CBSE Class 11 Chemistry Notes For Chapter 7 Equilibrium

CBSE Class 11 Chemistry Notes For Chapter 7 Equilibrium Introduction

Under a given set of experimental conditions, a system is said to be at equilibrium if the macroscopic properties of the system, such as temperature, pressure, concentration, etc. do not show any change with time.

There are two types of equilibria:

Static equilibrium and Dynamic equilibrium. Equilibrium involving physical and chemical changes is dynamic.

A dynamic equilibrium is established when two or more opposing processes occur in a system at the same rate. For example, if the decomposition of hydrogen iodide [2HI(g)⇌ H₂(g) +I₂(g)] is carried out in a closed vessel,  it is found that the reaction is never complete.

At the onset of the reaction, the system contains only hydrogen iodide (reactant) molecules. With time, the concentration of molecules gradually decreases.

In contrast, the concentrations of H₂ and I₂ (product) molecules gradually increase till a stage is reached at which no further change in concentrations of either the reactants or the products takes place.

At this stage, the reaction appears to have stopped. This state of the system at which no further change occurs is called a state of equilibrium. This state of equilibrium is not static, but it is dynamic because the forward and backward reactions are still going on at the same rate.

Due to this dynamic nature of equilibrium, no change in concentration and other properties of the system occurs at the equilibrium state.

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Equilibrium involving chemical reaction (i.e., chemical equilibrium) is represented as:

Reactants; F=± Products The double half arrows indicate that the reactions in both directions are going on simultaneously.

The mixture consisting of reactants and products in the equilibrium state is called an equilibrium mixture. Dynamic equilibrium is also observed in case of physical changes, particularly during the transition of state example melting of solids, vaporization of liquids, etc.

CBSE Class 11 Chemistry Notes For  Physical Equilibrium

Equilibrium involving physical processes is called physical equilibrium. Thus, the equilibria attained during the dissolution of a salt, the evaporation of a liquid, etc., are examples of physical equilibria. Different types of physical equilibria are briefly discussed in the following section.

Solid-liquid Equilibrium

At the melting point (or freezing point) of a pure substance, both its solid and liquid phases co-exist and a dynamic equilibrium develops between the two phases:

Solid-liquid When the system with the above equilibrium mixture is heated, the temperature of the system remains constant until the whole solid transforms into liquid. Similarly, if heat is withdrawn from this system, the temperature of the system remains constant until the whole liquid transforms into a solid.

Melting point Or freezing point

At Normal atmospheric pressure, the temperature at which the solid and the liquid states of a pure substance remain in equilibrium is called the normal melting point (or normal freezing point) of the substance.

When the solid and liquid phases of a pure substance are kept in contact with each other at its melting point in a closed insulated container, no exchange of heat takes place between the system and its surroundings.

However, a state of dynamic equilibrium is established between the solid and the liquid phases inside the container. It is also observed that the masses of solid and liquid phases do not change with time and the temperature of the system remains constant.

Example: Let us take some ice cubes together with some water inside a thermos flask at 0°C and 1 atm pressure and leave the mixture undisturbed. After some time it will be seen that the masses of ice and water are not changing with time and also the temperature remains unchanged. This represents an equilibrium between ice and water

Equilibrium: H₂O (s)⇌H₂O(l)

• Although we observed apparent change inside the thermos flask, a careful examination shows that some activity is still going on between the two phases of water.
• Some molecules of ice convert into water, while at the same time, the same number of molecules of water convert into ice.
• However, as the masses of ice and water remain unchanged, it can be concluded that the two opposite processes (i.e., melting of ice and freezing of water) occur at the same rate.

The rate of melting of ice = The rate of freezing of water

Thus, the equilibrium that is established in the solid-liquid system is dynamic.

Liquid-vapour equilibrium

• The equilibrium between a liquid and its vapor can be better understood if we consider the vaporization of water in a closed vessel. Let us take a closed vessel connected to a manometer and a vacuum pump as shown in
• The closed vessel is first evacuated. The levels of mercury are the same in both the limbs of the manometer.
• Then some pure water is introduced into the vessel and the whole apparatus is kept at room temperature (or any desired temperature by placing it in a thermostat).
• After some time it is seen that the level of mercury in the left limb begins to fall and the right begins to rise and eventually the levels of mercury in both limbs become fixed at two different levels. Under this condition, the system is said to have attained equilibrium

Equilibrium: H₂O(Z) ⇌H₂O(g)

Molecular interpretation:

At the initial stage of the experiment, as more and more water changes into vapor (by evaporation), the pressure inside the vessel gradually increases. This is indicated by the fall in mercury level in the left limb of the manometer.

• The molecules of water vapor, so produced, collide among themselves, with the walls of the vessel and also with the surface of the water.
• Water vapor molecules with lower kinetic energy get converted into liquid states when they come in contact with the surface of water. This is called condensation.
• At the beginning of the experiment, the rate of evaporation of water is greater than the rate of condensation of its vapor. However, with time, the rate of condensation increases, and that of evaporation decreases.
• After some time, the rates of evaporation and condensation become equal. It is said that a dynamic equilibrium is established between water and its vapor

At equilibrium: The rate of evaporation = The rate of condensation

• The difference in the levels of mercury in the two limbs gives a measure of the equilibrium vapor pressure or saturated vapor pressure of water at the experimental temperature.
• At a fixed temperature, if a liquid remains in equilibrium with its vapor, then the pressure exerted by the vapor is called the vapor pressure of the liquid at that temperature.
• An equilibrium between a liquid and its vapor is established only in a closed vessel. If the liquid is placed in an open vessel, then its vapor diffuses into the air. Consequently, no equilibrium is established between a liquid and its vapor in an open container.

Solid Vapour equilibrium

In general, solid substances have verylow vapor pressure compared with liquid substances at the same temperature.

• However, some substances, e.g., iodine, camphor, solid CO₂, naphthalene, etc. have high vapor pressure even at ordinary temperatures.
• Such substances can convert directly from the solid to the vapor state without passing through the liquid state.
• In this process of transformation of a solid directly to the When sublimation of volatile solids done in a closed vessel, equilibrium is established between the solid and its vapor.
• Example: If we take some solid iodine [I₂(s)] in a closed vessel and heat it below its melting point (113.6°C), it is found that the vessel gets filled with violet vapor of iodine.
• Initially the intensity of color increases and eventually it becomes constant. Under this condition, the rate of sublimation of solid iodine is equal to the rate of condensation of iodine vapor. This results in a state of dynamic equilibrium as below,

Equilibrium: I₂(s)⇌ I₂(g)

Other substances showing this kind of equilibrium are:

• NH₄Cl(s)⇌NH₄Cl(g)
• Camphor (s)⇌ Camphor (g)

Equilibrium involving dissolution of solid in liquid

• Suppose, at a fixed temperature, an excess amount of a solid substance, say sugar (solute), is added to a definite volume of a suitable solvent (say water) taken in a beaker and then the mixture is stirred well with a glass rod.
• The particles (i.e., molecules in case of non-electrolytes and ions in case of electrolytes) of solute gradually pass into the solvent, thereby increasing the concentration of the solute in the solution. This process is called dissolution of solute.
• Then a stage comes when no more solute dissolves in the solvent. Instead, the solute settles down at the bottom of the beaker i.e., a saturated solution is obtained.
• During the process of dissolution of solute, the reverse process also occurs simultaneously, i.e., the solute particles from the solution get deposited on the surface of the undissolved solute (a process called crystallization).
• Initially, the rate of dissolution of the solid solute is higher than the rate of crystallization of the dissolved solute.
• However, with time, as the solution becomes more and more concentrated, the rate of dissolution decreases, and that of crystallization increases. Finally, the rate of dissolution becomes equal to that of crystallization.
• Under this condition, a state of equilibrium is established between the dissolved solute particles and the undissolved solid solute. The equilibrium can be represented as,

Solute (solid)⇌Solute (in solution)

This state of equilibrium is said to be dynamic because the process of dissolution and crystallization continues as long as the temperature and other external conditions remain unchanged.

The solution obtained at equilibrium is called a saturated solution. The concentration of the saturated solution depends on the temperature.

Equilibrium involving the dissolution of gas in liquid

1. The solubility of a gas in a given liquid depends on the experimental temperature and pressure and also on the nature of the liquid and the gas under consideration.
2. When a gas (say CO₂) comes in contact with a liquid, the molecules of the gas begin to collide with the surface of the liquid. Consequently, some of the gas molecules get attracted by the molecules of the liquid and ultimately pass into the liquid phase.
3. % At a fixed temperature and pressure, if a gas is passed continuously through a fixed amount of a liquid kept in a closed vessel, gas molecules get dissolved in the liquid, and eventually, a saturated solution of the gas in the liquid is obtained.
4. In this solution, a dynamic equilibrium is established between the dissolved gas and the gas over the liquid surface.
5. Under this condition, the rate of dissolution of the gas molecules in the liquid is equal to the rate at which the dissolved gas molecules escape from the solution.
6. Thus, it is a dynamic equilibrium; Liquid + Gas Dissolved gas.

Taking CO₂ as the gaseous substance it can be represented as:

At a fixed temperature, the amount of gas dissolved in a liquid depends upon the pressure of the gas over the liquid. The concentration of the dissolved gas increases with the increase in the pressure of the gas. Henry’s law demonstrates how the solubility of a gas in a liquid varies with pressure at a given temperature.

Henry’s law:

The mass (or mole fraction) of a gas dissolved in a given mass of a solvent at a given temperature is directly proportional to the pressure of the gas over the solvent.

In a fixed amount of a liquid, if a gas with a pressure of p dissolves by an amount of w, then according to Henry’s law, wp or, w = kp [k is the proportionality constant]

The reason for fizzing out of CO₂ gas when a soda water uc is exposed:

In a sealed soda water bottle, CO₂ remains dissolved in liquid under high pressure, and an equilibrium exists between the dissolved CO₂ and CO₂ gas present over the liquid.

• As soon as the bottle is opened, the pressure of CO₂ gas over the liquid decreases and becomes equal to the atmospheric pressure.
• Since the solubility of a gas in a liquid is proportional to the pressure of the gas, the solubility decreases considerably because of the lowering of pressure.
• As a result, a large amount of dissolved CO₂ escapes from the solution until a new equilibrium is established.

This phenomenon of escaping CO₂ gas is associated with a fizzing sound. This is also the reason why a soda water bottle turns flat when left open in the air for some time.

CBSE Class 11 Chemistry Notes For Irrversible And Reversible Reactions

Irreversible reactions

A reaction in which the products formed do not react together to revert to the reactants despite the changes in reaction conditions is called an irreversible reaction.

Examples: When potassium chlorate (KClO₃) is heated in an open vessel, potassium chloride (KCl) and oxygen (O₂) are produced.

However, KCl and O₂ do not react with each other to regenerate KClO₃. So, the thermal decomposition of KClO₂ is an example of an irreversible reaction.

Most of the ionic reactions are irreversible. For example, when an aqueous solution of KCl is treated with an aqueous AgNO₃ solution, a curdy white precipitate of AgCl and KNO₃ is produced, but the precipitated AgCl and KNO₃ do not react back to AgNO₃ and KCl.

⇒ \(\mathrm{AgNO}_3(a q)+\mathrm{KCl}(a q) \rightarrow \mathrm{AgCl}(s) \downarrow+\mathrm{KNO}_3(a q)\)

Characteristics of an irreversible rea

ction:

1. The products in an irreversible reaction do not show any tendency to react together. So, the reaction in the opposite direction can never happen. For this reason, an irreversible reaction attains completion in course of time.
2. Since an irreversible reaction undergoes completion, the reactants participating in the reaction in equivalent amounts are completely exhausted.
3. Irreversible reactions are accompanied by a decrease in Gibbs free energy (i.e., ΔG<0).

Reversible reactions

A Reaction in which the products formed react together to regenerate the reactants, and an equilibrium is established between the reactants and products under the condition of the reaction is called a reversible reaction.

Example: The thermal decomposition of NH₄Cl vapor in a closed vessel is a reversible reaction.

⇒ \(\mathrm{NH}_4 \mathrm{Cl}(\text { vapour }) \rightleftharpoons \mathrm{NH}_3(g)+\mathrm{HCl}(g)\)

Explanation:

When NH₄Cl(vapor) is heated in a closed vessel at 350°C, it undergoes thermal decomposition, producing NH₃ and HCl gases. However, even after a long time, it is observed that the reaction mixture contains not only NH₃ and HCl but also NH₄Cl vapor.

This proves that the decomposition of NH₄Cl vapor in a closed container never gets completed. In another closed vessel, if an equimolar mixture of NH₃ and HCl gases is heated at 350°C for a long time, the vessel is found to contain NH₄Cl vapor along with NH₃ and HCl gases.

This means that the reaction between NH₃ and HCl in a closed vessel never gets completed. Thus, it can be concluded that on heating NH₄Cl vapor, it decomposes to produce NH₃ and HCl gases which again react partially to form NH₄Cl vapor.

Hence, the thermal decomposition of NH₄Cl vapor is a reversible reaction. The following reactions show reversibility when carried out in a closed vessel.

⇒ \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g) ; \mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g)\)

⇒ \(\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g) ; \mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)\)

Characteristics of reversible reaction:

In a reversible reaction, both the forward and the backward reactions occur simultaneously. In the forward reaction, the reactants react together to yield the products, while the products react together to produce the reactants in the backward reaction.

For example, when an equimolecular mixture of H₂ gas and I₂ vapor is heated in a closed container, the following reaction takes place—

⇒ \(\mathrm{H}_2(g)+\mathrm{I}_2(g) \rightleftharpoons 2 \mathrm{HI}(g) .\)

Here, the forward reaction is: H₂(g) + I₂(g) →2HI(g) and the backward reaction is: 2HI(g)→H₂(g) + I₂(g)

Since a reversible reaction is not complete, the reactants are not completely consumed in such reactions. Instead, a mixture containing both the reactants and the products is obtained.

• Such reactions achieve an equilibrium state when the rate of the forward reaction becomes equal to that of the backwaed reaction.
• At a given temperature and pressure, when a reversible reaction reaches equilibrum, the gibbs free energy change becomes zero i.e., ΔGp, T=0
• Reversibility and irreversibility of chemical reactions when carried out in open and closed containers
• Many chemical reactions that are found to be irreversible when carried out in open containers become reversible when they are carried outin closed containers.
• Different results are obtained when solid calcium carbonate (CaCO₃) is heated separately in a closed container and in an open container.

Explanation: On strong heating solid CaCO₂ decomposes to solid CaO and CO₂ gas. If CaCO₃ is decomposed in an open container, CO₂ gas escapes from the container into the air, and only solid CaO remains as residue. As the reactant (solid CaCO₃) in this reaction gets converted into products completely, the reaction is considered as an irreversible reaction.

H₂(g) +I₂(g) ⇌ 2HI(g).

If the same quantity of CaCO₃ is decomposed in a closed container, some quantity of CaCO₃ is still found to remain undecomposed. This is because CO₂ produced in the reaction cannot escape from the container.

As a result, a portion of CO₂ gas reacts with an equivalent amount of CaO to form CaCO₃ again. Hence, in the closed vessel, the reaction occurs reversibly. Consequently, a mixture of CaCO₃, CaO, and CO₂ is found to be present in the reaction vessel.

⇒ \(\mathrm{CaCO}_3(\mathrm{~s}) \stackrel{\Delta}{\rightleftharpoons} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

When steam is passed over the red-hot iron, ferrosoferric oxide (Fe₃O₄) and H₂ gas are produced.

Explanation: If the reaction is carried out in an open vessel, the H₂ gas produced diffuses into the air. So, Fe₃O₄ resulting from the reaction cannot have hydrogen gas to react with.

Consequently, the reverse reaction cannot take place. For this reason, at the end of the reaction, only Fe₃O₄ is left behind as residue in the reaction vessel.

⇒ \(3 \mathrm{Fe}(s)+4 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{Fe}_3 \mathrm{O}_4(s)+4 \mathrm{H}_2(\mathrm{~g}) \uparrow\)

However, if the reaction is carried out in a closed vessel, the H₂ gas produced cannot escape from the container. As a result, a certain amounts of H₂ gas and Fe₃O₄ together and regenerate Fe and H₂O.

Hence, in a closed vessel, the reaction occurs both in the forward and reserve. directions, giving a mixture of Fe(s), H₂O(g), Fe₃O₄(s) and H₂(g).

⇒ \(3 \mathrm{Fe}(s)+4 \mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_3 \mathrm{O}_4(s)+4 \mathrm{H}_2(g)\)

CBSE Class 11 Chemistry Notes For  Chemical Equilibrium

Let us consider a hypothetical reversible reaction A +B⇌C+D, Which is started with 1 mol of A and lmol of B in a closed container at a given temperature.

1. At the beginning, the reaction system does not contain C and D (products). It contains only A and B (reactants). So, the reaction occurs only in the forward direction (A + B→C+ D).

2. At the outset of the reaction, since the concentrations of the reactants are maximum, the rate of the forward reaction is also maximum. This is because the rate of a reaction is directly proportional to the concentrations of the reactants.

3. As there are no C and D molecules at the start, the backward reaction (C+D→A + B) does not occur. However, the backward reaction starts occurring with the formation of A and B in the forward reaction. As C and D accumulate is the reaction system, they begin to react together to form A and B.

4. With time, the concentrations of C and D increase, while the concentrations of A and B decrease. As a result, the rate of the backward reaction increases, while that of the forward reaction decreases.

5. Eventually, a moment comes when the rate of the forward reaction becomes equal to the rate ofthe backward reaction.

6. When the rate of the forward reaction is equal to that of the reverse reaction, the reaction is said to have reached the state of equilibrium. However, at equilibrium, the reaction does not stop; instead, both the forward and backward reactions occur simultaneously at the same rate

7. The mixture of reactants and products at the equilibrium of a reaction is called the equilibrium mixture. The concentrations of the reactants and products in the equilibrium mixture are called their equilibrium concentrations. If the conditions {i.e., temperature, pressure, etc.) of the reaction remain undisturbed the relative concentrations of the reactants and products in the equilibrium mixture do not change with time.

Chemical equilibrium

The state of a reversible chemical reaction at a given temperature and pressure when the rates of the forward and reverse reactions become the same, and the concentrations of the reactants and products remain constant with time then the particular state is called the state of chemical equilibrium.

Chemical equilibrium is a dynamic

After the attainment of equilibrium of a reversible chemical reaction, if the reaction system is left undisturbed for an indefinite period at constant temperature and pressure,

• Then the relative amounts of the reactants and the products are found to remain unaltered.
• This observation leads to the impression that a reaction stops completely at equilibrium. However, it has been proved experimentally that the reaction does not cease rather both the forward and the backward reactions continue at the same rate. This is the reason why chemical equilibrium is designated as a dynamic equilibrium.
• Experimental proof of the dynamic nature of chemical equilibrium:
• When some quantity of pure CaCO₃ is heated strongly above 827°Cin a closed vessel (A), it decomposes into CaO and CO_2, and an equilibrium is established

⇒ \(\mathrm{CaCO}_3(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\)

At equilibrium, the temperature and pressure of the reaction vessel remain unaltered with time.

• Now this vessel is connected with another closed vessel 14 containing CO₂ at the equilibrium pressure and temperature in such a way that there will be no effect on the equilibrium of the reaction occurring in vessel (A).
• After some time, a small quantity of solid is collected from the vessel (A) and analyzed. The analytical data indicates the presence of 14C in CaCO₃.
• This is possible only if some amount of 14CO₂ and CaO combine to form Ca14CO₃ at equilibrium. At the same time, some quantity of CaCO₃ decomposes to produce CaO and CO₂.
• So, the pressure on the reaction vessel remains constant. Thus, this experiment proves that even after the attainment of equilibrium, the reactions do not cease, both the forward and the reverse reactions proceed simultaneously at the same rate.

Characteristics of chemical equilibrium

Permanency of chemical equilibrium:

As long as the conditions under which a reaction attains equilibrium remain unaltered, no further change in equilibrium takes place, that is to say, the composition of the equilibrium mixture and other properties of it remains the same with time.

Dynamic nature of equilibrium:

Even after the attainment of equilibrium, a chemical reaction does not cease; both the forward and the reverse reactions continue at equal rates.

Incompleteness of the reaction at equilibrium:

At the equilibrium of a reaction, both the forward and reverse reactions take place simultaneously at the same rate. If any one of these reactions goes to completion, then the term equilibrium becomes irrelevant. Hence, for the equilibrium to exist, the reactions of both directions will have to be incomplete.

Approachability of equilibrium from either direction:

Under a given set of conditions, a reversible reaction attains the same equilibrium state irrespective of whether the reaction is started with its reactants or products.

Example: H₂ gas and I₂ vapor are allowed to react with each other in a closed vessel at 445°C. Eventually, the following equilibrium is established,

⇒ \(\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\)

The equilibrium mixture is found to contain H₁, H_2, and I₂ with mole percents of 80%, 10%, and 10%, respectively. In a separate container with the same volume, if 2 moles of H₁ gas are heated at 445°C, then H₁ gas decomposes to H₂ and I₂ gases, and eventually, the following equilibrium is established, 2HI(g) H₂(g) + I₂(g).

Here also, the equilibrium consists of H₁, H_2, and I₂ with mole percents of 80%, 10%, and 10%, respectively. Thus, the same equilibrium mixture is obtained, no matter whether we start the reaction with HI(g) or H₂(g) and l₂(g).

A catalyst cannot alter the state of equilibrium:

A catalyst is a substance that enhances the rate of a reaction without being used up in the reaction. A catalyst does not affect the position of equilibrium in a reaction. Its only function is to reduce the time that a reaction takes to reach an equilibrium state.

In the presence of a catalyst, the forward and reverse reactions of a reversible reaction are speeded up to the same extent. Under a given set of conditions, if a reaction is carried out in the presence or absence of a catalyst, then the same equilibrium mixture is obtained. That is, in both cases, the concentrations of the reactants and products are found to be the same.

Homogeneous and heterogeneous equilibria

Homogeneous equilibrium:

An equilibrium in which all the substances, Z.e., reactants, and products, are in the same phase is known as homogeneous equilibrium.

Examples 1. N₂(g) + 3H₂(g)⇌ 2NH₃(g)

2SO₂(g) + O₂(g) ⇌2SO₃(g)

CH₃COOH(Z) + C₂H₅OH(l) ⇌CH₃COOC₂H₅(Z) + H₂0(l)

Heterogeneous equilibrium:

An equilibrium in which the reactants and products are in different phases is known as heterogeneous equilibrium.

Examples CaCO₃(s) ⇌CaO(s) + CO₂(g)

2HgO(s)⇌ 2Hg(l) + O₂(g)

CBSE Class 11 Chemistry Notes For  The Law Of Mass Action

In 1864, C.W. Guldberg and P. Waage formulated a law regarding the dependence of the reaction rate on the concentration of the reactant. This law is known as the law of mass action.

At a constant temperature, the rate of a chemical reaction at any instant during the reaction is directly proportional to the active mass of each of the reactants at that instant.

So, the rate of a reaction increases with the increase in active masses of the reactants, while it decreases with the decrease in active masses of the reactants.

Active mass:

The active mass of a substance is generally considered as the same as its molar concentration. The active mass is expressed in different ways.

• In case of a dissolved substance is a solution, the active mass of the substance is taken to be the same as its molar concentration.
• If a VL solution contains n mol of a substance, then the active mass (or molar concentration) of the substance is n/v .
• In the case of a component gas in a gas mixture, the active mass of the component can be expressed either in terms of its molar concentration or partial pressure in the mixture.
• This is because the partial pressure of a component gas in a gas mixture is directly proportional to its molar concentration.

For a pure solid or liquid, the active mass is always taken as unity (1).

The molar concentration of a pure solid or liquid is directly proportional to its density:

⇒ \(\frac{\text { number of moles of the substance }}{\text { volume of the substance (in } \mathrm{L})}\)

= \(\frac{\text { mass of the substance }}{\text { molar mass of the substance }} \times \frac{1}{\text { volume of the substance (in L) }}\)

= \(\frac{\text { mass of the substance }}{\text { volume of the substance (in } \mathrm{L})} \times \frac{1}{\text { molar mass of the substance }}\)

= \(\frac{\text { density of the substance }}{\text { molar mass of the substance }}\)

As the molar mass of a pure substance is a fixed quantity, the molar concentration of a pure solid or liquid is directly proportional to its density. The density of a pure solid or liquid is constant at a given temperature, so its molar concentration.

Mathematical expression of the law of mass action: Let us consider the following simple chemical reaction in which one mole of A reacts with one mole of B, forming one:

Mole of C: A + B→C According to the law of mass action, at a particular moment during the reaction, the rate of the reaction, or, r = k(A) (B) Where (A) and (B) are the active masses or molar concentration + of A and B, respectively at that moment, and k is proportionality constant, known as the rate constant of the reaction. Equation (1) represents the rate equation of the said reaction. Here is a table is which some reactions and their rate equations are given.

General statement of the law of mass action:

At constant temperature, the rate of a chemical reaction at any instant is directly proportional to the product of molar concentrations (active masses) of the reactants at that instant, each concentration (active mass) term being raised to a power which appears as a stoichiometric coefficient of the species in the balanced chemical equation of the reaction.

Mathematical Form Of The Law Of Mass Action For A Reversible Reaction

Suppose, a reaction is started with ‘ a’ mol of A and ‘b 1 mol of B, and the reaction of A with B leads to the formation of C and D.

Let the reaction occur according to the following equation and form an equilibrium:

aA + bB cC + dD According to the law of mass action, at equilibrium, the rate of forward reaction

(r_f)=k_f [A]^a × [B]^b

Or rf = Kf

And that of backward reaction

(r_b)∝[D]^d × [E]e

Or,  r_b= k_b [D]^d× [E]^e

Where k and kb are the rate constants of the forward and the backward reactions respectively. (A), (B), (D), and (E) are the respective molar concentrations or active masses (mol.L^-1) of A, B, D, and E at equilibrium.

At equilibrium, the rate of the forward reaction (ry) = the rate of the backward reaction (r_b).

⇒ \(\text { So, } k_f[A]^a \times[B]^b=k_b[D]^d \times[E]^e\)

⇒ \(\text { or, } \frac{k_f}{k_b}=\frac{[D]^d \times[E]^e}{[A]^a \times[B]^b} \text { or, } \boldsymbol{K}=\frac{[\boldsymbol{D}]^d \times[E]^e}{[A]^a \times[B]^b}\)

The ratio of the two rate constants (fcy and kb) is a constant quantity at a given temperature.

So, K is a constant. The constant ‘K’ is called the equilibrium constant of the said reversible reaction. Equation (1) expresses the mathematical form of the law of mass action of the given reversible reaction.

In the expression of the equilibrium constant, the reaction. concentration terms of the reactants and the products represent their respective molar concentrations at equilibrium. They do not denote their initial concentrations

At constant temperature, the equilibrium constant (K) of a chemical reaction has a definite value. The value of K changes with temperature. This is because the changes of values of kJ- and kJ with the temperature change do not occur to the same extent.

Equilibrium constant

The equilibrium constant of a reaction is the ratio of the product of the active masses of products at equilibrium to the product of the active masses of reactants at equilibrium, with each active mass term raised to a power equal to its stoichiometric coefficient in the balanced equation of the reaction.

The equilibrium constant is also represented as K_c, K_p or K_x depending on whether the active mass is expressed in terms of molar concentrations partial pressure, or mole fraction.

Different kinds of equilibrium constants:

The law of chemical equilibrium:

This law states that when a reversible chemical reaction reaches equilibrium at a particular temperature, the ratio of the product of active masses of the products to that of the reactants, with each active mass term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation is constant.

The mathematical expression for the law of chemical equilibrium can be obtained by applying the law of mass action to a reversible reaction at equilibrium. For a general reversible reaction, aA + bB dD + cE; the mathematical expression can be written as, (constant at a particular temperature) [A]n[B];’

⇒ \(\frac{[D]^d[E]^e}{[\mathrm{~A}]^a[B]^b}=K\) = K (constant at a particular temperature

Expression of the equilibrium constant in case of a heterogeneous equilibrium

At a given temperature, the active mass or molar concentration of a pure solid or liquid is always taken as unity (1). For this reason, in the case of a heterogeneous equilibrium, the active mass or molar concentration term of a pure solid or liquid does not appear in the expression of the equilibrium constant.

Examples: The thermal decomposition of solid CaCO₃ in a closed container leads to the following equilibrium:

CaCO₃(s) ⇌ CaO(s) + CO₂(g)

⇒\(K_c=\frac{[\mathrm{CaO}(\mathrm{s})]\left[\mathrm{CO}_2(\mathrm{~g})\right]}{\left[\mathrm{CaCO}_3(\mathrm{~s})\right]}=\left[\mathrm{CO}_2(\mathrm{~g})\right]\)

[CaCO₃(s)] = 1, [CaO(s)] = 1 ] and Kp = PCO₂(g)

As the value of Kc or Kp is constant of a given temperature, the molar concentration or the partial pressure of CO₂(g) at the equilibrium formed on the decomposition of solid CaC03 at a given temperature is always constant.

The following equilibrium is established during the vaporization of water in a closed vessel:

H₂O(l) ⇌ H₂O(g)

Here, the equilibrium constant

⇒  \(K_c=\frac{\left[\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right]}{\left[\mathrm{H}_2 \mathrm{O}(l)\right]}=\left[\mathrm{H}_2 \mathrm{O}(\mathrm{g})\right]\)

Therefore, the molar concentration or the partial pressure of water vapor remaining in equilibrium with pure water at a particular temperature is always constant.

Expression of equilibrium constants of some chemical reactions:

Relation Between Different Equilibrium Constants

Relation between K_p and K_c

Let the following reversible gaseous reaction is at equiUbriumin a closed container at a certain temperature

aA(g) + bB(g);=± dD(g) + eE(g) If the molar concentrations of A(g), B(g), D(g) and E(g) at equilibrium be [A], [B], [D] and [£] respectively, and the partial pressures of A(g), B(g), D(g) and E(g) at equilibrium be pA, PiB pD and pp respectively, then

⇒ \(K_c=\frac{[D]^d \times[E]^G}{[A]^a \times[B]^b}\) ……………………(1)

⇒ \(K_p=\frac{p_D^d \times p_E^e}{p_A^a \times p_B^b}\) ……………………(2)

If the reaction mixture behaves as an ideal gas, then the ideal gas equation can be applied to each of the species present in the mixture.

If the pressure, temperature, and volume of n mole of an ideal gas are P, T, and V respectively,

Then \(P V=n R T \quad \text { or, } P=\left(\frac{n}{V}\right) R T=C R T\)

C= Molar concentration

Now by applying this equation to each species of the reaction mixture, we get

⇒ \(p_A=[A] R T, p_B=[B] R T, p_D=[D] R T \text { and } p_E=[E] R T \text {. }\)

Putting the values of pA, pB, PD, and pE into equation (2), we have

⇒ \(K_p=\frac{\{[D] R T\}^d \times\{[E] R T\}^e}{\{[A] R T\}^a \times\{[B] R T\}^b}\)

=\(\frac{[\mathrm{D}]^d \times[E]^e}{[A]^a \times[B]^b}(R T)^{(d+e)-(a+b)}\)

Or, \(K_p=K_c(R T)^{\Delta n}\)

Where, Δn = (d+ e)-(a + b) = the total number of moles of gaseous products – the total number of moles of gaseous reactants.

⇒ \(K_p=K_c(R T)^{\Delta n}\) ………………….(3)

Inequation (3), the value of Δn may be +ve, -ve, or zero. When the number of moles of the gaseous products is greater than, less than, or equal to the number of moles of the gaseous reactants, then the values of Δn become positive, negative, or zero, respectively.

If An is positive, K_p is greater than Kc. If Δn is negative, then the value of K_p is smaller than Kc. If Δn = 0, then K_p and K_c have the same value.

Relation between K_P and K_c in case of some chemical and physical changes:

Relation between K_P and K_x

Let, at a constant temperature, the following reversible gaseous reaction is at equilibrium in a closed vessel:

⇒ \(a A(g)+b B(g) \rightleftharpoons d D(g)+e E(g)\)

⇒ \(K_p=\frac{p_D^d \times p_E^e}{p_A^a \times p_B^b}\)……………….(1)

⇒ \(K_x=\frac{x_D^d \times x_E^e}{x_A^a \times x_B^b}\)……………….(2)

Where, P_A, P_B, P_D, and P_E are the partial pressures of A, B, D, and E respectively, at equilibrium, and xA, xB, xD and xE are their respective mole fractions at equilibrium. The partial pressure of a component gas in a gas mixture is its mole fraction times the total pressure of the mixture.

Hence, the relation between the partial pressures and mole fractions of the different components in the said gas mixture is

P_A = x_A xP, P_B = x_B xP, P_D = x_D xP, and P_E = x_E x P where P is the total pressure of the gas mixture at equilibrium.

Putting the values of P_A, P_B ,P_D, and P_E into equation (1), we have

⇒ \(K_p=\frac{\left(x_D \times P\right)^d \times\left(x_E \times P\right)^e}{\left(x_A \times P\right)^a \times\left(x_B \times P\right)^b}\)

= \(\frac{x_D^d \times x_E^e}{x_A^a \times x_B^b} \times(P)^{(d+e)-(a+b)}\)

= \(K_x \times(P)^{(d+e)-(a+b)}\)

∴ \(K_p=K_x \times P^{\Delta n}\) ……………….(3)

Here Δn = the total number of moles of gaseous products – the total number of moles of gaseous reactants. Equation (3) denotes the relation between Kp and Kx.

Relation between K_c and K_x

Let the following reversible gaseous reaction be at equilibrium at a temperature T and pressure P in a closed vessel

aA(g) + bB{g) ⇌  dD(g) + eE(g)

For this reaction, the relation between K_p and K_c is,

K_p = K_c (RT)^Δn………………………..(1)

And the relation between Kp and Kx is

, K_p = K_x (P)^Δn………………………..(2)

Where Δn = total number of moles of the gaseous products- total number of moles of the gaseous reactants. From equations (1) and (2), we have, Kc(RT)ÿn = Kx(P)ÿn

∴ , K_c = \(K_x\left(\frac{P}{R T}\right)^{\Delta n}\)………………………(3)

Equation (3) gives the relation between K_P and K_k.

Relation Between  K_P, K_c & K_x

K_P = K_c(RT)Δn = K_X(P)Δn When Δn = 0 for a reaction

Example: H₂(g) +I₂(g) ⇌ 2HI(g)

Or, N₂(g) + O₂(g) 2NO(g)], ⇌ then, K_P= K_c = K_x

Characteristics of the equilibrium constant

At constant temperature, the value of the equilibrium constant for each chemical reaction has a definite value. Temperature change brings about an increase or decrease in the value equilibrium constant.

• In the case of an endothermic reaction, the value of the equilibrium constant increases with the rise in temperature, while in the case of an exothermic reaction, the value of the equilibrium constant decreases with the rise in temperature
• The values of K_P and K_c are not influenced by pressure. If the temperature remains constant, the increase or decrease in pressure does not alter the values of K_P and K_P. However, except for a reaction for which Δn = 0, the value of depends upon pressure.
• The value of the equilibrium constant for any reaction neither increases nor decreases in the presence of a catalyst because the rate of both the forward and the backward reactions increase equally.
• The value of the equilibrium constant of a chemical reaction at a given temperature does not depend on the initial concentration of the reactants.

Example:

PCl₅(g)⇌ PCl₃(g) + Cl₂(g)

In this reaction, K_p at 450°C is 0.19. At 450°C, if the reaction is started with PCl₅(g) at any concentration, the value of K_p for the reaction is always 0.19.

The value of the equilibrium constant of any reaction depends on how the balanced equation for the reaction is written.

Consider the reaction in which NH₃(g) is synthesized from N₂ and H₂ gases. The balanced equation for this reaction is:

N₅(g) + 3H₂(g) ⇌ 2NH₃(g)

One can also write the equation as

⇒  \(\frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g})\)

In case of(1), equilibrium constant

⇒ \(K_{c_1}=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{H}_2\right]^3},\)

While in case of (2), equilibrium constant, \(K_{c_2}=\frac{\left[\mathrm{NH}_3\right]}{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{H}_2\right]^{3 / 2}}\)

Thus, the values of Kcl and Kc are not the same. Comparingÿ andÿ gives =

⇒ \(K_{c_1}=K_{c_2}^2 \text {, i.e., } K_{c_2}=\sqrt{K_{c_1}}\)

Suppose, the equilibrium constant of the reaction,

aA + bB dD+eE is K.

If the coefficients of the reactants and products are multiplied by then the equation becomes:

maA + mbB mdD + meE For this equation, the equilibrium constant, K = Km. Again, if the coefficients of the reactants and products are divided by m then the equation becomes:

⇒ \(\frac{1}{m} a A+\frac{1}{m} b B \rightleftharpoons \frac{1}{m} d D+\frac{1}{m} e E\)

For this equation, the equilibrium constant, K” = K^1/m

For any reversible chemical reaction, the values of ΔT equilibrium constants for the forward and the reverse reactions are reciprocal to each other.
Example:

⇒ \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\) for this reaction equilibrium constant

⇒ \(K_c=\frac{\left[\mathrm{NH}_3\right]^2}{\left[\mathrm{~N}_2\right] \times\left[\mathrm{H}_2\right]^3}\)……………………(1)

If the reaction is started with NH₃ gas (product), then the reaction is:

2NH₃(g) ⇌ N₂(g) + 3H₂(g)

= \(\frac{\left[\mathrm{N}_2\right] \times\left[\mathrm{H}_2\right]^3}{\left[\mathrm{NH}_3\right]^2}\) ……………………(2)

From equations (1) and (2), we get \(K_c=\frac{1}{K_c{ }^{\prime}}\)

if a given reaction is expressed as the sum of two or more individual reactions, then the equilibrium constant of the given reaction equals the product of the equilibrium constants of the individual reactions.

If reaction (3) =reaction (2) + reaction (1), then equilibrium constant of reaction [3] = equilibrium constant of reaction [2] x equilibrium constant of reaction [1].

Example:

Reaction 1: \(A+B \rightleftharpoons C ; K_1=\frac{[C]}{[A][B]}\)

Reaction 2: \(C \rightleftharpoons D ; K_2=\frac{[D]}{[C]}\)

Reaction 3. \(A+B \rightleftharpoons D ; K_3=\frac{|D|}{[A][B]} .\)

Reaction 1+ reaction 2: \(A+B \rightleftharpoons D\)

∴ \(K_1 \times K_2=\frac{[C]}{[A][B]} \times \frac{[D]}{[C]}=\frac{[D]}{[A][B]}=K_3\)

Unit of the equilibrium constant

The unit of equilibrium constant depends on the difference between the sum of the exponents of the concentration (or partial pressure) terms in the numerator and that is the denominator of the equilibrium constant expression. Suppose, this difference is Δx. If Cl Δx = 0, then neither K_c nor K_p has a unit

Δx = 0, then both K_p and K_c have units.

K_p and K_c  for different values of Δx

Values of Δx and the unit of equilibrium constant for some reactions:

The equilibrium constant is unitless:

The term ‘active mass’ mentioned in the law of mass action is unitless. Consequently, Kp or Kc is also unitless. With the help of thermodynamics, it can be shown that the partial pressure of any species present in the expression of Kp is the ratio of measured pressure (P⁰) of that species at equilibrium to its standard pressure (P⁰). Since (P/P⁰) is unitless, Kp is also unitless.

In the case of a pure gas, standard pressure (P⁰) is taken as 1 atm. Similarly, the concentration of any species present in the expression of Kc is the ratio of the measured concentration (C) of that constituent at equilibrium to its standard concentration (C⁰). Since (C/C⁰) is unitless, Kc Is also unitless. The standard concentration (C⁰) of a solute dissolved in a solution is taken as 1(M) or 1 mol. L^-1

Graphical representations of some reversible reactions

1. Reaction: H₂(g)  +I₂ (g) ⇌ 2 HI(g)

Initialconcn.(mol.L^-1): 1.0 + 1.0 ⇌ 0

The equilibrium concentration of H₂(g) = 0.22mol.L^-1

2. Reaction:  2HI(g)+ ⇌ H₂(g)  +I₂ (g)

Initialconcn.(mol.L^-1): 2.0  ⇌ 0+  0

The equilibrium concentration of HI(g) = 1.56 mol. L^-1

3. Reaction:  N₂(g) +3H₂(g)  ⇌  2NH₃(g)

Initialconcn.(mol.L^-1): 1.0 + 3.0  ⇌ 0

The equilibrium concentration of NH₃(g) = 1.56 mol. L^-1

4. Reaction:  H₂(g) + CO₂(g)  ⇌ H₂O(g) + CO(g)

Initialconcn.(mol.L^-1): 2.0 + 1.0  ⇌ 0 + 0

The equilibrium concentration of H₂(g)= 1.6 mol. L^-1

Numerical Examples

Question 1. At a particular temperature, the values of rate constants of forward and backward reactions are 1.5× 10^-2 L -mol^-1 .s^-1 and 1.8 × 10^-3 L-mol^-1.s^-1 respectively for the reaction A + B — C + D. Determine the equilibrium constant of the reaction at that temperature
Answer:

Equilibrium constant,

⇒ \(K=\frac{\text { Rate constant of forward reaction }}{\text { Rate constant of backward reaction }}\)

⇒ \(=\frac{1.5 \times 10^{-2} \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}}{1.8 \times 10^{-3} \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}}=8.33\)

Question 2. At a particular temperature, the equilibrium constant of the reaction 2A + B 2C is 8.0 × 10 L-mol^-1. If the rate constant of the reverse reaction is 1.24 L.mol^-1 .s^-1, then find the value of the rate constant of the forward reaction at that temperature.
Answer:

Equilibrium constant

⇒ \(K=\frac{\text { Rate constant of forward reaction }}{\text { Rate constant of reverse reaction }}\)

∴ The rate constant of the forward reaction = K x rate constant of the reverse reaction =

8 × 10⁴ ×1.24 L^-2.mol^-2.s^-1

= 9. 92 × 10⁴ L^-2.mol^-2.s^-1

Question 3. For the reaction 2SO₂(g) +O₂(g)⇌ 2SO₃(g), Kp = 3×1024 at 25°C. Find the value of ICc.
Answer:

For the given reaction, Δn = 2-(2 + 1) = -1

As given, Kp = 3 × 1024, T = (273 + 25) = 298K and

R = 0.0821 L-atm-mol^-1.K^-1

We know Kp = Kc(RT)n

Or, 3 × 10²⁴ = Kc(0.0821 × 298)^-1

K_c = 3 × 10²⁴ × 0.0821 × 298

= 7.34 × 10²⁵

Question 4. At 1500 K, Kc = 2.6 × 10-9 for the reaction 2BrFg(g) Br2(g) + 5F2(g). Determine the Kp of the reaction at that temperature.
Answer:

For the given reaction, Δn = (1+5)-2 = +4.

As per given data,

K_c = 2.6 × 10^-9

T = 1500K

Using R = 0.0821 L-atm-mol^-1.K^-1

In the equation Kp = Kc{RT)^Δn,

K_p = 2.6 × 10^-9× (0.0821 × 1500)4

= 0.598

Question 5. Find the temperature at which the numerical values of Kp and Kc will be equal to each other for the reaction \(\frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{3}{2} \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_3(\mathrm{~g})\)
Answer:

In the case of the given reaction,

⇒ \(\Delta n=1-\left(\frac{1}{2}+\frac{3}{2}\right)=-1\).

If the numerical values of K_p and K_c be x, then for the above reaction, K_p = x atm^-1 and = x(mol. L^-1)-1=x L-mol^-1

Therefore, Kp = Kc(RT)^-1

Or, \(x \mathrm{~atm}^{-1}=x \mathrm{~L} \cdot \mathrm{mol}^{-1} \times \frac{1}{0.0821 \mathrm{~L} \cdot \mathrm{atm} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \times T}\)

∴ T=12.18K

∴ The numerical values of Kp and Kc for the given reaction will be equal to 12.18K.

Question 6. At 400°C, H₂(g) and I₂(g) are allowed to react in a closed vessel of 5 L capacity to produce in HI(g). At equilibrium, the mixture in the flask Is found to consist of 0.6 mol H₂(g), 0.6 mol I₂(g), and 3.5 mol I₂(g). Determine the value of Kc of the reaction.
Answer: Equilibrium of the reaction is:

H₂(g) + I₂(g)⇒2HI(g)

⇒ \(\text { Therefore, } K_c=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right] \times\left[\mathrm{I}_2\right]}\)

The capacity of the vessel = 5 L.

Hence, molar concentrations of H₂, I₂ and HI are:

⇒  \(\left[\mathrm{H}_2\right]=\frac{0.6}{5}=0.12 \mathrm{~mol} \cdot \mathrm{L}^{-1} ;\)

⇒   \(\left[\mathrm{I}_2\right]=\frac{0.6}{5}=0.12 \mathrm{~mol} \cdot \mathrm{L}^{-1} \text { and }[\mathrm{HI}]=\frac{3.5}{5}=0.7 \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

∴ \(K_c=\frac{(0.7)^2}{(0.12) \times(0.12)}=34.03\)

Question 7. At a particular temperature, CO(g) reacts with Cl₂(g) in a closed container to produce COCI₂(g). In the equilibrium mixture, partial pressures of CO(g), Cl₂(g), and COCl₂(g) are found to be 0.12, 1.2, and 0.58 atm respectively. Find the value of Kp of the reaction, \(\mathrm{CO}(g)+\mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}_2(\mathrm{~g}).\)
Answer:

For the given equilibrium \(K_p=\frac{p_{\mathrm{COCl}_2}}{p_{\mathrm{CO}} \times p_{\mathrm{Cl}_2}}\)

As given, P_CO =o 0.12 atm,P_Cl2= 1.2 atm, and

P_COCl = 0.58 atm at equilibrium.

∴ \(K_p=\frac{0.58}{0.12 \times 1.2}\)

= 4.03

Question 8. In a closed vessel of 1 dm₃ capacity, 1 mol N₂(g) and 2 mol H₂(g) interact to produce 0.8 mol NH₃(g) in the equilibrium mixture. What is the concentration of H₂(g) in the equilibrium mixture?
Answer:

Equation of the equilibrium reaction:

⇒ \(\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)\)

It is observed from the reaction that 1 mol N₂(g) and 3 mol H₂(g) are necessary for the production of 2 mol NH₃(g).

Therefore, for the production of 0.8 mol NH₃(g), the number of moles of H2(g) required \(=\frac{3}{2} \times 0.8=1.2 \mathrm{~mol}\)

Hence, the number of moles of H2(g) remaining in the equilibrium mixture

= 2-1.2 = 0.8

And its molar concentration =0.8 mol.L^-1

Since 1 dm³=1L

Question 9. At 20°C, 0.258mol A(g) and 0.592 mol 5(g) are mixed in a closed vessel of capacity to conduct the following reaction: A(g) + 2B(g) C(g). If 0.035 mol C(g) remains in the equilibrium mixture, then determine the partial pressure of each constituent at equilibrium.
Answer:

According to the equation, 1 mol A(g) reacts with 2 mol 5(g) to produce 1 mol C(g).

Hence, 0.035 mol A(g) and 2 × 0.035 = 0.07 mol 5(g) are required to produce 0.035 mol C(g).

Therefore, equilibrium molar concentrations of different constituents will be as follows:

As given, T = (273 + 20) K = 293 K

∴ At equilibrium,

P_A = [A]BT =0.0446 × 0.0821 × 293 = 1.072 atm

P_B = [B]5T = 0.1044 × 0.0821 × 293 = 2.511 atm

P_C = [C]5T = 7 × 10^-3×  0.0821 × 293 = 0.168 atm

Question 10. 2 mol of were heated in a sealed tube at 440°C until the equilibrium was reached. HI was found to be 22% dissociated. Calculate the equilibrium constant for the reaction

2HI(g) ⇌  H₂(g) +I₂(g)

Answer:

For the given reaction \(K_c=\frac{\left[\mathrm{H}_2\right] \times\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}\)

As given in the question, HI(g) undergoes 22% dissociation. Hence, out of 2 mol HI(g),

2 × 0.22 = 0.44 mol HI(g) dissociates.

As obtained from the equation, 2 mol HI(g) dissociates to produce 1 mol H₂(g) and 1 mol I₂(g). Therefore, 0.44 mol HI(g) dissociates to form 0.22mol of each of H₂(g) and I₂(g).

If the volume of the container is V L, then the equilibrium molar concentrations of different constituents will be as follows:

= 1.56 /V

∴ \(K_c=\frac{\left[\mathrm{H}_2\right] \times\left[\mathrm{I}_2\right]}{[\mathrm{HI}]^2}=\frac{\frac{0.22}{V} \times \frac{0.22}{V}}{\left(\frac{1.56}{V}\right)^2}\)

= 0.0198

Question 11. 1 mol PCl₅(g) is heated in a closed container of 2-litre capacity. If at equilibrium, the quantity of PCl₅g) is 0.2 mol then calculate the value of the equilibrium constant for the given reaction

⇒ \(\mathrm{PCl}_5(g)\rightleftharpoons\mathrm{PCl}_3(g)+\mathrm{Cl}_2(g)\)

Answer:

For the above reaction \(K_c=\frac{\left[\mathrm{PCl}_3\right] \times\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}\)

As per the given data, the number of moles of PCl₅(g) at equilibrium =0.2. Hence, number of moles of PCl₅(g) dissociated =(1- 0.2)

= 0.8.

According to the equation, 1 mol PCl₅(g) dissociates to produce 1 mol of each of PCl₃(g) and Cl₂(g). Therefore, 0.8 mol PCl₅(g) will dissociate to give 0.8 mol of each ofPCl₃(g) and Cl₂(g).

As given, the volume of the container = 2 L. So, the equilibrium concentrations of different constituents are as follows:

∴ \(K_c=\frac{\left[\mathrm{PCl}_3\right] \times\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}=\frac{0.4 \times 0.4}{0.1}\)

=1.6

Question 12. The following reaction is carried out at a particular temperature in a closed vessel of definite volume: CO₂(g) +H₂(g)s=± CO(g) + H₂O(g). Initially, partial pressures of CO₂(g) and H₂(g) are 2 atm and 1 atm respectively and that of CO₂(g) at equilibrium is 1.4 atm. Calculate the equilibrium constant of the reaction.

Answer:

For the given reaction \(K_p=\frac{p_{\mathrm{CO}} \times p_{\mathrm{H}_2 \mathrm{O}}}{p_{\mathrm{CO}_2} \times p_{\mathrm{H}_2}}\)

As given, partial pressure of CO₂(g) at equilibrium (Pco₂) = 1-4 atm. Hence, decrease in pressure of CO₂(g) until the equilibrium is reached =(2- 1.4)atm = 0.6atm According to the equation, 1 mol CO₂(g) reacts with 1 mol H₂(g) to form 1 mol CO(g) and 1 mol H₂O(g).

Therefore, if the pressure of CO₂(g) is reduced by 0.6 atm, then the pressure of H₂(g) will also be reduced by 0.6 atm and the pressure of each of CO(g) and H₂O(g) will be 0.6 atm Hence, partial pressures of different constituents at equilibrium will be as follows:

∴ \(K_p=\frac{p_{\mathrm{CO}} \times p_{\mathrm{H}_2 \mathrm{O}}}{p_{\mathrm{CO}_2} \times p_{\mathrm{H}_2}}\)

= \(\frac{0.6 \times 0.6}{1.4 \times 0.4}\)

= 0.64

Question 13. B(g) + C(g) ⇌  A(g). At constant temperature, a mixture of 1 mol A(g), 2 mol E(g), and 3 mol C(g) is left to stand in a closed vessel of 1 L capacity. The equilibrium mixture is found to contain B(g) of 0.175 molar concentration (mol.L^-1). Find the value of the equilibrium constant at that temperature.
Answer:

For the above reaction \(K_c=\frac{[A]}{[B] \times[C]}\)

As given, at equili-brium, [B] =0.175 mol.L^-1.

So, 2-0.175=1.825 mol.L^-1 of B participated in the reaction.

Now according to the equation, 1 mol B and 1 mol C combine to form 1 mol A. So, 1.825 mol of B and 1.825 mol of C combine to produce 1.825 mol of A. Thus at equilibrium, the concentration of A, B, and C will be

Since the volume of the container = 1L

∴ \(K_c=\frac{[A]}{[B] \times[C]}=\frac{2.825}{0.175 \times 1.175}=13.74\)

Reaction Quotient, Q

The reaction quotient of a reaction has the same form as the equilibrium constant expression. However, the values of molar concentrations (or partial pressures) in the expression of the reaction quotient are the values at any instant of the reaction, whereas these values in the equilibrium constant expression represent the equilibrium values.

Hence, at a particular temperature, the value of the equilibrium constant of a reaction is fixed but that of the reaction quotient is not.

Reaction Quotient:

At constant temperature, the reaction quotient of a reaction at any instant may be defined as the ratio of the product of molar concentrations (or partial pressures) of the products to that of the reactants, with each concentration (or partial pressure) term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation.

The reaction quotient is denoted by Q. When the reaction quotient is expressed in terms of the molar concentration of the reactants and the products, then it is represented by Qc.

The reaction quotient expressed by the partial pressures of the reactants and the products is represented by Q_P.

• At the start of the reaction, only reactants are present in the reaction system. So, the value of the numerator in the expression of Q (reaction quotient) is zero, and consequently Q = 0.
• If a reaction goes to completion, then only the products are present in the reaction system. So, the value of the denominator in the expression of Q is zero, and hence Q→∞.

If the reaction system contains both the reactants and products, then Q assumes a value in between zero and ∞.

1. For the reaction,  aA + bB ⇌  dD + eE,  the reaction quotient at any moment

 ⇒ \(Q_c=\frac{[D]^d \times[E]^e}{[A]^a \times[B]^b}\).

Where (A), (B), (D), and (E) are molar concentrations of A, B, D, and E respectively, at that moment.

For the same reaction, the expression of Kc:

 ⇒ \(K_c=\frac{[D]_{e q}^d \times[E]_{e q}^e}{[A]_{e q}^a \times[B]_{e q}^b}\)

Where [A]_eq, [B]_eq, [D]_eq, and [E]_eq are the molar concentrations of A, B, D, and E respectively, at equilibrium.

2. For the reaction, aA(g) + bB(g) ⇌  dD(g) + eE(g)

The reaction quotient at any instant,

⇒  \(Q_p=\frac{p_D^d \times p_E^e}{p_A^a \times p_B^b}\)

P_A, P_B, P_D, and P_E are the partial pressures of A, B, D, and E respectively, at that instant.

For the same reaction, the expression of Kp:

⇒  \(K_p=\frac{\left(p_D\right)_{e q}^d \times\left(p_E\right)_{e q}^e}{\left(p_A\right)_{e q}^a \times\left(p_B\right)_{e q}^b}\)

Where (P_A)_eq, (P_B)_eq, (P_C)_eq, and (Pp)eq are the partial pressures of, B, D, and E, respectively, at equilibrium.

Significance of reaction quotient:

At a particular temperature, by comparing the values of the reaction quotient (Q) of a chemical reaction at any instant and the equilibrium constant (K) of the reaction at that temperature, one can predict the extent to which the reaction has proceeded.

From the values of Q and AT, it is possible to predict whether the reaction has already reached or will reach the state of equilibrium. If equilibrium is not attained, then it can also be predicted whether the reaction will proceed in the forward or backward direction to achieve equilibrium.

Predicting the direction of reaction from the values of Q_c and K_P

Example: At 700 K, for the following reaction which is carried out in a closed vessel: H₂(g) + I₂(g) ⇌  2HI(g). Kc = 55.0 at 700K.

Analysis of the reaction mixture at a given moment during the reaction shows that the molar concentrations of H₂(g), I₂(g), and HI(g) are 1.8, 2.8, and 0.4 mol-L^-1, respectively. Is the reaction at equilibrium at that moment? If not, in which direction will we proceed to attain equilibrium?
Answer:

For the given reaction, \(Q_c=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right] \times\left[\mathrm{I}_2\right]}\)

At the moment of analysis \(Q_c=\frac{(0.4)^2}{1.8 \times 2.8}=0.0317\)

Hence, Q_c< K_c. Therefore, the reaction is not at equilibrium. To attain equilibrium, the value of Q_c will increase until it becomes equal to K_c.

Again, the value of Q_c will increase if, in its expression, the value of the Numerator increases and that of the denominator decreases. This is possible if the forward reaction occurs to a greater extent Therefore, the reaction will proceed more in the forward direction to attain equilibrium.

Applications Of Equilibrium Constant

Application-1:

At a given temperature, the value of the equilibrium constant indicates the extent to which a reaction has proceeded before it attains equilibrium.

Explanation:

In the expression equilibrium constant (K) of a reaction, the concentrations of the products appear in the numerator, while that of the reactants appears in the denominator.

Hence, a larger value of equilibrium constant (K) for any reversible reaction signifies higher concentrations of the products than reactants in the equilibrium mixture. This means reactants convert into products to a large extent before attaining equilibrium. Hence, the position of equilibrium lies far to the right.

Example:

Application 2: If the value of the equilibrium constant of a reaction at a given temperature is known, then the concentrations of the reactants and products at equilibrium can be calculated from the known initial concentrations of the reactants

Question 1. At 550 K, the value of the equilibrium constant (K£) is 0.08 for the given

PCl₅(g)⇌ PCl₃(g)+Cl₂(g) occurring in a closed container.

If the equilibrium concentration of PClg(g) and Cl₂(g) are 0.75 and 0.32 mol-L^-1 respectively, then find the concentration of PCl₃(g).
Answer:

For the given reaction \(K_c=\frac{\left[\mathrm{PCl}_3\right] \times\left[\mathrm{Cl}_2\right]}{\left[\mathrm{PCl}_5\right]}\)

As given, [PCl₅] = 0.75mol.L^-1

[Cl₅] = 0.32 mol.L^-1 and K_C = 0.08.

∴ \(\left[\mathrm{PCl}_3\right]=K_c \times \frac{\left[\mathrm{PCl}_5\right]}{\left[\mathrm{Cl}_2\right]}\)

= \(0.08 \times \frac{0.75}{0.32}=0.187 \mathrm{~mol} \cdot \mathrm{L}^{-1}\)

∴ The equilibrium concentration of PCl₃ = 0.187mol.L^-1.

Question 2. At a given temperature, Kp is 0.36 for the reaction, 2SO₂(g) + O₂(g)⇌ 2SO₃(g) occurring in a closed vessel. If at equilibrium, the partial pressures of SO₂(g) & O₂(g) are 0.15 atm & 0.8 atm respectively, then calculate the partial pressure of SO₃(g).
Answer:

For the given reaction, Kp \(K_p\)

= \(\frac{\left(p_{\mathrm{SO}_3}\right)^2}{\left(p_{\mathrm{SO}_2}\right)^2 \times p_{\mathrm{O}_2}}\)

As Given K_p =0.36, P_SO2= 0.15 atm and Po₂= 0.8 atm

∴ (P_SO3)2 = K_p ×(P_SO2×2 ×Po₂ = 036 × (0.15)2 × (0.8)

= 6.48 × 10^-3. atm or PSO₃= 0.08 atm

Question 3. For the reaction, A₂(g)+B₂(G) ⇌ 2AB (g), the value of the equilibrium constant is 50 at 100°C. If a flask of 1 L capacity containing 1 mol A₂ is connected with another flask of 2 L capacity containing 2 mol B₂, then calculate the number of moles of AB at equilibrium.
Answer:

Equilibrium constant, \(K_c=\frac{[\mathrm{AB}]^2}{\left[\mathrm{~A}_2\right] \times\left[\mathrm{B}_2\right]}\)

The equation of the reaction shows that the reduction of x mol of A₂ leads to the reduction of x mol of B₂ and the formation of 2x mol of AB. Hence, the number of moles of A₂, B₂, and AB at equilibrium will be as follows.

Now if the two flasks are connected, then the total volume of the reaction system becomes (1 + 2)L = 3L.

Hence, at equilibrium \(\left[\mathrm{A}_2\right]=\left(\frac{1-x}{3}\right) \mathrm{mol} \cdot \mathrm{L}^{-1}\)

⇒ \([B]=\left(\frac{2-x}{3}\right) \mathrm{mol}^{-1} \mathrm{~L}^{-1} \text { and }[\mathrm{AB}]=\left(\frac{2 x}{3}\right) \mathrm{mol} \cdot \mathrm{L}^{-1} \text {. }\)

Therefore,

⇒ \(K_c=\frac{\left(\frac{2 x}{3}\right)^2}{\left(\frac{1-x}{3}\right) \times\left(\frac{2-x}{3}\right)}=50 \text { or, } \frac{4 x^2}{(1-x)(2-x)}=50\)

or, 4x² =  50x²– 1 50x + 100

Or,  46x²– 150x+ 100 = 0

Or, \(x=\frac{150 \pm \sqrt{(150)^2-4 \times 46 \times 100}}{2 \times 46}=2.32 \text { or, } 0.934\)

∴ 2x = 4.64 or, 1.868

By the Initial number of moles of A₂ and B₂, the number of moles of AB cannot be 4.64. Therefore, the number of moles of AB at equilibrium = 1 .868.

Question 4. At a particular temperature, the value of Kp is 100 for the reaction, \(2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_2(g)+\mathrm{O}_2(g)\)occurringin a closed container. If the initial pressure of NO(g) is 25 atm then calculate the partial pressures of NO, N₂, and O₂ at equilibrium.
Answer:

For the given reaction \(K_p=\frac{p_{\mathrm{N}_2} \times p_{\mathrm{O}_2}}{\left(p_{\mathrm{NO}}\right)^2}\)

After the attainment of equilibrium, if the reduction in pressure of (g) is p atm, then partial pressures of different constituents at equilibrium will be as follows:

[The equation shows that at constant temperature and pressure if the pressure reduction of NO(g) is p atm then the increase in pressure of each of N₂(g) and O₂(g) will be P/2 atm.

⇒ \(\text { Hence, } K_p=\frac{\left(\frac{p}{2}\right) \times\left(\frac{p}{2}\right)}{(25-p)^2}=100 \text { or, }\left(\frac{p}{25-p}\right)^2=400\) \(\text { or, } \frac{p}{25-p}=20 \text { or, } 21 p=25 \times 20\) ∴ P= 23.8 atm

Therefore, at equilibrium, partial pressure of NO(g) = (25- 23.8) atm = 1.2 atm andpartial pressure of N₂(g) = partial pressure Of

⇒  \(\mathrm{O}_2(\mathrm{~g})=\frac{23.8}{2}=11.9 \mathrm{~atm}\)

Application-3: If the value of the equilibrium constant of any reaction at a constant temperature is known, then it is possible to predict whether the mixture of reactants and products is in equilibrium and if not, in which direction the reaction will proceed Formore discussion, to attains equilibrium article number at that 7.6. temperature.

Free Energy Change (ΔG) Of A Reaction and Equilibrium Constant

Suppose, at a temperature of TK, A and B react together to produce D and E according to the following equation: \(a A+b B \rightleftharpoons d D+e E\)

⇒ \(K=\frac{[D]_{e q}^d \times[E]_{e q}^e}{[A]_{e q}^a \times[B]_{e q}^b}[e q=\text { equilibrium }]\)…………………………..(1)

Where (A)_eq, (B)_eq,  (D) _eq, and (E )_eq are the molar concentrations of A, B, D, and E respectively, at equilibrium.

If ΔG is the from the energy of the system, then with the help of can show that

⇒ \(\Delta G=\Delta G^0+N T \ln \frac{[D]^a \times[B]^b}{[A]^a \times[B]^b}\)

or, ΔG = ΔG⁰+RT in Q…………………………..(2)

Where,

⇒ \(Q=\frac{[D]^d \times[B]^c}{[A]^a \times[B]^b}\) and (A),(B),(D ), and (E)

Represent the active masses or molar concentrations of A, B, I) and respectively at a given moment during the reaction. Is the reaction quotient.

ΔG° Is the standard free energy change of the reaction. If In a reaction, (lie molar concentration of each of the reactants and products. If unity, then the free energy change of the reaction is called (IK; standard free energy change (ΔG⁰).

For a reaction at constant temperature and pressure, if—

1. ΔG is negative, the reaction is spontaneous as written.
2. ΔG is positive, the reaction Is non-spontane O us as written.
3. ΔG is zero, the reaction Is at equilibrium.

Now, at the equilibrium of a reaction, ΔG = 0 and

⇒ \(Q=\frac{[D]_{e q}^a \times[B]_{e q}^e}{[A]_{e q}^a \times[B]_{e q}^b}=K \text { [equilibrium constant] }\)

Hence, from equation (2) we get, 0 = ΔG° + RTlnK or, ΔG° = -RTlnK …………………………..(3)

Or,ΔG° = -2.303RTlogK …………………………..(4)

Or, K= e^-ΔG0/Rt…………………………..(5)

Equations (3), (4), and (5) represent the relation between equilibrium constant {K) and the standard free energy change (ΔG°) of a reaction at a given temperature.

From equations (2) and (3), we get, ΔG = -RTlnK +RTlnQ

or ΔG = RT In \(\frac{Q}{K}\) …………………………..(6)

Equation (6) is called reaction isotherm.

If the equilibrium constant expression of a reaction is written in terms of the molar concentrations of reactants and products then K=Ke. This gives

⇒ \(\Delta G^0=-R T \ln K_c \text { or, } K_c=e^{-\Delta G^0 / R T}\)

For gaseous reactions, equilibrium constant expression is written in terms of the partial pressures of reactants and products. So, for such reactions K=K and AG° = -RTlnK p or, Kp = e-ΔG⁰/RT.

In the equation, reactant products taken the standards I molstatel, of 1, the equation the; -RT in K p, the standard state of each reactant and products Is considered to be 1 atm.

Significance Of the reaction ΔG°=-RT in K

1. At a particular temperature If the value of AG° is known, then the value of the equilibrium constant (K) can be calculated by using this equation. Similarly, If the equilibrium constant of a reaction at a given temperature is known, then the value of A <7° can be determined by using this reaction.
2. If ΔG° < 0, i.e., ΔG° = – ve, then K will be greater than [K> 1]. Under such conditions, the amount of products will be relatively more than that of the reactants present in the equilibrium mixture, i.e., the forward reaction predominates.
3. If ΔG° >0, i.e., ΔG = + ve, then K will be less than 1 [K < 1]. In such cases, the concentration of the reactants is greater than that of the products, i.e., the backward reaction predominates.

CBSE Class 11 Chemistry Notes For Free Energy Change (ΔG) Of A Reaction and Equilibrium Constant Numerical Examples

Question 1. For the reaction A(g) + 2B(g) ⇌ 2D(g), ΔG⁰– 2kJ.mol^-1 at 500 K. What is the value of Kp for the reaction ½A(g) + B(g) ⇌  D(g) at that temperature?
Answer:

From the equation ΔG⁰ = -RTlnKp, we get

⇒ \(\ln K_p=-\frac{\Delta G^0}{R T}=-\frac{2000 \mathrm{~J} \cdot \mathrm{mol}^{-1}}{8.314 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1} \times 500 \mathrm{~K}}\)

=-0.4811

∴ K_p= 0.6181

∴ For the given reaction \(K_p=\frac{\left(p_D\right)^2}{p_A \times\left(p_B\right)^2}\)

If the equilibrium constant for the reaction

⇒ \(\frac{1}{2} A(g)+B(g) \rightleftharpoons D(g) \text { be } K_p^{\prime} \text {, then } K_p^{\prime}=\frac{p_D}{p_A^{1 / 2} \times p_B}\)

∴ \(K_p^{\prime}=\sqrt{K_p} \quad K_p^{\prime}=\sqrt{0.6181}=0.7862\).

Question 2. Find the value of ΔG⁰ and ICc for the following reaction at 298K. \(\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{NO}_2(g)\) Given: standard free energy of formation (ΔG) of NO₂ and NO are 52.0 and 87.0 kj.mol^-1 respectively.
Answer:

ΔG° of a reaction = ∑ΔG_f⁰ of products -∑ΔG_f⁰ of reactants. For the given reaction,

⇒ \(\Delta G^0=\Delta G_f^0\left(\mathrm{NO}_2\right)-\Delta G_f^0(\mathrm{NO})-\frac{1}{2} \Delta G_f^0\left(\mathrm{O}_2\right)\)

= 52.0- 87.0- 0.0 = -35.0 kl -mol^-1

We know that, ΔG⁰=-RT In K_c

Or, -35 × 10³ J-mol^-1

= – 8.314 K^-1  mol^-1 ×  298K ln Kc

or; lnKc = 14.126

∴ Kc = 1.365 × 10⁶

Question 3. At 298K, for the attainment of equilibrium of the reaction N₂O₄(g) ⇌ 2NO₂5mol of each of the constituents is taken. Due to this, the total pressure of the mixture turns 20 atm. If ΔG_f⁰ (N₂O₄)= 100 kJ-mol^-1 and ΔG_f⁰ (NO₂) = 50 k(J-moI_1) then

1. Give the value of ΔG of the reaction.
2. In which direction will the reaction proceed to equilibrium?

Answer:

For the reaction,

⇒ \(\Delta G^0 & =2 \Delta G_f^0\left(\mathrm{NO}_2\right)-\Delta G_f^0\left(\mathrm{~N}_2 \mathrm{O}_4\right)\)

= \((2 \times 50-100)=0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\)

Total number of moles in the reaction mixture =5 + 5 =10

⇒ \(\text { So, } p_{\mathrm{N}_2 \mathrm{O}_4}=\frac{5}{10} \times 20=10 \mathrm{~atm} \& p_{\mathrm{NO}_2}\)

= \(\frac{5}{10} \times 20=10 \mathrm{~atm}\)

Therefore, the Q_p of the reaction

⇒ \(\frac{\left(p_{\mathrm{NO}_2}\right)^2}{p_{\mathrm{N}_2 \mathrm{O}_4}}=\frac{(10)^2}{10}=10\)

1. We know, ΔG = ΔG⁰ + RT]nQp

∴ ΔG =0 + 8.314 ×  298 In 10 = 5.706 kJ

Since ΔG⁰ =0

Again, ΔG⁰ =-RT In Kp;

So, 0 = -RT In Kp [since ΔG⁰ = 0]

or, Kp = 1

Since, Qp > Kp, the reaction wall proceeds more toward the left to attain equilibrium.

CBSE Class 11 Chemistry Notes For Determination Of Equilibrium Constants Of Some Reactions

Esterification of alcohol:

CH₃COOH(I) + C₂H₅OH(I) ⇌  CH₃COOC₂H₅(Z) + H₂O(Z)

Let at a particular temperature, a mol of acetic acid (CH₃COOH) reacts with b mol of ethyl alcohol (C₂H₅OH) to produce x mol of ethyl acetate (CH₃COOC₂H₅) and x mol of water (H₂O) at equilibrium.

According to the balanced equation, for the formation of x mol of ester and x mol of H_2O, x mol of CH₃COOH and x mol of C₂H₅OH are required. If the volume of the reaction mixture is by V L, the die concentration of the different species at equilibrium as well as

The expression of the equilibrium constant will be as given in the following table:

Formation of NO(g) from N₂(g) and O₂(g): For the reaction:

⇒ \(\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\);

Let us assume, at a constant temperature a mol of N₂ reacts with b mol of O₂ in a closed container of VL to produce 2xmol of NO at equilibrium.

According to the equation,ifx mol of N₂ and O₂ reacts with each other, then 2x mol of NO are formed.

Let, the total pressure of the reaction mixture at equilibrium = P. The molar concentrations and partial pressures of the reactants and products at equilibrium and

The expressions of K_p and K_c are given in the table:

Dissociation of PCl₅ gas: For the reaction:

PCl₅(g)⇌PCl₃(g) + Cl₂(g);

Let us assume, at a particular temperature, 1 mol of PCl₅ gas is heated in a closed vessel of V L capacity so that x mol of PCI5 gets dissociated at equilibrium.

According to the given equation, x mol of PCl₅ on dissociation produces x mol of PCl₃ and x mol of Cl₂.

1. Let us assume that die total pressure of the reaction mixture at equilibrium is P. Hence, the molar concentrations and partial pressures of the reactant and products at equilibrium

The expression equilibrium constants are given in the following table:

Formation of NH₃ from N₂ and H₂:

For the reaction: N₂(g) + 3H₂(g) v=± 2NH₃(g);

Let us assume, at a particular temperature, 1 mol of N₂(g) is allowed to react with 3 mol of H₂(g) in a closed vessel of V L capacity. After the attainment of equilibrium, if 2x mol of NH₃(g) is produced, then according to the equation x mol of N₂(g) and 3x mol of H₂(g) would be consumed.

Let, the total pressure of the reaction mixture be at equilibrium. Therefore, the molar concentrations and partial pressures of the reactants and product at equilibrium and the expression of equilibrium constants are given in the following table

Thermal decomposition of ammonium carbamate:

Reaction:

⇒\(\mathrm{NH}_2 \mathrm{CO}_2 \mathrm{NH}_4(\mathrm{~s}) \Rightarrow 2 \mathrm{NH}_3(\mathrm{~g})+\mathrm{CO}_2(\mathrm{~g})\).

Let us assume, the l mol of NH₂CO₂NH₄(s) is heated at a particular temperature in a closed vessel of volume VL.

At equilibrium, if x mol of NH₂CO₂NH₄(g) undergoes decomposition, then, according to the equation, 2x mol of NH₃ and x mol of CO₂, will be produced.

The total pressure of the reaction mixture at equilibrium = P, then the molar concentrations and partial pressures of products and the expression of equilibrium constants are given in the table:

CBSE Class 11 Chemistry Notes For Relation Between Degree Of Dissociation And Degree Of Association With Vapour Density

When a compound undergoes incomplete dissociation, an equilibrium is established between the undissociated molecules of the die compound with the species formed on dissociation.

The extent of dissociation of a compound is usually quantized in terms of the degree of dissociation. It is defined as the ratio of the number of moles of the compound that dissociates to the initial number of moles of the compound.

Similarly, if a compound undergoes association, its extent of association is quantized in terms of degree of association. It is defined as the ratio of the number of moles of die compound associates to the initial number of moles of the compound.

Relation between the vapor density and degree of dissociation

Suppose, each molecule of gas A₂ on dissociation forms n molecules of gas B. If the reaction is initiated with 1 mol of gas A in a closed vessel of volume V at constant temperature and pressure, then the die number of moles of A₂ and B at equilibrium are as follows-

Where = the degree of dissociation of A₂

∴Total number of moles of the mixture at equilibrium =1- α + nα =1 + α(n- 1) mol

At constant temperature and pressure, the total volume of [1 + o(n- 1)] mol of the gas mixture is [1 + α(n- 1)× V].

Let the actual density and vapor density of gas A₂ before dissociation be d and D, respectively. Since the volume of the system increases on the dissociation of gas A₂ the density and vapour density of the gas mixture at equilibrium will be different from the actual values of these two quantities for gas A₂.

Suppose, the observed density and vapor density of the gas mixture at equilibrium be d’ and D’, respectively.

Since the mass of the system remains the same, we can write

d ×  V = d’ ×  V[1 + α(n- 1)]

Or, \(\frac{d}{d^{\prime}}=1+\alpha(n-1)\)……………………………….(1)

\(\frac{D}{D!}=1+\alpha(n-1)\)……………………………….(1)

∴ \(\frac{d}{d^{\prime}}=\frac{D}{D^{\prime}}\)

From equation (1) we get \(\alpha=\frac{d-d^{\prime}}{d^{\prime}(n-1)}\)

From equation (2) we get \(\alpha=\frac{d-d^{\prime}}{d^{\prime}(n-1)}\)

Relation between the vapor density and degree of association

Let us consider n mol of gas A associated to form 1 mol of gas B. If the reaction is initiated with 1 mol of gas A in a closed vessel of volume V at constant temperature and pressure,

Then the number of moles of A and B at equilibrium are as follows:

where a = the degree of association of A

∴ Total number of moles of the mixture at equilibrium

⇒ \(=1-\alpha+\frac{\alpha}{n}=1-\alpha\left(1-\frac{1}{n}\right) \mathrm{mol}\)

At constant temperature and pressure, the total volume of

⇒ \(\left[1-\alpha\left(1-\frac{1}{n}\right)\right]\) mol gas

⇒ \(\left[1-\alpha\left(1-\frac{1}{n}\right) \times v\right]\)

Now, let actual density and vapor density of gas A before association be d and D respectively.

If the observed density and vapor density of the gas mixture at equilibrium are d’ and D’, respectively, then—

⇒ \(d \times V=d^{\prime} \times V\left[1-\alpha\left(1-\frac{1}{n}\right)\right]\)

Or,  \(\frac{d}{d^{\prime}}=1-\alpha\left(1-\frac{1}{n}\right)\)…………………………(1)

⇒ \(\frac{D}{D^{\prime}}=1-\alpha\left(1-\frac{1}{n}\right)\) …………………………(2)

Since ⇒ \(\frac{d}{d^{\prime}}=\frac{D}{D^{\prime}}\)

From equation (1) we get

⇒ \(\alpha=\frac{d^{\prime}-d}{d^{\prime}\left(1-\frac{1}{n}\right)}\)

From equation (2) we get

⇒ \(\alpha=\frac{D^{\prime}-D}{D^{\prime}\left(1-\frac{1}{n}\right)}\)

CBSE Class 11 Chemistry Notes For Chapter 7 Equilibrium Ionic Equilibrium Introduction

The compounds that conduct electricity in a molten state or solution and dissociate chemically into new substances are called electrolytes.

Various acids Example: HCl, HNO₃, H₂SO₄),

Bases Example: NaOH, KOH)

Salts Example: NaCl, KCl, CuSO₄, AgNO₃)

Dissolve in water to conduct electricity. Hence, these are electrolytes.

On the both er hand, substances that are unable to conduct electricity either in a molten state or in solution are known as non-electrolytes. Glucose, sugar, alcohol, benzene, etc., are examples of non-electrolytes. In an aqueous solution (or in a molten state), electrolytes undergo spontaneous dissociation or ionization to produce positively and negatively charged particles or ions.

This is known as electrolytic dissociation or ionization. The ions can conduct electricity in the solution. In a particular solvent and at a certain temperature and concentration, the degree of dissociation or ionization of any electrolytic substance depends on the nature of that substance. The fraction of the total amount of dissolved electrolyte that exists in a dissociated or ionized state is called the degree of dissociation or ionization.

Depending on the degree of ionization in an aqueous solution, electrolytes can be classified into two categories Strong electrolytes and Weak electrolytes. The electrolytes which dissociate almost completely in aqueous solution at all concentrations are called strong electrolytes.

Strong acids Example: HCl, HNO₃, H₂SO₄ ),

Strong bases Example: NaOH, KOH

Most of the salts Example: NaCl, KC1, NH₄Cl, CuSO₄ )

Are strong electrolytes. Electrolytes that dissociate partially aqueous solution are called weak electrolytes.

Some salts Example: BaSO₄, HgCl₂

Most of the organic acids Example: HCOOH, CHgCOOH

A few in organic bases Example: Fe(OH)₃, NH₃ ),

Some inorganic acids Example: HCN, and H₂CO₃ ) are weak electrolytes. In aqueous solutions, strong electrolytes undergo almost complete ionization.

Hence, such ionisations are represented by a single arrow(→)

For example:

NaCl(aq)→ Na⁺(aq) + Cl^–(aq); HCl(aq)→H⁺(aq) + Cl^– (aq).

On the other hand, weak electrolytes undergo partial ionization in aqueous solution. Hence, such ionisations are reversible. Consequently, in an aqueous solution of weak electrolytes, a dynamic equilibrium exists between the dissociated ions and unionized molecules. This is known as ionic equilibrium.

Due to its irreversible nature, such ionisations are represented by double arrows(→).

For example:

HCN(aq) ⇌  H+(ag) + CN^–(aq); CH₃COOH(aq) ⇌  CH₃COO^–(aq) + H⁺(aq)