1205:Chemical Equilibrium

Chemical Equilibrium

Reversible Reactions

Not all chemical reactions proceed to completion. In most reactions, two or more substances react to form products which themselves react to give back the original substances. 
Thus, A and B may react to form C and D which react together to reform A and B.

A + B → C + D (Forward reaction)

A + B ← C + D (Reverse reaction)

   

Definition: A reaction which can go in the forward and backward direction simultaneously is called a reversible reaction. Such a reaction is represented by writing a pair of 
arrows between the reactants and products.

A + B ⇌C + D

The arrow pointing right indicates the forward reaction, while that pointing left shows the reverse reaction.

A few common examples of reversible reactions are listed below:

2NO2 (g) ⇌ N2O4 (g)

H2 (g) + I2 (g) ⇌2HI (g)

PCl5 (s) ⇌ PCl3 (s) + Cl2 (g)

CaCO3 (s) ⇌ CaO (s) + CO2 (g)   

Chemical Equilibrium

Let us consider the reaction,

A + B ⇌ C + D

If we start with reactants A and B in a closed vessel, the forward reaction will proceed to form products C and D. The concentrations of A and B decrease and those of C and D
 increase continuously. As a result, the rate of forward reaction also decreases and the rate of the reverse reaction increases. Eventually, the rate of the two opposing reactions 
equals and the system attains a state of equilibrium.

Thus, chemical equilibrium may be defined as: the state of a reversible reaction when the two opposing reactions occur at the same rate and the concentrations of 
reactants and products do not change with time.

Figure: At equilibrium the forward reaction rate equals the reverse reaction rate.

Types of Chemical Equilibrium

Depending on the phases of reaction species, chemical equilibrium can be divided into two types. They are-

1. Homogeneous Equilibria: The equilibrium reaction in which all the reactants and products remain in the same phase is called a homogeneous equilibrium. For example,

i. H2 (g) + I2 (g) ⇌ 2HI (g),         ii. N2O4(g) ⇌ 2NO2(g)

iii. CH3COOH(l) + C2H5OH(l) ⇌CH3COOC2H5(l) + H2O(l)

2. Heterogeneous Equilibria: The equilibrium reaction in which all the reactants and products are not in the same phase is called a heterogeneous equilibrium. For example,

i. CaCO3 (s) ⇌ CaO (s) + CO2 (g)         ii. NH4Cl(s) ⇌NH3(g) + HCl(g)

iii. 3Fe(s) + 4H2O(g) ⇌Fe3O4(s) + 4H2(g)

A more formal look at dynamic equilibria

Thinking about reaction rates

This is the equation for a general reaction which has reached dynamic equilibrium:

A + B ⇌ C + D

How did it get to that state? Let's assume that we started with A and B. At the beginning of the reaction, the concentrations of A and B were at their maximum. That 
means that the rate of the reaction was at its fastest.

As A and B react, their concentrations fall. That means that they are less likely to collide and react, and so the rate of the forward reaction falls as time goes on.

In the beginning, there isn't any C and D, so there can't be any reaction between them. As time goes on, though, their concentrations in the mixture increase and they 
are more likely to collide and react.

With time, the rate of the reaction between C and D increases:

Eventually, the rates of the two reactions will become equal. A and B will be converting into C and D at exactly the same rate as C and D convert back into A and B 
again.

At this point there won't be any further change in the amounts of A, B, C and D in the mixture. As fast as something is being removed, it is being replaced again by the 
reverse reaction. We have reached a position of dynamic equilibrium.

Furthermore, the true equilibrium of a reaction can be attained from both sides. Thus, the equilibrium concentrations of the reactants and products are the same 
whether we start with A and B or, C and D.

Chemical Equilibrium is Dynamic Equilibrium

We know that when a reaction say  A + B ⇌ C + D attains equilibrium, the concentrations of A and B as also of C and D remain constant with time. Apparently it appears that 
the equilibrium is dead. But it is not so. The equilibrium is dynamic. Actually, the forward and the reverse reactions are taking place at equilibrium but the concentrations remain 
unchanged.

Figure: Molecules of A and B colliding to give C and D, and those of C and D colliding to give A and B.

The dynamic nature of chemical equilibrium can be easily understood on the basis of the kinetic molecular model. The molecules of A and B in the equilibrium mixture collide 
with each other to form C and D. Likewise, C and D collide to give back A and B. The collision of molecules in a closed system is a ceaseless phenomenon. Therefore, 
collisions of A and B giving C and D (Forward reaction) and collisions of C and D giving back A and B (reverse reaction) continue to occur even at equilibrium while concentrations 
remain unchanged.

Characteristics of Chemical Equilibrium

(1) Constancy of concentrations

When a chemical equilibrium is established in a closed vessel at constant temperature, concentrations of the various species in the reaction mixture become constant.

The reaction mixture at equilibrium is called equilibrium mixture.

The concentrations at equilibrium are called equilibrium concentrations. The equilibrium concentrations are represented by square brackets [ ]. Thus, [A] denotes the equilibrium 
concentration of substance A in moles per liter.

(2) Equilibrium can be initiated from either side

The state of equilibrium of a reversible reaction can be approached whether we start with reactants or products. For example, the equilibrium is established if we start the reaction 
with H2 and I2 or, 2HI.

H2(g) + I2(g) ⇌ 2HI(g)

Figure: The left graph depicts the attainment of equilibrium for start with 2HI. The right graph shows how equilibrium is attained for start with H and I. Equilibrium concentrations in both 
cases are the same.

(3) Equilibrium cannot be attained in an open vessel

The equilibrium can be established only if the reaction vessel is closed and no part of the reactants or products is allowed to escape out. In an open vessel, the gaseous reactants 
and/or products may escape into the atmosphere leaving behind no possibility of attaining equilibrium.

However, the equilibrium can be attained when all the reactants and products are in the same phase. e.g. ethanol and ethanoic acid.

(4) A catalyst cannot change the equilibrium point

When a catalyst is added to a system in equilibrium, it speeds up the rate of both the forward and the reverse reaction to an equal extent. Therefore, a catalyst cannot change the 
equilibrium point except that it is achieved earlier. This enhances the rate of the reaction.

(5) Value of equilibrium constant does not depend upon the initial concentration of reactants

It has been found that equilibrium constant must be the same when the concentrations of reacting species are varied over a wide range.

(The equilibrium constant of a chemical reaction is the value of the reaction quotient when the reaction has reached equilibrium. The equilibrium constant always has the 
same value if the temp.is constant, irrespective of the amounts of reactants)    

Law of Mass Action

Two Norwegian chemists, Guldberg and Waage, studied experimentally a large number of equilibrium reactions. In 1864, they postulated a generalisation called the Law of Mass 
action.
It states that: the rate of a chemical reaction is proportional to the active masses of the reactants.

Here the term ‘active mass’ means the molar concentration i.e., number of moles per litre.
It is expressed by enclosing the formula of the substance in square brackets.

Mathematical Expression of Law of Mass Action

Let us consider a general reaction

A + B ⇌ C + D

And let [A], [B], [C] and [D] represent the molar concentrations of A, B, C and D at the equilibrium point.

According to the Law of Mass action,

Rate of forward reaction ∝ [A][B] = k1[A] [B]

Rate of reverse reaction  ∝ [C][D] = k2[C][D]

Where, k1 and k2 are rate constants for the forward and reverse reactions.

At equilibrium, rate of forward reaction = rate of reverse reaction.

Therefore,  k1[A][B] = k2[C][D]

At any specific temperature k1/k2 is constant since both k1 and k2 are constants. The ratio k1/k2 is called equilibrium constant and is represented by the symbol Kc or simply k. The 
subscript ‘c’ indicates that the value is in terms of concentrations of reactants and products.

The equation (1) may be written as;
Consider the following reaction,

        Coefficient of A      2A ⇌C+D

Here, the forward reaction is dependent on the collisions of each of two A molecules.
Therefore, for writing the equilibrium constant expression, each molecule is regarded as a separate entity i.e.,

A + A ⇌ C + D

Then the equilibrium constant expression is

As a general rule, if there are two or more molecules of the same substance in the chemical equation, its concentration is raised to the power equal to the numerical coefficient of 
the substance in the equation.

Equilibrium Constant Expression for a Reaction in General Terms

The general reaction may be written as

aA + bB ⇌ cC + dD

Where, a, b, c and d are numerical coefficients of the substance, A, B, C and D respectively.
The equilibrium constant expression is

Where, Kc is the equilibrium constant. The general definition of the equilibrium constant may thus be stated as: the product of the equilibrium concentrations of the products 
divided by the product of the equilibrium concentrations of the reactants, with each concentration term raised to a power equal to the coefficient of the substance in the 
balanced equation.   

PROBLEM: Give the equilibrium constant expression for the reaction, N2(g) + H2(g) ⇌NH3(g)

PROBLEM: Write the equilibrium constant expression for the reaction,    N2O5(g) ⇌ NO2(g) + O2 (g)

Equilibrium Constant Expression in Terms of Partial Pressures

When all the reactants and products are gases, we can also formulate the equilibrium constant expression in terms of partial pressure. The relationship between the partial pressure
 (p) of any one gas in the equilibrium mixture and the molar concentration follows from the general ideal gas equation

The quantity n/V is the number of moles of the gas per unit volume and is simply the molar concentration. Thus,  p = (Molar concentration)☓RT

i.e., the partial pressure of a gas in the equilibrium mixture is directly proportional to its molar concentration at a given temperature. Therefore, we can write the equilibrium 
constant expression in terms of partial pressure instead of molar concentrations.

For a general reaction,

    l L(g) + mM(g) ⇌ yY(g) + zZ(g)

   

Here, Kp is the equilibrium constant, the subscript p referring to partial pressure. Partial pressures are expressed in atmospheres.

Relationship between Kp And Kc

Let us consider a general reaction,

jA + kB ⇌ l C + mD

Where, j, k, l and m are numerical coefficients of the substance, A, B, C and D respectively.
And let [A], [B], [C] and [D] represent the molar concentrations of A, B, C and D at the equilibrium point.

The equilibrium constant expression is

Where, all reactants and products are gases. We can write the equilibrium constant expression in terms of partial pressures as-

Assuming that all these gases constituting the equilibrium mixture obey the ideal gas equation, the partial pressure (p) of a gas is-

Where, n/V is the molar concentration. Thus, the partial pressures of individual gases A, B, C and D are: PA = [A] RT;     PB = [B] RT;     PC = [C] RT;     PD = [D] RT

Substituting these values in equation (2), we have

                        

Where, ∆n = (l + m) – (j + k), is the difference in the sums of the co-efficients for the gaseous products and reactants. From the expression (3) it is clear that when ∆n = 0, Kp = Kc.

Mathematical Problem

At 500°C, the reaction between N2 and H2 to form ammonia has Kc = 6.0☓10–2. What is the numerical value of Kp for the reaction?

Units of Equilibrium Constant

In the equilibrium expression for a particular reaction, the concentrations are given in units of moles/litre or mol/L, and the partial pressure are given in atmospheres (atm).
The units of Kc and Kp, depend on the specific reaction.

(1) When the total number of moles of reactants and products are equal.

In the equilibrium expression of these reactions, the concentration or pressure terms in the numerator and denominator exactly cancel out. Thus, Kc or Kp for such a reaction is 
without units.

Taking example of the reaction,

H2(g) + I2(g) ⇌2HI(g)

(2) When the total number of moles of the reactants and products are unequal.

In such reactions, KC will have units (mol/litre)n and Kp will have units (atm)n, where n is equal to the total number of moles of products minus the total number of moles of reactants. 
Thus for the reaction,

N2O4(g) ⇌ 2NO2(g)

Thus units for KC are mol/L and KP units are atm.

For the reaction

        N2(g) + 3H2(g) ⇌2NH3(g)

Units for KC are and KP may be found as follows

Thus, units of KC are mol–2 L2 and units of Kp are atm–2.

It may be noted, however, that the units are often omitted when equilibrium constants are listed in tables.

Homogeneous equilibria

Liquid Systems (Liquid equilibria)

The chemical equilibrium in which all the reactants and products are in the liquid phase, are referred to as the liquid equilibria.

Like the gas-phase equilibria, the liquid equilibria are also called homogeneous equilibria.
For example, alcohols and acids react to form esters and water.

    CH3COOH(l) + C2H5OH(l) ⇌  CH3COOC2H5(l) + H2O(l)

Let us start with a mole of acetic acid and b moles of alcohol.

If x moles of acetic acid react with x moles of ethyl alcohol, x moles of ester and x moles of water are produced when the equilibrium is established.

Now the moles present at equilibrium are:

                    CH3COOH = (a – x) moles

                    C2H5OH = (b – x) moles

                    CH3COOC2H5 = x moles

                    H2O = x moles

CH3COOH(l) + C2H5OH(l) ⇌  CH3COOC2H5(l) + H2O(l)

(a – x)        (b – x)            x        x

If V litter be the total volume of the equilibrium mixture, the concentrations of the various species are:

The equilibrium constant expression may be written as                                                   

It may be noted that the volume terms V in the numerator and denominator cancel out. In liquid systems when there is a change in the number of moles as a result of the reaction,
 it is necessary to consider the volume V while calculating the equilibrium constant K.

Heterogeneous  Equilibria

Chemical equilibria in which the reactants and products are not all in the same phase is called heterogeneous equilibria.

Example: The decomposition of calcium carbonate upon heating to form calcium oxide and carbon dioxide.

If the reaction is carried in a closed vessel, the following equilibrium is established.

CaCO3(s) ⇌ CaO(s) + CO2(g)

Le Chatelier’s Principle

In 1884, the French Chemist Henry Le Chatelier proposed a general principle which applies to all systems in equilibrium. The principle states that-

“When a system is at equilibrium a change in any one of the factors upon which the equilibrium depends will cause the equilibrium to shift in a direction such that the effect of
 the change is diminished.”

There are three ways in which the change can be caused on a chemical equilibrium:

(1) Changing the concentration of a reactant or product.

(2) Changing the pressure (or volume) of the system.

(3) Changing the temperature.

Thus, when applied to a chemical reaction in equilibrium, Le Chatelier’s principle can be
stated as: if a change in concentration, pressure or temperature is caused to a chemical reaction in equilibrium, the equilibrium will shift to the right or the left so as to minimize
 the change.

Effect of a Change in Concentration

We can restate Le Chatelier’s principle for the special case of concentration changes: when concentration of any of the reactants or products is changed, the equilibrium shifts in 
a direction so as to reduce the change in concentration that was made.

A change in the concentration of a reactant or product can be effected by the addition or removal of that species. Let us consider a general reaction,

A + B ⇌ C

When a reactant, say, A is added at equilibrium, its concentration is increased. The forward reaction alone occurs momentarily. According to Le Chatelier’s principle, a new 
equilibrium will be established so as to reduce the concentration of A. Thus, the addition of A causes the equilibrium to shift to right. This increases the concentration (yield) of the 
product C.

Following the same line of argument, a decrease in the concentration of A by its removal from the equilibrium mixture, will be undone by shift to the equilibrium position to the left. 
This reduces the concentration (yield) of the product C.

Let us illustrate the effect of change of concentration on a system at equilibrium by taking example of the ammonia synthesis reaction:

N2(g) + 3H2(g) ⇌ 2NH3(g)

When N2 (or H2) is added to the equilibrium already in existence (equilibrium I), the equilibrium will shift to the right so as to reduce the concentration of N2 (Le Chatelier’s principle). 
The concentration of NH3 at the equilibrium II is more than at equilibrium I. The results in a particular case after the addition of one mole/litre are given below.

Obviously, the addition of N2 (a reactant) increases the concentration of NH3, while the concentration of H2 decreases. Thus, to have a better yield of NH3, one of the reactants should
 be added in excess.

Effect of a Change in Pressure

To predict the effect of a change of pressure, Le Chatelier’s principle may be stated as: when pressure is increased on a gaseous equilibrium reaction, the equilibrium will shift in a 
direction which tends to decrease the pressure.

The pressure of a gaseous reaction at equilibrium is determined by the total number of molecules it contains. If the forward reaction proceeds by the reduction of molecules, it will
 be accompanied by a decrease of pressure of the system and vice versa.

Let us consider a reaction,

A + B ⇌ C

The combination of A and B produces a decrease of number of molecules while the decomposition of C into A and B results in the increase of molecules. Therefore, by the increase 
of pressure on the equilibrium it will shift to right and give more C. A decrease in pressure will cause the opposite effect. The equilibrium will shift to the left when C will decompose 
to form more of A and B.

The reactions in which the number of product molecules is equal to the number of reactant molecules, are unaffected by pressure changes.

For example, H2 (g) + I2 (g)⇌ 2HI (g)

In such a case, the system is unable to undo the increase or decrease of pressure.

In light of the above discussion, we can state a general rule to predict the effect of pressure changes on chemical equilibria.

The increase of pressure on a chemical equilibrium shifts it in that direction in which the number of molecules decreases and vice-versa.

Effect of a Change in Temperature

Chemical reactions consist of two opposing reactions. If the forward reaction proceeds by the evolution of heat (exothermic), the reverse reaction occurs by the absorption of 
heat (endothermic).

Both these reactions take place at the same time and equilibrium exists between the two. If temperature of a reaction is raised, heat is added to the system. The equilibrium shifts in a 
direction in which heat is absorbed in an attempt to lower the temperature. Thus, the effect of temperature on an equilibrium reaction can easily be predicted by the following version 
of the Le Chatelier’s principle.

When temperature of a reaction is increased, the equilibrium shifts in a direction in which heat is absorbed.

Let us consider an exothermic reaction,

A + B ⇌ C + heat

When the temperature of the system is increased, heat is supplied to it from outside. According to Le Chatelier’s principle, the equilibrium will shift to the left which involves the 
absorption of heat. This would result in the increase of the concentration of the reactants A and B.

In an endothermic reaction the increase of temperature will shift the equilibrium to the right as it involves the absorption of heat.

X + Y + heat ⇌ Z

This increases the concentration of the product Z.

In general, we can say that the increase of temperature favors the reverse change in an exothermic reaction and the forward change in an endothermic reaction.

Formation of Ammonia from N2 and H2

The synthesis of ammonia from nitrogen and hydrogen is an exothermic reaction.

N2 + 3H2 ⇌ 2NH3 + 22.2 kcal

When the temperature of the system is raised, the equilibrium will shift from right-to-left which absorbs heat (Le Chatelier’s principle).  This results in the lower yield of ammonia. On 
the other hand, by lowering the temperature of the system, the equilibrium will shift to the right which evolves heat in an attempt to raise the temperature. This would increase the 
yield of ammonia. But with decreasing temperature, the rate of reaction is slowed down considerably and the equilibrium is reached slowly. Thus, in the commercial production of 
ammonia, it is not feasible to use temperature much lower than 500°C. At lower temperature, even in the presence of a catalyst, the reaction proceeds too slowly to be practical.

Conditions for Maximum Yield in Industrial Processes

With the help of Le Chatelier’s principle we can work out the optimum conditions for securing the maximum yield of products in industrial processes.

Synthesis of Ammonia (Haber Process)

The manufacture of ammonia by Haber process is represented by the equation-

N2(g) + 3H2(g) ⇌ 2NH3(g) + 22.0 kcal

A look at the equation provides the following information:

(a) The reaction is exothermic

(b) The reaction proceeds with a decrease in the number of moles.

(1) Low temperature. By applying Le Chatelier’s principle, low temperature will shift the equilibrium to the right. This gives greater yield of ammonia. In actual practice a temperature
 of about 450°C is used when the percentage of ammonia in the equilibrium mixture is 15.

(2) High pressure. High pressure on the reaction at equilibrium favors the shift of the equilibrium to the right. This is so because the forward reaction proceeds with a decrease in the 
number of moles. A pressure of about 200 atmospheres is applied in practice.

(3) Catalyst. As already stated, low temperature is necessary for higher yield of ammonia. But at relatively low temperatures, the rate of reaction is slow and the equilibrium is attained
 in a long time.
To increase the rate of reaction and thus quicken the attainment of equilibrium, a catalyst is used. Finely divided iron containing molybdenum is employed in actual practice.

Molybdenum acts as a promoter that increases the life and efficiency of the catalyst.

Pure N2 and H2 gases are used in the process. Any impurities in the gases would poison the catalyst and decrease its efficiency.

Manufacture of Sulphuric Acid (Contact Process)

The chief reaction used in the process is-

2SO2(g) + O2(g) ⇌ 2SO3(g) + 42 kcal

Following information is revealed by the above equation:

(a) The reaction is exothermic.

(b) The reaction proceeds with a decrease in number of moles.

On the basis of Le Chatelier’s principle, the conditions for the maximum yield can be worked out as below:

(1) Low temperature. Since the forward reaction is exothermic, the equilibrium will shift on the right at low temperature. An optimum temperature between 400-450°C is required 
for the maximum yield of sulphur trioxide.

(2) High pressure. Since the number of moles is decreased in the forward reaction, increase of pressure will shift the equilibrium to the right. Thus, for maximum yield of SO3,  
2 to 3 atmosphere pressure is used.

(3) Catalyst. At the low temperature used in the reaction, the rate of reaction is slow and the equilibrium is attained slowly. A catalyst is, therefore, used to speed up the establishment 
of the equilibrium. Vanadium pentoxide, V2O5, is commonly used and it has replaced the earlier catalyst platinum asbestos which was easily poisoned by the impurities present in the
 reacting gases. All samples, SO2 and O2 used for the manufacture of sulphuric acid must be pure and dry.

Manufacture of Nitric Acid (Birkeland-Eyde Process)

Nitric acid is prepared on a large scale by making use of the reaction

N2 (g) + O2 (g) ⇌ 2NO (g) – 43.2 kcal

The equation tells us that:

(a) The reaction proceeds with no change in the number of moles.

(b) The reaction is endothermic and proceeds by absorption of heat.

The favorable conditions for the maximum yield of NO are:

(1) High temperature. Since the forward reaction is endothermic, increase of temperature will favor it (Le Chatelier’s principle). Thus, a high temperature of the order of 3000°C is 
employed to get high yield of nitric acid.

(2) No effect of pressure. Since the forward reaction involves no change in the number of moles, a change in pressure has no effect on the equilibrium.

(3) High concentration. The formation of nitric oxide is favored by using high concentrations of the reactants i.e. N2 and O2.

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