Chemical Equilibrium

Definition of Chemical Equilibrium

Chemical equilibrium applies to reactions that can occur in both directions. In a reaction such as:

CH₄(g) + H₂O(g) <--> CO(g) + 3H₂(g)

The reaction can happen both ways. So after some of the products are created the products begin to react to form the reactants. At the beginning of the reaction, the rate that the reactants are changing into the products is higher than the rate that the products are changing into the reactants. Therefore, the net change is a higher number of products.

Even though the reactants are constantly forming products and vice-versa the amount of reactants and products does become steady. When the net change of the products and reactants is zero the reaction has reached equilibrium. The equilibrium is a dynamic equilibrium. The definition for a dynamic equilibrium is when the amount of products and reactants are constant. (They are not equal but constant. Also, both reactions are still occurring.)

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Chemical equilibria are important in numerous biological and environmental processes. For example, equilibria involving O₂ molecules and the protein hemoglobin play a crucial role in the transport and delivery of O₂ from our lungs to our muscles. Similar equilibria involving CO molecules and hemoglobin account for the toxicity of CO.

When a liquid evaporates in a closed container, molecules with relatively higher kinetic energy escape the liquid surface into the vapour phase and number of liquid molecules from the vapour phase strike the liquid surface and are retained in the liquid phase. It gives rise to a constant vapour pressure because of an equilibrium in which the number of molecules leaving the liquid equals the number returning to liquid from the vapour. We say that the system has reached equilibrium state at this stage. However, this is not static equilibrium and there is a lot of activity at the boundary between the liquid and the vapour. Thus, at equilibrium, the rate of evaporation is equal to the rate of condensation. It may be represented by

The double half arrows indicate that the processes in both the directions are going on simultaneously. The mixture of reactants and products in the equilibrium state is called an equilibrium mixture.

Equilibrium can be established for both physical processes and chemical reactions. The reaction may be fast or slow depending on the experimental conditions and the nature of the reactants. When the reactants in a closed vessel at a particular temperature react to give products, the concentrations of the reactants keep on decreasing, while those of products keep on increasing for some time after which there is no change in the concentrations of either of the reactants or products. This stage of the system is the dynamic equilibrium and the rates of the forward and reverse reactions become equal. It is due to this dynamic equilibrium stage that there is no change in the concentrations of various species in the reaction mixture. Based on the extent to which the reactions proceed to reach the state of chemical equilibrium, these may be classified in three groups.

(i) The reactions that proceed nearly to completion and only negligible concentrations of the reactants are left. In some cases, it may not be even possible to detect these experimentally.

(ii) The reactions in which only small amounts of products are formed and most of the reactants remain unchanged at equilibrium stage.

(iii) The reactions in which the concentrations of the reactants and products are comparable, when the system is in equilibrium.

The extent of a reaction in equilibrium varies with the experimental conditions such as concentrations of reactants, temperature, etc. Optimisation of the operational conditions is very important in industry and laboratory so that equilibrium is favorable in the direction of the desired product. Some important aspects of equilibrium involving physical and chemical processes are dealt in this unit along with the equilibrium involving ions in aqueous solutions which is called as ionic equilibrium.

7.1 EQUILIBRIUM IN PHYSICAL PROCESSES

The characteristics of system at equilibrium are better understood if we examine some physical processes. The most familiar examples are phase transformation processes, e.g.,

7.1.1 Solid-Liquid Equilibrium

Ice and water kept in a perfectly insulated thermos flask (no exchange of heat between its contents and the surroundings) at 273K and the atmospheric pressure are in equilibrium state and the system shows interesting characteristic features. We observe that the mass of ice and water do not change with time and the temperature remains constant. However, the equilibrium is not static. The intense activity can be noticed at the boundary between ice and water. Molecules from the liquid water collide against ice and adhere to it and some molecules of ice escape into liquid phase. There is no change of mass of ice and water, as the rates of transfer of molecules from ice into water and of reverse transfer from water into ice are equal at atmospheric pressure and 273 K.

It is obvious that ice and water are in equilibrium only at particular temperature and pressure. For any pure substance at atmospheric pressure, the temperature at which the solid and liquid phases are at equilibrium is called the normal melting point or normal freezing point of the substance. The system here is in dynamic equilibrium and we can infer the following:

(i) Both the opposing processes occur simultaneously.

(ii) Both the processes occur at the same rate so that the amount of ice and water remains constant.

7.1.2 Liquid-Vapour Equilibrium

This equilibrium can be better understood if we consider the example of a transparent box carrying a U-tube with mercury (manometer). Drying agent like anhydrous calcium chloride (or phosphorus penta-oxide) is placed for a few hours in the box. After removing the drying agent by tilting the box on one side, a watch glass (or petri dish) containing water is quickly placed inside the box. It will be observed that the mercury level in the right limb of the manometer slowly increases and finally attains a constant value, that is, the pressure inside the box increases and reaches a constant value. Also the volume of water in the watch glass decreases (Fig. 7.1). Initially there was no water vapour (or very less) inside the box. As water evaporated the pressure in the box increased due to addition of water molecules into the gaseous phase inside the box. The rate of evaporation is constant. However, the rate of increase in pressure decreases with time due to condensation of vapour into water. Finally it leads to an equilibrium condition when there is no net evaporation. This implies that the number of water molecules from the gaseous state into the liquid state also increases till the equilibrium is attained i.e.,

rate of evaporation= rate of condensation

At equilibrium the pressure exerted by the water molecules at a given temperature remains constant and is called the equilibrium vapour pressure of water (or just vapour pressure of water); vapour pressure of water increases with temperature. If the above experiment is repeated with methyl alcohol, acetone and ether, it is observed that different liquids have different equilibrium vapour pressures at the same temperature, and the liquid which has a higher vapour pressure is more volatile and has a lower boiling point.

If we expose three watch glasses containing separately 1mL each of acetone, ethyl alcohol, and water to atmosphere and repeat the experiment with different volumes of the liquids in a warmer room, it is observed that in all such cases the liquid eventually disappears and the time taken for complete evaporation depends on (i) the nature of the liquid, (ii) the amount of the liquid and (iii) the temperature. When the watch glass is open to the atmosphere, the rate of evaporation remains constant but the molecules are dispersed into large volume of the room. As a consequence the rate of condensation from vapour to liquid state is much less than the rate of evaporation. These are open systems and it is not possible to reach equilibrium in an open system.

Water and water vapour are in equilibrium position at atmospheric pressure (1.013 bar) and at 100°C in a closed vessel. The boiling point of water is 100°C at 1.013 bar pressure. For any pure liquid at one atmospheric pressure (1.013 bar) the temperature at which the liquid and vapours are at equilibrium is called normal boiling point of the liquid. Boiling point of the liquid depends on the atmospheric pressure. It depends on the altitude of the place; at high altitude the boiling point decreases.

7.1.3 Solid — Vapour Equilibrium

Let us now consider the systems where solids sublime to vapour phase. If we place solid iodine in a closed vessel, after sometime the vessel gets filled up with violet vapour and the intensity of colour increases with time. After certain time the intensity of colour becomes constant and at this stage equilibrium is attained. Hence solid iodine sublimes to give iodine vapour and the iodine vapour condenses to give solid iodine. The equilibrium can be represented as,

Other examples showing this kind of equilibrium are,

7.1.4 Equilibrium Involving Dissolution of Solid or Gases in Liquids Solids in liquids

We know from our experience that we can dissolve only a limited amount of salt or sugar in a given amount of water at room temperature. If we make a thick sugar syrup solution by dissolving sugar at a higher temperature, sugar crystals separate out if we cool the syrup to the room temperature. We call it a saturated solution when no more of solute can be dissolved in it at a given temperature. The concentration of the solute in a saturated solution depends upon the temperature. In a saturated solution, a dynamic equilibrium exits between the solute molecules in the solid state and in the solution:

and the rate of dissolution of sugar = rate of crystallisation of sugar.

Equality of the two rates and dynamic nature of equilibrium has been confirmed with the help of radioactive sugar. If we drop some radioactive sugar into saturated solution of non-radioactive sugar, then after some time radioactivity is observed both in the solution and in the solid sugar. Initially there were no radioactive sugar molecules in the solution but due to dynamic nature of equilibrium, there is exchange between the radioactive and non-radioactive sugar molecules between the two phases. The ratio of the radioactive to nonradioactive molecules in the solution increases till it attains a constant value.

Gases in liquids

When a soda water bottle is opened, some of the carbon dioxide gas dissolved in it fizzes out rapidly. The phenomenon arises due to difference in solubility of carbon dioxide at different pressures. There is equilibrium between the molecules in the gaseous state and the molecules dissolved in the liquid under pressure i.e.,

This equilibrium is governed by Henry’s law, which states that the mass of a gas dissolved in a given mass of a solvent at any temperature is proportional to the pressure of the gas above the solvent. This amount decreases with increase of temperature. The soda water bottle is sealed under pressure of gas when its solubility in water is high. As soon as the bottle is opened, some of the dissolved carbon dioxide gas escapes to reach a new equilibrium condition required for the lower pressure, namely its partial pressure in the atmosphere. This is how the soda water in bottle when left open to the air for some time, turns ‘flat’. It can be generalised that:

(i) For  equilibrium, there is only one temperature (melting point) at 1 atm (1.013 bar) at which the two phases can coexist. If there is no exchange of heat with the surroundings, the mass of the two phases remains constant.

(ii) For  equilibrium, the vapour pressure is constant at a given temperature.

(iii) For dissolution of solids in liquids, the solubility is constant at a given temperature.

(iv) For dissolution of gases in liquids, the concentration of a gas in liquid is proportional to the pressure (concentration) of the gas over the liquid. These observations are summarised in Table 7.1

Table 7.1 Some Features of Physical Equilibria

7.1.5 General Characteristics of Equilibria Involving Physical Processes

For the physical processes discussed above, following characteristics are common to the system at equilibrium:

(i) Equilibrium is possible only in a closed system at a given temperature.

(ii) Both the opposing processes occur at the same rate and there is a dynamic but stable condition.

(iii) All measurable properties of the system remain constant.

(iv) When equilibrium is attained for a physical process, it is characterised by constant value of one of its parameters at a given temperature. Table 7.1 lists such quantities.

(v) The magnitude of such quantities at any stage indicates the extent to which the physical process has proceeded before reaching equilibrium.

7.2 EQUILIBRIUM IN CHEMICAL PROCESSES ? DYNAMIC EQUILIBRIUM

Analogous to the physical systems chemical reactions also attain a state of equilibrium. These 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. This equilibrium is dynamic in nature as it consists of a forward reaction in which the reactants give product(s) and reverse reaction in which product(s) gives the original reactants.

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

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 (page 191) 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 D₂ (deuterium) in place of H₂. The reaction mixtures starting either with H₂ or D₂ reach equilibrium with the same composition, except that D2 and ND₃ are present instead of H₂ and NH₃. After equilibrium is attained, these two mixtures (H₂, N₂, NH₃ and D₂, N₂, ND₃) 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 (NH₃, NH₂D, NHD₂ and ND₃) and dihydrogen and its deutrated forms (H₂, HD and D₂) are present. Thus one can conclude that scrambling of H and 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, H₂(g) and N₂(g) and get NH₃(g) or by taking NH₃(g) and decomposing it into N₂(g) and H₂(g).

Similarly let us consider the reaction,  . If we start with equal initial concentration of H₂ and I₂, the reaction proceeds in the forward direction and the concentration of H₂ and I₂ decreases while that of HI increases, until all of these become constant at equilibrium (Fig. 7.5). We can also start with HI alone and make the reaction to proceed in the reverse direction; the concentration of HI will decrease and concentration of H₂ and I₂ will increase until they all become constant when equilibrium is

reached (Fig.7.5). If total number of H and I atoms are same in a given volume, the same equilibrium mixture is obtained whether we start it from pure reactants or pure product.

Dynamic Equilibrium — A Student’s Activity

Equilibrium whether in a physical or in a chemical system, is always of dynamic nature. This can be demonstrated by the use of radioactive isotopes. This is not feasible in a school laboratory. However this concept can be easily comprehended by performing the following activity. The activity can be performed in a group of 5 or 6 students. Take two 100mL measuring cylinders (marked as 1 and 2) and two glass tubes each of 30 cm length. Diameter of the tubes may be same or different in the range of 3-5mm. Fill nearly half of the measuring cylinder-1 with coloured water (for this purpose add a crystal of potassium permanganate to water) and keep second cylinder (number 2) empty. Put one tube in cylinder 1 and second in cylinder 2. Immerse one tube in cylinder 1, close its upper tip with a finger and transfer the coloured water contained in its lower portion to cylinder 2. Using second tube, kept in 2nd cylinder, transfer the coloured water in a similar manner from cylinder 2 to cylinder 1. In this way keep on transferring coloured water using the two glass tubes from cylinder 1 to 2 and from 2 to 1 till you notice that the level of coloured water in both the cylinders becomes constant. If you continue intertransferring coloured solution between the cylinders, there will not be any further change in the levels of coloured water in two cylinders. If we take analogy of ?level? of coloured water with ?concentration? of reactants and products in the two cylinders, we can say the process of transfer, which continues even after the constancy of level, is indicative of dynamic nature of the process. If we repeat the experiment taking two tubes of different diameters we find that at equilibrium the level of coloured water in two cylinders is different. How far diameters are responsible for change in levels in two cylinders? Empty cylinder (2) is an indicator of no product in it at the beginning.

 

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Equilibrium Constant

To determine the amount of each compound that will be present at equilibrium you must know the equilibrium constant. To determine the equilibrium constant you must consider the generic equation:

aA + bB <--> cC + dD

The upper case letters are the molar concentrations of the reactants and products. The lower case letters are the coefficients that balance the equation. Use the following equation to determine the equilibrium constant (K_c).

For example, determining the equilibrium constant of the following equation can be accomplished by using the K_c equation.

Using the following equation, calculate the equilibrium constant.

N₂(g) + 3H₂(g) <--> 2NH₃(g)

A one-liter vessel contains 1.60 moles NH₃, .800 moles N₂, and 1.20 moles of H₂. What is the equilibrium constant?

Answer: 1.85

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 LAW OF CHEMICAL EQUILIBRIUM AND EQUILIBRIUM CONSTANT

A mixture of reactants and products in the equilibrium state is called an equilibrium mixture. In this section we shall address a number of important questions about the composition of equilibrium mixtures: What is the relationship between the concentrations of reactants and products in an equilibrium mixture? How can we determine equilibrium concentrations from initial concentrations?

What factors can be exploited to alter the composition of an equilibrium mixture? The last question in particular is important when choosing conditions for synthesis of industrial chemicals such as H₂, NH₃, CaO etc.

To answer these questions, let us consider a general reversible reaction:

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,

K_c= [C][D]/[A][B] ———————————————————————————————- (7.1)

where 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 H₂ and I₂ carried out in a sealed vessel at 731K.

Six sets of experiments with varying initial conditions were performed, starting with only gaseous H₂ and I₂ in a sealed reaction vessel in first four experiments (1, 2, 3 and 4) and only HI in other two experiments (5 and 6). Experiment 1, 2, 3 and 4 were performed taking different concentrations of H₂ and / or I₂, 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.

Table 7.2 Initial and Equilibrium Concentrations of H₂, I₂ and HI

Experiment number	Initial concentration/mol L-1			Equilibrium concentration/mol L-1
	[ H2 (g) ]	[ I2 (g) ]	[ HI (g) ]	[ H2 (g) ]	[ I2 (g) ]	[ HI (g) ]
1	2.4 x 10-2	1.38 x 10-2	0	1.14 x 10-2	0.12 x 10-2	2.52 x 10-2
2	2.4 x 10-2	1.68 x 10-2	0	0.92 x 10-2	0.20 x 10-2	2.96 x 10-2
3	2.44 x 10-2	1.98 x 10-2	0	0.77 x 10-2	0.31 x 10-2	3.34 x 10-2
4	2.46 x 10-2	1.76 x 10	0	0.92 x 10-2	0.22 x 10-2	3.08 x 10-2
5	0	0	3.04 x 10-2	0.345 x 10-2	0.345 x 10-2	2.35 x 10-2
6	0	0	7.58 x 10-2	0.86 x 10-2	0.86 x 10-2	5.86 x 10-2

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

[H₂(g)]_eq = [I₂(g)]_eq

Knowing the above facts, in order to establish a relationship between

concentrations of the reactants and products, several combinations can be tried. Let us consider the simple expression,

[HI(g)]_eq / [H₂(g)] _eq [I₂(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,

[HI(g)]_eq² / [H₂(g)]_eq [I₂(g)]_eq

Table 7.3 Expression Involving the Equilibrium Concentration of Reactants

Experiments Number	[HI(g)]eq [H2(g)]eq[I2(g)]eq	[HI(g)]2eq [H2(g)]eq[I2(g)]eq
1	1840	46.4
2	1610	47.6
3	1400	46.7
4	1520	46.9
5	1970	46.4
6	790	46.4

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  , following equation 7.1, the equilibrium constant K_c is written as,

K_c = [HI(g)]_eq² / [H₂g)]_eg [I₂(g)]_eg ———————————————————(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,

K_c = [HI(g)]² / [H₂(g)] [I₂(g)] —————————————————————-(7.3)

The subscript ‘c’ indicates that K_c is expressed in concentrations of mol L^-1.

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,

is expressed as,

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,

is written as K_c = [NO]⁴[H₂O]⁶ / [NH₃]⁴ [O₂]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,

———————————————————————————————–(7.5)

as, K_c = [HI]² / [H₂] [I₂] = x ————————————————————————————————–(7.6)

The equilibrium constant for the reverse reaction,  at the same temperature is,

K′_c = [H₂] [I₂] / [HI]² = 1/ x = 1 / K_c————————————————————————————(7.7)

Thus, 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,

the equilibrium constant for the above reaction is given by

K″_c = [HI] / [H₂]^1/2[I₂]^1/2 = {[HI]² / [H₂][I₂]}^1/2 = x^1/2 = K_c^1/2————————————————————————————(7.10)

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

Therefore, equilibrium constant for the reaction is equal to K_cⁿ. These findings are summarised in Table 7.4. It should be noted that because the equilibrium constants K_c and 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.

Table 7.4 Relations between Equilibrium Constants for a General Reaction and its Multiples.

Problem

The following concentrations were obtained for the formation of NH₃ from N₂ and H₂ at equilibrium at 500K. [N₂] = 1.5 x10^-2M. [H₂] = 3.0 x10^-2 M and [NH₃] = 1.2 x 10^-2M. Calculate equilibrium constant.

Solution

The equilibrium constant for the reaction,  can be written as,

K_c = [NH₃(g)]²/[N₂(g)][H₂(g)]³

= (1.2 x 10^-2)²/(1.5×10^-2)(3.0×10^-2)³

= 0.106 x 10⁴ = 1.06 x 10³

Problem 7.2

At equilibrium, the concentrations of N₂=3.0 x10–^-3M, O₂ = 4.2 x 10^-3M and NO= 2.8 x10^-3M in a sealed vessel at 800K. What will be K_c for the reaction

Solution

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

K_c = [NO]²/[N₂][O₂]

= (2.8 x 10^-3)²/(3.0×10^-3M)(4.2×10^-3M)

= 0.622

HOMOGENEOUS EQUILIBRIA

In a homogeneous system, all the reactants and products are in the same phase. For example, in the gaseous reaction,  , reactants and products are in the homogeneous phase. Similarly, for the reactions,

all the reactants and products are in homogeneous solution phase. We shall now consider equilibrium constant for some homogeneous reactions.

 Equilibrium Constant in Gaseous Systems

So far we have expressed equilibrium constant of the reactions in terms of molar concentration of the reactants and products, and used symbol, Kc for it. For reactions involving gases, however, it is usually more convenient to express the equilibrium constant in terms of partial pressure.

The ideal gas equation is written as,

pV = nRT

⇒ p = (n/v) RT

Here, p is the pressure in Pa, n is the number of moles of the gas, V is the volume in m³and T is the temperature in Kelvin

Therefore,

n/V is concentration expressed in mol/m³

If concentration c, is in mol/L or mol/dm³, and p is in bar then

p = cRT,

We can also write p = [gas]RT.

Here, R= 0.0831 bar litre/mol K

At constant temperature, the pressure of the gas is proportional to its concentration i.e., p ∝ [gas]

For reaction in equilibrium

We can write either

K_c = [HI(g)]²/[H₂(g)][I₂(g)]

or K_c = (P_HI)²/(P_H2)(P_I2) ———————————————————————————————(7.12)

Further, since P_HI = [HI(g)]RT

P_I2 = [I₂(g)] RT

P_H2 = [H₂(g)] RT

Therefore,

K_p = (P_HI)²/(P_H2)(P_I2) = [HI(g)]² [RT]²/[H₂(g)] RT. [I₂(g)] RT

= [HI(g)]²/[H₂(g)][I₂(g)] = K_c—————————————————————————————-(7.13)

In this example, K_p = K_c i.e., both equilibrium constants are equal. However, this is not always the case. For example in reaction

K_p = (P_NH3)²/(P_N2)(P_H2)³

= [NH₃(g)]²[RT]²/[N₂(g)]RT . [H₂(g)]³ (RT)³

=[NH₃(g)]²[RT]^-2/[N₂(g)]RT . [H₂(g)]³ = K_c(RT)^-2

or K_p = K_c (RT)^-2 ————————————————————————-(7.14)

Similarly, for a general reaction

K_p = (p_C^c)(p_D^d)/(p_A^a)(p_B^b)

= [C]^c [D]^d (RT)^c+d/[A]^a [B]^b (RT)^a+b

= [C]^c [D]^d /[A]^a [B]^b (RT)^{(c+d) – (a+b)}

= [C]^c [D]^d/[A]^a [B]^b (RT)^Δn = K_c (RT)^Δn—————————————————————————————(7.15)

where Δn = (number of moles of gaseous products) – (number of moles of gaseous reactants) in the balanced chemical equation. (It is necessary that while calculating the value of K_p, pressure should be expressed in bar as standard state is 1bar). We have known from Unit 1,

1pascal, Pa=1Nm^-2, and 1bar = 105 Pa

K_p values for a few selected reactions at different temperatures are given in Table 7.5

Table 7.5 Equilibrium Constants, K_p for a Few Selected Reactions

Problem

PCl₅, PCl₃ and Cl₂ are at equilibrium at 500 K and having concentration 1.59M PCl₃, 1.59M Cl₂ and 1.41 M PCl₅. Calculate K_c for the reaction,

Solution

The equilibrium constant K_c for the above reaction can be written as,

K_c = [PCl₅][Cl₂]/[PCl₅] = (1.59)²/(1.41) = 1.79

Problem

The value of K_c = 4.24 at 800K for the reaction,

Calculate equilibrium concentrations of CO₂, H₂, CO and H₂O at 800 K, if only CO and H₂O are present initially at concentrations of 0.10M each.

Solution

For the reactio

where x is the amount of CO₂ and H₂ at equilibrium.

Hence, equilibrium constant can be written as,

K_c = x²/(0.1-x)² = 4.24

x² = 4.24(0.01 + x²-0.2x)

x² = 0.0424 + 4.24x²-0.848x

3.24x² – 0.848x + 0.0424 = 0

a = 3.24, b = – 0.848, c = 0.0424

(for quadratic equation ax² + bx + c = 0,

x = ( -b ± √(b2 – 4ac) )/2a

x = 0.848±√(0.848)²- 4(3.24)(0.0424)/(3.24 x 2)

x = (0.848 ± 0.4118)/ 6.48

x₁ = (0.848 – 0.4118)/6.48 = 0.067

x₂ = (0.848 + 0.4118)/6.48 = 0.194

the value 0.194 should be neglected because it will give concentration of the

reactant which is more than initial concentration.

Hence the equilibrium concentrations are,

[CO₂] = [H_2-] = x = 0.067 M

[CO] = [H₂O] = 0.1 – 0.067 = 0.033 M

Problem

For the equilibrium,

the value of the equilibrium constant, K_c is 3.75 x 10^-6 at 1069 K. Calculate the K_p for the reaction at this temperature?

Solution

We know that,

K_p = K_c(RT)^Δn

For the above reaction,

Δn = (2+1) – 2 = 1

K_p = 3.75 x 10^-6 (0.0831 x 1069)

K_p = 0.033

HETEROGENEOUS EQUILIBRIA

Equilibrium in a system having more than one phase is called heterogeneous equilibrium. The equilibrium between water vapour and liquid water in a closed container is an example of heterogeneous equilibrium.

In this example, there is a gas phase and a liquid phase. In the same way, equilibrium between a solid and its saturated solution,

is a heterogeneous equilibrium.

Heterogeneous equilibria often involve pure solids or liquids. We can simplify equilibrium expressions for the heterogeneous equilibria involving a pure liquid or a pure solid, as the molar concentration of a pure solid or liquid is constant (i.e., independent of the amount present). In other words if a substance ‘X’ is involved, then [X(s)] and [X(l)] are constant, whatever the amount of ‘X’ is taken. Contrary to this, [X(g)] and [X(aq)] will vary as the amount of X in a given volume varies. Let us take thermal dissociation of calcium carbonate which is an interesting and important example of heterogeneous chemical equilibrium.

On the basis of the stoichiometric equation, we can write,

K_c = [CaO(s)][CO₂(g)]/[CaCO₃(s)]

Since [CaCO₃(s)] and [CaO(s)] are both constant, therefore modified equilibrium constant for the thermal decomposition of calcium carbonate will be

K´_c = [CO₂(g)] ———————————————————————————–(7.17)

or K_c = p_CO2 ———————————————————————————–(7.18)

Units of Equilibrium Constant

The value of equilibrium constant Kc can be calculated by substituting the concentration terms in mol/L and for Kp partial pressure is substituted in Pa, kPa, bar or atm. This results in units of equilibrium constant based on molarity or pressure, unless the exponents of both the numerator and denominator are same. For the reactions,  , Kc and Kp have no unit.  , Kc has unit mol/L and Kp has unit bar Equilibrium constants can also be expressed as dimensionless quantities if the standard state of reactants and products are specified. For a pure gas, the standard state is 1bar. Therefore a pressure of 4 bar in standard state can be expressed as 4 bar/1 bar = 4, which is a dimensionless number. Standard state (c0) for a solute is 1 molar solution and all concentrations can be measured with respect to it. The numerical value of equilibrium constant depends on the standard state chosen. Thus, in this system both Kp and Kc are dimensionless quantities but have different numerical values due to different standard states.

This shows that at a particular temperature, there is a constant concentration or pressure of CO₂ in equilibrium with CaO(s) and CaCO₃(s). Experimentally it has been found that at 1100 K, the pressure of CO₂ in equilibrium with CaCO₃(s) and CaO(s), is 2.0 105 Pa. Therefore, equilibrium constant at 1100K for the above reaction is:

K_p = P_CO2 = 2 ×10⁵ Pa/10⁵Pa = 2.00

Similarly, in the equilibrium between nickel, carbon monoxide and nickel carbonyl (used in the purification of nickel),

the equilibrium constant is written as

K_c = [Ni(CO)₄]/[CO]⁴

It must be remembered that in heterogeneous equilibrium pure solids or liquids must be present (however small the amount may be) for the equilibrium to exist, but their concentrations or partial pressure do not appear in the expression of the equilibrium constant. In the reaction,

K_c = [AgNO₃]²/[HNO₃]²

Problem

The value of K_p for the reaction,

is 3.0 at 1000 K. If initially P_CO2= 0.48 bar and P_CO = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO₂.

Solution

For the reaction,

let ‘x’ be the decrease in pressure of CO₂, then

K_p = P²_CO/P_CO²

K_p = (2x)²/(0.48 – x) = 3

4x² = 3(0.48 – x)

4x² = 1.44 – x

4x² + 3x – 1.44 = 0

a = 4, b = 3, c = -1.44

x = ( -b ± √(b2 – 4ac) )/2a

= [-3 ± √(3)² - 4(4)(-1.44)]/2 x 4

= (-3 ± 5.66)/8

= (-3 + 5.66)/ 8 (as value of x cannot be negative hence we neglect that value)

x = 2.66/8 = 0.33

The equilibrium partial pressures are,

P_CO = 2x = 2 0.33 = 0.66 bar

P_CO2 = 0.48 – x = 0.48 – 0.33 = 0.15 bar

APPLICATIONS OF EQUILIBRIUM CONSTANTS

Before considering the applications of equilibrium constants, let us summarise the important features of equilibrium constants as follows:

1. Equilibrium constant is applicable only when concentrations of the reactants and products have attained their equilibrium state.

2. The value of equilibrium constant is independent of initial concentrations of the reactants and products.

3. Equilibrium constant is temperature dependent having one unique value for a particular reaction represented by a balanced equation at a given temperature.

4. The equilibrium constant for the reverse reaction is equal to the inverse of the equilibrium constant for the forward reaction.

5. The equilibrium constant K for a reaction is related to the equilibrium constant of the corresponding reaction, whose equation is obtained by multiplying or dividing the equation for the original reaction by a small integer.

Let us consider applications of equilibrium constant to:

• predict the extent of a reaction on the basis of its magnitude,

• predict the direction of the reaction, and

• calculate equilibrium concentrations.

 Predicting the Extent of a Reaction

The numerical value of the equilibrium constant for a reaction indicates the extent of the reaction. But it is important to note that an equilibrium constant does not give any information about the rate at which the equilibrium is reached. The magnitude of K_c or K_pis directly proportional to the concentrations of products (as these appear in the numerator of equilibrium constant expression) and inversely proportional to the concentrations of the reactants (these appear in the denominator). This implies that a high value of K is suggestive of a high concentration of products and vice-versa.

We can make the following generalisations concerning the composition of equilibrium mixtures:

• If K_c > 10³, products predominate over reactants, i.e., if K_c is very large, the reaction proceeds nearly to completion. Consider the following examples:

(a) The reaction of H₂ with O₂ at 500 K has a very large equilibrium constant ,

K_c = 2.4 x 10⁴⁷.

(b)  at 300K has K_c = 4.0 x 10³¹.

(c)  at 300 K, K_c = 5.4 c 10¹⁸

• If K_c < 10^-3, reactants predominate over products, i.e., if K_c is very small, the reaction proceeds rarely. Consider the following examples:

(a) The decomposition of H₂O into H₂ and O₂ at 500 K has a very small equilibrium constant, K_c = 4.1 x 10^-48

(b)  , at 298 K has K_c = 4.8 x 10³¹.

• If K_c is in the range of 10^-3 to 10³, appreciable concentrations of both reactants and products are present. Consider the following examples:

(a) For reaction of H₂ with I₂ to give HI, K_c = 57.0 at 700K.

(b) Also, gas phase decomposition of N₂O₄ to NO₂ is another reaction with a value of K_c = 4.64 x 10^-3 at 25°C which is neither too small nor too large. Hence, equilibrium mixtures contain appreciable concentrations of both N₂O₄ and NO₂.

These generarlisations are illustrated in Fig. 7.6

7.6.2 Predicting the Direction of the Reaction

The equilibrium constant helps in predicting the direction in which a given reaction will proceed at any stage. For this purpose, we calculate the reaction quotient Q. The reaction quotient, Q (Q_c with molar concentrations and Q_p with partial pressures) is defined in the same way as the equilibrium constant K_c except that the concentrations in Q_c are not necessarily equilibrium values. For a general reaction:

Q_c = [C]^c[D]^d / [A]^a[B]^b——————————————————————————————–(7.20)

Then,

If Q_c > K_c, the reaction will proceed in the direction of reactants (reverse reaction).

If Q_c < K_c, the reaction will proceed in the direction of the products (forward reaction).

If Q_c = K_c, the reaction mixture is already at equilibrium.

Consider the gaseous reaction of H₂ with I₂,

 ; K_c = 57.0 at 700 K.

Suppose we have molar concentrations [H₂]_t=0.10M, [I₂]_t = 0.20 M and [HI]_t = 0.40 M. (the subscript t on the concentration symbols means that the concentrations were measured at some arbitrary time t, not necessarily at equilibrium).

Thus, the reaction quotient, Q_c at this stage of the reaction is given by,

Q_c = [HI]²_t/[H₂]_t [I₂]_t = (0.40)²/ (0.10)x(0.20) = 8.0

Now, in this case, Q_c (8.0) does not equal K_c (57.0), so the mixture of H₂(g), I₂(g) and HI(g) is not at equilibrium; that is, more H₂(g) and I₂(g) will react to form more HI(g) and their concentrations will decrease till Q_c = K_c.

The reaction quotient, Q_c is useful in predicting the direction of reaction by comparing the values of Q_c and K_c.

Thus, we can make the following generalisations concerning the direction of the reaction (Fig. 7.7) :

• If Q_c < K_c, net reaction goes from left to right

• If Q_c > K_c, net reaction goes from right to left.

• If Q_c = K_c, no net reaction occurs.

Problem

The value of K_c for the reaction  is 2 x 10^-3. At a given time, the composition of reaction mixture is [A] = [B] = [C] = 3 x 10^-4 M. In which direction the reaction will proceed?

Solution

For the reaction the reaction quotient Q_c is given by,

Q_c = [B][C]/ [A]

as [A] = [B] = [C] = 3 x10^-4M

Q_c = (3 x 10^-4)(3 x 10^-4) / (3 x 10^-4)2 = 1

as Q_c > K_c so the reaction will proceed in the reverse direction.

 Calculating Equilibrium Concentrations

In case of a problem in which we know the initial concentrations but do not know any of the equilibrium concentrations, the following three steps shall be followed:

Step 1. Write the balanced equation for the reaction.

Step 2. Under the balanced equation, make a table that lists for each substance involved in the reaction:

(a) the initial concentration,

(b) the change in concentration on going to equilibrium, and

(c) the equilibrium concentration.

In constructing the table, define x as the concentration (mol/L) of one of the substances that reacts on going to equilibrium, then use the stoichiometry of the reaction to determine the concentrations of the other substances in terms of x.

Step 3. Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve for x. If you are to solve a quadratic equation choose the mathematical solution that makes chemical sense.

Step 4. Calculate the equilibrium concentrations from the calculated value of x.

Step 5. Check your results by substituting them into the equilibrium equation.

13.8g of N₂O₄ was placed in a 1L reaction vessel at 400K and allowed to attain equilibrium

The total pressure at equilbrium was found to be 9.15 bar. Calculate K_c, K_p and partial pressure at equilibrium.

Solution

We know pV = nRT

Total volume (V ) = 1 L

Molecular mass of N₂O₄ = 92 g

Number of moles = 13.8g/92 g = 0.15 of the gas (n)

Gas constant (R) = 0.083 bar L mol^-1K^-1

Temperature (T ) = 400 K

pV = nRT

p 1L = 0.15 mol x 0.083 bar L mol^-1K^-1 x 400 K

p = 4.98 bar

Initial pressure: 4.98 bar 0

At equilibrium: (4.98 – x) bar 2x bar

Hence,

p_total at equilibrium = p_N2O4 + NO₂

9.15 = (4.98 – x) + 2x

9.15 = 4.98 + x

x = 9.15 – 4.98 = 4.17 bar

Partial pressures at equilibrium are,

p_N2O4 = 4.98 – 4.17 = 0.81bar

p_NO2 = 2 x 4.17 = 8.34 bar

K_p=(p_NO2)²/p_N2O4

= (8.34)²/0.81 = 85.87

K_p = K_c(RT)^Δn

85.87 = K_c(0.083 x 400)¹

K_c = 2.586 = 2.6

3.00 mol of PCl₅ kept in 1L closed reaction vessel was allowed to attain equilibrium at 380K. Calculate composition of the mixture at equilibrium. K_c= 1.80

Solution

K_c = [PCl₃][Cl₂]/[PCl₅]

1.8 = x²/ (3 – x)

x² + 1.8x – 5.4 = 0

x = [-1.8 ± √(1.8)2 - 4(-5.4)]/2

x = [-1.8 ± √3.24 + 21.6]/2

x = [-1.8 ± 4.98]/2

x = [-1.8 + 4.98]/2 = 1.59

[PCl₅] = 3.0 – x = 3 -1.59 = 1.41 M

[PCl₃] = [Cl₂] = x = 1.59 M

RELATIONSHIP BETWEEN QUILIBRIUM CONSTANT K, REACTION QUOTIENT Q AND GIBBS ENERGY G

The value of K_c for a reaction does not depend on the rate of the reaction. However, as you have studied in Unit 6, it is directly related to the thermodynamics of the reaction and in particular, to the change in Gibbs energy, ΔG. If,

• ΔG is negative, then the reaction is spontaneous and proceeds in the forward

direction.

• ΔG is positive, then reaction is considered non-spontaneous. Instead, as reverse reaction would have a negative ΔG, the products of the forward reaction shall be converted to the reactants.

• ΔG is 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.

A mathematical expression of this thermodynamic view of equilibrium can be

described by the following equation:

ΔG = ΔG^Θ + RT lnQ —————————————————————————————-(7.21)

where, G^Θ is standard Gibbs energy.

At equilibrium, when ΔG = 0 and Q = K_c, the equation (7.21) becomes,

ΔG = ΔG^Θ + RT ln K = 0

ΔG^Θ = – RT lnK —————————————————————————————(7.22)

lnK = – ΔG^Θ / RT

Taking antilog of both sides, we get,

K = e^{– ΔGΘ} / RT ————————————————————————————– (7.23)

Hence, using the equation (7.23), the reaction spontaneity can be interpreted in terms of the value of ΔG^Θ.

• If ΔG^Θ < 0, then -ΔG^Θ/RT is positive, and e^{– ΔGΘ} /RT >1, making K >1, which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.

• If ΔG^Θ > 0, then -ΔG^Θ/RT is negative, and e^{– ΔGΘ} /RT < 1, that is , K < 1, which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.

The value of ΔG^Θfor the phosphorylation of glucose in glycolysis is 13.8 kJ/mol. Find the value of Kc at 298 K.

Solution

ΔG^Θ = 13.8 kJ/mol = 13.8 x 10³J/mol

Also, ΔG^Θ = – RT lnK_c

Hence, ln K_c = -13.8 x 10³J/mol (8.314 J mol^-1K^-1 298 K)

ln K_c = – 5.569

K_c = e^-5.569

K_c = 3.81 x 10^-3

Problem 7.11

Hydrolysis of sucrose gives,

Equilibrium constant K_c for the reaction is 2 x 10¹³ at 300K. Calculate ΔG^Θ at 300K.

Solution

ΔG^Θ = – RT lnKc

ΔG^Θ = – 8.314J mol^-1K^-1 x 300K ln(21013)

ΔG^Θ = – 7.64 104 J mol^-1

Le Chatelier's Principle

Le Chatelier's principle states that when a system in chemical equilibrium is disturbed by a change of temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that tends to counteract this change of variable. The three ways that Le Chatelier's principle says you can affect the outcome of the equilibrium are as follows:

• Changing concentrations by adding or removing products or reactants to the reaction vessel.
• Changing partial pressure of gaseous reactants and products.
• Changing the temperature.

These actions change each equilibrium differently, therefore you must determine what needs to happen for the reaction to get back in equilibrium.

Example involving change of concentration:

In the equation

2NO_(g) + O_2(g) <--> 2NO_2(g)

If you add more NO_(g) the equilibrium shifts to the right producing more NO_2(g)

If you add more O_2(g) the equilibrium shifts to the right producing more NO_2(g)

If you add more NO_2(g) the equilibrium shifts to the left producing more NO_(g) and O_2(g)

Example involving pressure change:

In the equation

2SO_2(g) + O_2(g) <--> 2SO_3(g),

an increase in pressure will cause the reaction to shift in the direction that reduces pressure, that is the side with the fewer number of gas molecules. Therefore an increase in pressure will cause a shift to the right, producing more product. (A decrease in volume is one way of increasing pressure.)

Example involving temperature change:

In the equation

N_2(g) + 3H_2(g) <--> 2NH₃ + 91.8 kJ,

an increase in temperature will cause a shift to the left because the reverse reaction uses the excess heat. An increase in forward reaction would produce even more heat since the forward reaction is exothermic. Therefore the shift caused by a change in temperature depends upon whether the reaction is exothermic or endothermic.