EQUILIBRIUM

### Topic to be covered

star EQUILIBRIUM IN CHEMICAL PROCESSES – DYNAMIC EQUILIBRIUM.
star LAW OF CHEMICAL EQUILIBRIUM AND EQUILIBRIUM CONSTANT

### EQUILIBRIUM IN CHEMICAL PROCESSES – DYNAMIC EQUILIBRIUM

The chemical reactions can occur both in forward and backward directions. When the rates of the forward and reverse reactions become equal, the concentrations of the reactants and the products remain constant. This is the stage of chemical equilibrium( dynamic equilibrium).

For a better comprehension, let us consider a general case of a reversible reaction,

color{red}(A + B ⇌ C + D)

With passage of time, there is accumulation of the products C and D and depletion of the reactants A and B (Fig. 7.2). This leads to a decrease in the rate of forward reaction and an increase in he rate of the reverse reaction

Eventually, the two reactions occur at the same rate and the system reaches a state of equilibrium.
Similarly, the reaction can reach the state of equilibrium even if we start with only C and D; that is, no A and B being present initially, as the equilibrium can be reached from either direction.

The dynamic nature of chemical equilibrium can be demonstrated in the synthesis of ammonia by Haber’s process. In a series of experiments, Haber started with known amounts of dinitrogen and dihydrogen maintained at high temperature and pressure and at regular intervals determined the amount of ammonia present. He was successful in determining also the concentration of unreacted dihydrogen and dinitrogen. Fig. 7.4 shows that after a certain time the composition of the mixture remains the same even though some of the reactants are still present. This constancy in composition indicates that the reaction has reached equilibrium. In order to understand the dynamic nature of the reaction, synthesis of ammonia is carried out with exactly the same starting conditions (of partial pressure and temperature) but using color{red}(D_2) (deuterium) in place of color{red}(H_2). The reaction mixtures starting either with color{red}(H_2) or color{red}(D_2) reach equilibrium with the same composition, except that color{red}(D_2) and color{red}(ND_3) are present instead of color{red}(H_2) and color{red}(NH_3). After equilibrium is attained, these two mixtures (color{red}(H_2, N_2, NH_3) and color{red}(D_2, N_2, ND_3)) are mixed together and left for a while. Later, when this mixture is analysed, it is found that the concentration of ammonia is just the same as before. However, when this mixture is analysed by a mass spectrometer, it is found that ammonia and all deuterium containing forms of ammonia (color{red}(NH_3, NH_2D, NHD_2) and color{red}(ND_3)) and dihydrogen and its deutrated forms (color{red}(H_2, HD) and color{red}(D_2)) are present. Thus one can conclude that scrambling of color{red}(H) and color{red}(D) atoms in the molecules must result from a continuation of the forward and reverse reactions in the mixture. If the reaction had simply stopped when they reached equilibrium, then there would have been no mixing of isotopes in this way.

Use of isotope (deuterium) in the formation of ammonia clearly indicates that chemical reactions reach a state of dynamic equilibrium in which the rates of forward and reverse reactions are equal and there is no net change in composition.

Equilibrium can be attained from both sides, whether we start reaction by taking, color{red}(H_2(g)) and color{red}(N_2(g)) and get color{red}(NH_3(g)) or by taking
color{red}(NH_3(g)) and decomposing it into color{red}(N_2(g)) and color{red}(H_2(g)).

color{red}(N_2 (g) +3H_2 (g) ⇌ 2NH_3 (g))

color{red}(2NH_3 (g) ⇌ N_2 (g) +3H_2 (g))

Similarly let us consider the reaction, color{red}(H_2(g) + I_2(g) ⇌ 2HI(g)). If we start with equal initial concentration of color{red}(H_2) and color{red}(I_2), the reaction proceeds in the forward direction and the concentration of color{red}(H_2) and color{red}(I_2) decreases while that of color{red}(HI) increases, until all of these become constant at equilibrium (Fig. 7.5). We can also start with color{red}(HI) alone and make the reaction to proceed in the reverse direction; the concentration of color{red}(HI) will decrease and concentration of color{red}(H_2) and color{red}(I_2) will increase until they all become constant when equilibrium is reached (Fig.7.5). If total number of color{red}(H) and color{red}(I) atoms are same in a given volume, the same equilibrium mixture is obtained whether we start it from pure reactants or pure product.

### LAW OF CHEMICAL EQUILIBRIUM AND EQUILIBRIUM CONSTANT

Let us consider a general reversible reaction:

color{red}(A+B ⇌ C+ D)

where A and B are the reactants, C and D are the products in the balanced chemical equation. On the basis of experimental studies of many reversible reactions, the Norwegian chemists Cato Maximillian Guldberg and Peter Waage proposed in 1864 that the concentrations in an equilibrium mixture are related by the following equilibrium equation,

color{red}(K_c = ([ C] [D])/([A] [B])) ...........(7.1)

where color{red}(K_c) is the equilibrium constant and the expression on the right side is called the equilibrium constant expression.

The equilibrium equation is also known as the law of mass action because in the early days of chemistry, concentration was called “active mass”. In order to appreciate their work better, let us consider reaction between gaseous color{red}(H_2) and color{red}(I_2) carried out in a sealed vessel at 731K.

color{red}(underset(1 mol)(H_2 (g)) +underset(1 mol)(I_2 (g)) ⇌ underset(2 mol)(2HI (g)))

Six sets of experiments with varying initial conditions were performed, starting with only gaseous color{red}(H_2) and color{red}(I_2) in a sealed reaction vessel in first four experiments (1, 2, 3 and 4) and only color{red}(HI) in other two experiments (5 and 6). Experiment 1, 2, 3 and 4 were performed taking different concentrations of color{red}(H_2) and / or color{red}(I_2), and with time it was observed that intensity of the purple colour remained constant and equilibrium was attained. Similarly, for experiments 5 and 6, the equilibrium was attained from the opposite direction.

Data obtained from all six sets of experiments are given in Table 7.2.

It is evident from the experiments 1, 2, 3 and 4 that number of moles of dihydrogen reacted = number of moles of iodine reacted =
½ (number of moles of HI formed). Also, experiments 5 and 6 indicate that,

color{red}([H_2 (g)]_(eq) = [I_2 (g) ]_(eq))

Let us consider the simple expression

color{red}([HI (g)]_(eq) // [H_2 (g) ]_(eq) [ I_2 (g)]_(eq))

It can be seen from Table 7.3 that if we put the equilibrium concentrations of the reactants and products, the above expression is far from constant. However, if we consider the expression,

color{red}([HI (g)]_(eq)^2 // H_2 [H_2 (g)]_(eq) [I_2 (g) ]_(eq))

we find that this expression gives constant value (as shown in Table 7.3) in all the six cases. It can be seen that in this expression the power of the concentration for reactants and products are actually the stoichiometric coefficients in the equation for the chemical reaction. Thus, for the reaction color{red}(H_2(g) + I_2(g) ⇌ 2HI(g)), following equation 7.1, the equilibrium constant color{red}(K_c) is written as,

color{red}(K_c = [HI (g) ]_(eq)^2 // [H_2 (g)]_(eq) [I_2 (g) ]_(eq)) .............(7.2)

Generally the subscript ‘eq’ (used for equilibrium) is omitted from the concentration terms. It is taken for granted that the
concentrations in the expression for K_c are equilibrium values. We, therefore, write.

color{red}(K_c = [HI (g) ]^2 // [ H_2 (g) ] [I_2 (g) ]) ...........(7.3)

The subscript color{red}(‘c’) indicates that color{red}(K_c) is expressed in concentrations of color{red}(mol L^(–1)).

color{purple}(✓✓)color{purple} " DEFINITION ALERT"

At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the Equilibrium Law or Law of Chemical Equilibrium.

The equilibrium constant for a general reaction,

 color{red}(a A + b B ⇌ c C + d D)

is expressed as,

color{red}(K_c = [C]^c [D]^d // [A]^a [B]^b) ...........(7.4)

where [A], [B], [C] and [D] are the equilibrium concentrations of the reactants and products. Equilibrium constant for the reaction,

color{red}(4NH_3 (g) +5O_2 (g) ⇌ 4NO(g) +6H_2O (g))  is written as

color{red}(K_c = [NO]^4 [H_2O]^6 // [NH_3]^4 [O_2]^5)

Molar concentration of different species is indicated by enclosing these in square bracket and, as mentioned above, it is implied that these are equilibrium concentrations. While writing expression for equilibrium constant, symbol for phases (s, l, g) are generally ignored

Let us write equilibrium constant for the reaction, color{red}(H_2(g) + I_2(g) ⇌ 2HI(g)) .......................... (7.5)

as color{red}(K_c = [HI]^2 // [H_2] [I_2] = x) ....................(7.6)

The equilibrium constant for the reverse reaction, color{red}(2HI(g) ⇌ H_2(g) + I_2(g)), at the same temperature is,

color{red}(K_c = [H_2] [I_2] // [HI]^2 = 1/x = 1/K_c) ...................(7.7)

Thus color{red}(K_c = 1/K_c) .....................(7.8)

Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward
direction.

If we change the stoichiometric coefficients in a chemical equation by multiplying throughout by a factor then we must make
sure that the expression for equilibrium constant also reflects that change. For example, if the reaction (7.5) is written as,

color{red}(1/2 H_2 (g) +1/2 I_2 (g) ⇌ HI (g)) ............(7.9)

the equilibrium constant for the above reaction is given by

color{red}(K_c^('') = [HI] // [H_2]^(1/2) [I_2]^(1/2) = { [ HI]^2 // [ H_2] [I_2]}^(1/2))

 =color{red}( x^(1/2) = K_c^(1/2)) ...............(7.10)

On multiplying the equation (7.5) by n, we get

color{red}(nH_2 (g) + n I_2 (g) ⇌ 2 n HI (g))  ........(7.11)

Therefore, equilibrium constant for the reaction is equal to color{red}(K_c^n). These findings are summarised in Table 7.4. It should be noted
that because the equilibrium constants color{red}(K_c) and color{red}(K_c^(')) have different numerical values, it is important to specify the form of the balanced chemical equation when quoting the value of an equilibrium constant.
Q 3079580416

The following concentrations were obtained for the formation of NH_3 from N_2 and H_2 at equilibrium at 500K. [N_2] = 1.5 × 10^(–2)M. [H_2] = 3.0 ×10^(–2) M and [NH_3] = 1.2 ×10^(–2)M. Calculate equilibrium constant.

Solution:

The equilibrium constant for the reaction, N_2(g) + 3H_2(g) ⇌ 2NH_3(g) can be written as,

K_c = ([ NH_3 (g) ]^2)/( [ N_2(g) ] [H_2 (g)]^3)

 = { (1.2xx10^(-2))^2}/{(1.5xx10^(-2) ) (3.0xx10^(-2))^3}

 = 0.106 xx10^4 = 1.06xx 10^3
Q 3059680514

At equilibrium, the concentrations of N_2=3.0 × 10^(–3)M, O_2 = 4.2 × 10^(–3)M and NO= 2.8 × 10^(–3)M in a sealed vessel at 800K. What will be K_c for the reaction N_2(g) + O_2(g) ⇌ 2NO(g)

Solution:

For the reaction equilibrium constant, K_c can be written as,

K_c = ([ NO]^2)/([N_2 ] [O_2])

 = (2.8 xx 10^(-3) M)^2/{(3.0xx10^(-3)) (4.2 xx 10^(-3)) }

 = 0.662