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Dynamic nature of chemical equilibrium :

i. Consider a chemical reaction : A ⇌ B.

K_C = [B]/[A]

At equilibrium,

The ratio of concentration of the product to that of the concentration of the reactant is constant and this is equal to K_C.

ii. At this stage reaction takes place in both the directions with same speed although the reaction appears to have stopped. 

Thus, 

The chemical equilibrium is dynamic in nature. 

Dynamic means moving and at a microscopic level, the system is in motion.

iii. For example, in the reaction between H₂ and I₂ to form HI, the colour of the reaction mixture becomes constant because the concentrations of H₂, I and HI become constant at equilibrium.

H₂ + I₂ ⇌ 2HI

Thus, 

When equilibrium is reached,

The reaction appears to have stopped. 

However, 

This is not the case. 

The reaction is still going on in the forward and backward direction but the rate of forward reaction is equal to the rate of backward reaction. 

Hence, 

Chemical equilibrium is dynamic in nature and not static.

(D) 0.41 × 10^-47 

• Reversible reactions are the reactions which do not go to completion and occur in both the directions simultaneously.
• If such a reaction is allowed to take place for a long time, so that the concentrations of the reactants and products do not vary with time, then the reaction will attain equilibrium.
• Since, both the forward and backward reactions continue to take place in opposite directions in the same speed, the equilibrium achieved is dynamic in nature.

Thus,

If the reaction is not reversible then it cannot attain dynamic equilibrium.

a. Reversible reaction :

i. Reactions which do not go to completion and occur in both the directions simultaneously are called reversible reactions.

ii. Reversible reactions proceed in both directions. The direction from reactants to products is the forward reaction, whereas the opposite reaction from products to reactants is called the reverse or backward reaction.

iii. A reversible reaction is denoted by drawing in between the reactants and product a double arrow, one pointing in the forward direction and other in the reverse direction (⇌ or ⇄).

iv. At high temperature in an open container, the CO₂ gas formed will escape away. 

Therefore,

It is not possible to obtain back 

e.g.,

a. H_2(g) + I_2(g) ⇌ 2HI_(g) 

b. CH₃COOH_(aq) + H₂O_(I) ⇌ CH₃COO^-_(aq) + H₃O⁺_(aq)

b. Rate of reaction :

Rate of a chemical reaction :

i. The rate of a chemical reaction can be determined by measuring the extent to which the concentration of a reactant decreases in the given time interval, or extent to which the concentration of a product increases in the given time interval.

ii. Mathematically, the rate of reaction is expressed as :

Rate = \(-\frac{d[Reactant]}{dT}\) 

= \(\frac{d[Product]}{dT}\)

Where, d[reactant] and d[product] are the small decrease or increase in concentration during the small time interval dT.

The change of pressure does not affect the value of equilibrium constant.

i. Concentration : 

Addition of H₂ or N₂ both favours forward reaction. This increases the yield of NH₃.

ii. Temperature : 

The formation NH₃ is exothermic. 

Hence,

Low temperature should favour the formation of NH₃. 

However, 

At low temperatures, 

The rate of reaction is small. 

At high temperatures, 

The reaction occurs rapidly but decomposition of NH₃ occurs. 

Hence,

Optimum temperature of about 773 K is used.

iii. Pressure : 

The forward reaction is favoured with high pressure as it proceeds with decrease in number of moles. 

At high pressure,

The catalyst becomes inefficient. 

Therefore,

Optimum pressure needs to be used. 

The optimum pressure is about 250 atm.

Irreversible reaction :

1. Products are not converted back to reactants. 

2. Reaction stops completely and almost goes to completion. 

3. It can be carried out in an open or closed vessel. 

4. It takes place only in one direction. It is represented by →

5. e.g. C_(s) + O_2(g) → CO_2(g)

Reversible reaction :

1. Products arc converted back to reactants. 

2. Reaction appears to have stopped but does not undergo completion. 

3. It is generally carried out in a closed vessel. 

4. It takes place in both directions. It is represented by ⇌

5. e.g. N_2(g) + O_2(g) ⇌ 2NO_(g)

A system in which there is no exchange of matter with the surroundings is called a closed system.

(A) Catalyst lowers activation energy for the forward and reverse reactions by exactly the same amount.

(D) Addition of Fe₂ O₃

Given: i. K_c = 54.0 at 700 K

ii. k_r = 1.16 × 10^-3 at 700 K

To find: i. Whether k_f is larger or smaller than k_r

ii. Value of k_f

Formula : K_c = \(\frac{K_f}{K_r}\)

Calculation: 

i. As K_c = \(\frac{K_f}{K_r}\) = 54.0

K_f is greater than k_r by a factor of 54.0.

ii. K_f = K_c x K_r = 54.0 x (1.16 x 10^-3)

= 62.64 x 10^-3

i. K_f is greater than K_r,

ii. Value of K_r is 62.64 x 10^-3

(B) favours the decomposition of N₂O₄  

Correct option is (C) Fe + Mo 

Natural waterfall, spreading of smoke from burning incense stick and diffusion of fragrance of flowers are irreversible physical changes.

The rate equation is written by applying the law of mass action. 

i. The reactants are C and O₂

Rate ∝ [C] [O₂] 

∴ Rate = k [C] [O₂] 

ii. The reactant is KClO₃ and its 2 molecules appear in the balanced equation.

∴ Rate ∝ [KClO₃]²

∴ Rate = k [KClO₃]²

Option : a. dynamic

Option : b. temperature is increased

Characteristics of equilibrium constant:

• The value of equilibrium constant is independent of initial concentrations of either the reactants or products.
• Equilibrium constant is temperature dependent. Hence, K_c and K_p change with change in temperature.
• Equilibrium constant has a characteristic value for a particular reversible reaction represented by a balanced equation at a given temperature.
• Higher value of K_c or K_p means more concentration of products is formed and the equilibrium point is more towards right hand side and vice versa.

Option : a. adding Cl₂

Option : a. K_c = K_p/RT

i. For the given equilibrium mixture:

K_p = \(\frac{P_{PQ}}{P_p \times P_Q}\) = \(\frac7{4 \times 6}\) = 0.29

ii. Increasing the volume will shift the equilibrium position to the side with higher number of gaseous moles. In the given reaction, the equilibrium will shift to the left (toward reactant) resulting in an increase in the concentration of P and Q accompanied by a corresponding decrease in concentration of PQ.

Ideal gas equation is PV = nRT. 

Where, 

P = Pressure of the gas 

V = Volume of the gas 

n = Number of moles of the gas 

R = universal gas constant 

T = Absolute temperature of the gas

PV = nRT.  

Where,  P = Pressure of the gas  

V = Volume of the gas  

n = Number of moles of the gas  

R = universal gas constant  = 0.082L atm K^-1 mol^-1

T = Absolute temperature of the gas 

For any pure liquid at 1 atm pressure, the temperature at which its saturated vapour pressure equals to atmospheric pressure is called the normal boiling point of that liquid. e.g. The boiling point of ethyl alcohol is 78 °C i.e., the saturated vapour pressure of ethyl alcohol at 78 °C is 1 atm (1.013 bar).

A mixture of ice and water in a perfectly insulated thermos flask at 273 K is an example of solid-liquid equilibrium.

H₂O_(s) ⇌ H₂O_(l)

i. Solid – vapour equilibrium 

ii. Solid – vapour equilibrium Solid 

iii. Solid – vapour equilibrium

iv. Solid – liquid equilibrium

Camphor, ammonium chloride.

4NO + 6H₂O ⇄ 4NH₃ +5O₂

Statement: The law of mass action states that the rate of a chemical reaction at each instant is proportional to the product of concentrations of all the reactants. Explanation: A rate equation can be written for a reaction by applying the law of mass action as follows: Consider a reaction, A + B → C Here A and B are the reactants and C is the product. The concentrations of chemical species are expressed in mol L^-1 and denoted by putting the formula in square brackets. On applying the law of mass action to this reaction, a proportionality expression can be written as: Rate ∝ [A] [B]

This proportionality expression is transformed into an equation by introducing a proportionality constant, k, as follows: Rate = k [A] [B]

This equation is called the rate equation and the proportionality constant, k, is called the rate constant of the reaction.

Option : a. shift from left to right.

 (A) \(\frac{[N_2][H_2]^3}{[NH_3]^2}\)

 (D) \(\frac{[C]}{[A]^3[B]^2}\)

Law of mass action :

The law of mass action states that the rate of a chemical reaction at each instant is proportional to the product of concentrations of all the reactants.

Consider a hypothetical reversible reaction A + B ⇌ C + D. 

Two reactions, namely, forward and reverse reactions occur simultaneously in a reversible chemical reaction. The rate equations for the forward and reverse reactions are:

Rate_forward ∝ [A][B]

∴ Rate_forward = k_f [A] [B] …… (1)

∴ Rate_reverse ∝ [C] [D]

∴ Rate_reverse = k_r [C] [D] …. (2)

At equilibrium, the rates of forward and reverse reactions are equal. Thus,

Rate_forward  = Rate_reverse 

∴ k_f [A] [B] = k_r [C] [D]

∴ \(\frac{K_f}{K_r}\) = K_C = \(\frac{[C][D]}{[A][B]}\).............(3)

K_c is called the equilibrium constant.

For a reversible chemical reaction at equilibrium, 

aA + bB ⇌ cD + dD

Equilibrium constant (K_c) = \(\frac{[C]^c[D]^d}{[A]^a[B]^b}\)

When the concentrations of reactants and products in gaseous reactions are expressed in terms of their partial pressure, then the equilibrium constant is represented as K_p.

∴ For the reaction,

A_(g) + B_(g) ⇌ C_(g) + D_(g)

The equilibrium constant (K_c) can be expressed using partial pressure as : 

K_p = \(\frac{P_C \times P_D}{P_A \times P_B}\)

Where P_A, P_B, P_C and P_D are equilibrium partial pressures of A, B, C and D respectively.

As there is the same number of molecules of gas on both sides, change of pressure has no effect on the equilibrium.

i. H_2(g) + I_2(g) ⇌ 2HI_(g), K_C = 20 at 550 K

For the reaction, 

K_C = 20 at 550 K 

If the value of K_C is the range of 10^-3 to 10³,

The forward and reverse proceed to equal extents.

Hence,

The given reaction will form appreciable concentrations of both reactants and the product at equilibrium.

ii. H_2(g) + Cl_2(g) ⇌ 2HCl_(g), K_C= 10¹⁸ at 550 K

For the reaction,

K_C= 10¹⁸ at 550 K

If the value of K_C >>> 10³, forward reaction is favoured.

Hence,

The given reaction will proceed in the forward direction and will nearly go to completion.

i. In a homogeneous equilibrium, the reactants and products are in the same phase. 

e.g. Dissociation of HI:

2HI_(g) ⇌ H_2(g) + I_2(g)

ii. In a heterogeneous equilibrium, the reactants and products exist in different phases, 

e.g. Formation of NH₄Cl:

NH_3(g) + HCl_(g) ⇌ NH₄Cl_(s)