In this article we shall study the application of Le-Chatelier’s principle in Haber’s process, contact proces, etc.

Statement of Le-Chatelier’s Principle:

This principle is given by, a French chemist Le-Chatelier in 1888. It states that “If an external stress is applied to a reacting system at equilibrium, the system will adjust itself in such a way that the effect of the stress is nullified”.

Application of Le-Chatelier’s Principle to Haber’s process (Synthesis of Ammonia):

Ammonia is manufactured by using Haber’s process. In this reaction Nitrogen and Hydrogen in ratio 1:3 by volume are made to react at 773 K and 200 atm. Pressure.

The chemical reaction is

N_2(g)  +   3H_2(g)   ⇌   2 NH_3(g)  + 96.3 kJ

1 Vol         3 Vol              2 Vol

4 Vol                2 Vol

From this reaction it is clear that, the reaction is exothermic and accompanied by the decrease in volume.

Effect of Concentration:

By the law of mass action; the increase in concentration of one of the reactant will shift equilibrium towards right.  And here increase in concentration of hydrogen (as more moles of it are used) in preference to Nitrogen has more effect.

Effect of Pressure:

Above reaction indicate that formation of ammonia takes place with decrease in volume.  Hence increase in pressure will favour forward reaction. Optimum pressure for maximum yield of ammonia is about 200 atm.

Effect of Temperature:

The reaction is exothermic, so lowering the temperature will favour forward reaction. But decrease in temperature results in the decrease in the rate of reaction. Hence temperature of 773K is maintained and iron is used as catalyst.

Application of Le-Chatelier’s Principle to Contact process (Synthesis of Sulphur Trioxide):

H₂SO₄ is manufactured by contact process. In this reaction SO₂ is oxidized to SO₃. Sulphur trioxide is further used for manufacturing of sulphuric acid.

2SO_2(g) + O_2(g)   ⇌   2 SO_3(g) +  189 kJ

2 Vol           1 Vol          2 Vol

From this reaction it is clear that, the reaction is exothermic and accompanied by the decrease in volume.

Effect of Concentration:

By the law of mass action; increase in concentration of one of the reactant will shift equilibrium towards right.  And here increase in concentration of sulphur dioxide in preference to oxygen has more effect.

Effect of Pressure:

Above reaction indicate that formation of sulphur trioxide takes place with decrease in volume.  Hence increase in pressure will favour forward reaction. Optimum pressure for maximum yield of sulphur trioxide is about 1.5 atm to 1.7 atm

Effect of Temperature:

The reaction is exothermic, so lowering the temperature will favour forward reaction. But decrease in temperature results in the decrease in the rate of reaction. Hence temperature of 723 K is maintained and vanadium pentoxide is used as catalyst.

Application of Le-Chatelier’s Principle to  Manufacture of Ozone:

Ozone is manufactured by passing silent electric discharge through pure oxygen.

3O_2(g)   ⇌  2 O_3(g) –  288.56 kJ

3 Vol               2 Vol

From this reaction it is clear that, the reaction is endothermic and accompanied by the decrease in volume.

Effect of Concentration:

By the law of mass action; increase in concentration of one of the reactant will shift equilibrium towards right.  And here increase in concentration of oxygen increases the rate of forward reaction.

Effect of Pressure:

Above reaction indicate that formation of ozone takes place with decrease in volume.  Hence increase in pressure will favour forward reaction.

Effect of Temperature:

The reaction is endothermic, so increasing the temperature will favour forward reaction. Due to increase in temperature heat will be absorbed by the reaction.

Application of Le-Chatelier’s Principle to  Manufacture of Nitric oxide:

The reaction is

N_2(g) +  O_2(g)   ⇌  2 NO_(g) –  181 kJ

1Vol  + 1 Vol        →       2 Vol

From this reaction it is clear that, the reaction is endothermic and accompanied by no change in volume.

Effect of Concentration:

By the law of mass action; increase in concentration of one of the reactant will shift equilibrium towards right.  And here increase in concentration of nitrogen or oxygen increases the rate of forward reaction.

Effect of Pressure:

Above reaction indicate that formation of nitric oxide takes place with no change in volume.  Hence pressure has no effect on the equilibrium.

Effect of Temperature:

The reaction is endothermic, so increasing the temperature will favour forward reaction. Due to increase in temperature heat will be absorbed by the reaction.

Application of Le-Chatelier’s Principle to Manufacture of Nitrogen dioxide:

The reaction is

2NO_(g) +  O_2(g)   ⇌  2 NO_2(g) +  116.4 kJ

2Vol  + 1 Vol        →       2 Vol

From this reaction it is clear that, the reaction is exothermic and accompanied by decrease in volume.

Effect of Concentration:

By the law of mass action; increase in concentration of one of the reactant will shift equilibrium towards right.  And here increase in concentration of nitrogen or oxygen increases the rate of forward reaction. Due to use of more number of moles, the increase in concentration of nitrogen oxide has a prominent effect.

Effect of Pressure:

Above reaction indicate that formation of nitrogen dioxide takes place with decrease in volume.  Hence increase in pressure favour forward reaction.

Effect of Temperature:

The reaction is exothermic, so decreasing the temperature will favour forward reaction. Due to the decrease in temperature heat will be removed from the reaction.

Application of Le-Chatelier’s Principle to Dissociation of Phosphorous Pentachloride:

The reaction is

PCL_5(g)   ⇌  PCl_3(g)  +  Cl_2(g)    –  62.8kJ

1Vol  → 1 Vol      +    1 Vol

From this reaction it is clear that, the reaction is endothermic and accompanied by increase in volume.

Effect of Concentration:

By the law of mass action; increase in concentration of one of the reactant will shift equilibrium towards right.  And here increase in concentration of phosphorous pentachloride increases the rate of forward reaction.

Effect of Pressure:

Above reaction indicate that formation of nitrogen dioxide takes place with increase in volume.  Hence decrease in pressure favour forward reaction.

Effect of temperature:

The reaction is endothermic, so increasing the temperature will favour forward reaction. Due to the increase in temperature heat will be absorbed by the reaction.

Applications of Le-Chatelier’s Principle to Physical Equilibria:

Melting of Ice:

The reaction is

_Ice(s)   ⇌  Water_(l)     –  6.01 kJ

more vol  → less vol

From this reaction, it is clear that the reaction is endothermic and accompanied by decrease in volume.

Effect of Pressure:

Above reaction indicate that formation of liquid water takes place with decrease in volume.  Hence increase in pressure favour forward reaction.

Effect of Temperature:

The reaction is endothermic, so increasing the temperature will favour forward reaction. Due to the increase in temperature heat will be absorbed by the reaction.

Boiling of Water:

The reaction is

_Water(l)   ⇌  Water vapours_(g)     –  40.84 kJ

less vol  → more vol

From this reaction, it is clear that the reaction is endothermic and accompanied by increase in volume.

Effect of Pressure:

Above reaction indicate that formation of liquid water takes place with increase in volume.  Hence increase in pressure favour forward reaction. (Principle of pressure cooker).

Effect of Temperature:

The reaction is endothermic, so increasing the temperature will favour forward reaction. Due to the increase in temperature heat will be absorbed by the reaction.

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In this article, we shall study Le-Chatelier’s principle with examples.

Statement:

This principle is given by, a French chemist Le-Chatelier in 1888. It states that “If an external stress is applied to a reacting system at equilibrium, the system will adjust itself in such a way that the effect of the stress is nullified”.

Explanation:

Let us consider a general reversible reaction.

A  +  B  ⇌   C + D

The equilibrium constant for the reaction is given by

If the concentration of any one reactant say A is increased then by Le-Chatelier’s Principle the forward reaction should be favoured so that the increase in the concentration of A is nullified. It can be explained as follows. As the concentration of reactant, A increases the denominator of mass equation increases. To keep the value of the equilibrium constant the same the value of K_c constant the numerator should increase. This is possible only if the concentration of C and D is increased. It is possible if more and more C and D are formed thus the reaction proceeds in the forward direction. i.e. forward reaction is favoured.

If the concentration of any one product, say C is increased then by Le-Chatelier’s Principle the backward reaction should be favoured so that the increase in the concentration of C is nullified. It can be explained as follows. As the concentration of product C increases the numerator of mass equation increases. To keep the value of K_c, the denominator should increase. This is possible only if the concentration of A and B is increased. It is possible if more and more A and B are formed thus the reaction proceeds in the backward direction. i.e. backward reaction is favoured.

Effect of the Change of Concentration on the chemical Equilibrium:

According to Le-Chatelier’s principle, when the concentration of one of the substance in a system in equilibrium is increased, then the equilibrium will shift so as to use up the substance added.

Explanation Using Le-Chatelier’s Principle:

Let us consider a general reversible reaction.

A  +  B   ⇌   C + D

The equilibrium constant for the reaction is given by

If the concentration of any one reactant say A is increased then by Le-Chatelier’s Principle the forward reaction should be favoured so that the increase in the concentration of A is nullified. It can be explained as follows. As the concentration of reactant, A increases the denominator of mass equation increases. To keep the value of the equilibrium constant the same the value of K_c constant the numerator should increase. This is possible only if the concentration of C and D is increased. It is possible if more and more C and D are formed thus the reaction proceeds in the forward direction. i.e. forward reaction is favoured.

If the concentration of any one product say C is increased then by Le-Chatelier’s Principle the backward reaction should be favoured so that the increase in the concentration of C is nullified. It can be explained as follows. As the concentration of product C increases the numerator of mass equation increases. To keep the value of K_c, the denominator should increase. This is possible only if the concentration of A and B is increased. It is possible if more and more A and B are formed thus the reaction proceeds in the backward direction. i.e. backward reaction is favoured.

The following graph shows the variation in concentration of the species on increasing the concentration of hydrogen in a reaction to produce ammonia from nitrogen and hydrogen.

Everyday Life Examples to Explain the Effect of the Change of Concentration on the Equilibrium:

Clothes dry quicker when there is a breeze.

Due to breeze or by shaking clothes in the air, the water vapours in the nearby air are removed or carried away. To establish the equilibrium the water from wet clothes starts evaporating. Thus they get dried fast.

On a humid day, we sweat more

Our body is losing water continuously by forming sweat. On a normal day, this sweat gets evaporated as soon as it is formed on the surface of the body. On a humid day the surrounding air contains a large amount of water vapours. Hence our body cannot lose water in the form of vapours. Thus the water remains on our body as sweat.

Transfer of oxygen by haemoglobin in the blood.

Haemoglobin is a protein present in red blood corpuscles which act as an oxygen carrier. The equilibrium is represented as

Hb_(s) +  O_2(g)  ⇌    HbO_2(s)

In lungs, there is an equilibrium of this reaction, when  HbO₂ (oxyhaemoglobin) reaches the site of tissues where the partial pressure is low. Now the equilibrium adjusts itself by shifting towards the right by releasing oxygen from oxyhaemoglobin. When the blood returns back to the lungs, where the partial pressure is higher, more oxyhaemoglobin is formed.

Removal of CO₂ from tissues

At the site of the tissues, the partial pressure of carbon dioxide is high. It dissolves into the blood and carried to lungs. In lungs partial pressure of carbon dioxide is low and it gets released from the blood.

Effect of the Pressure on the Chemical Equilibrium:

Change in pressure plays an important role in gaseous reactions.  The change of pressure has effect only on those equilibria which involves gaseous substances. There can be three types of gaseous reactions:

• Chemical reactions accompanied by an increase in volume
• Chemical reactions accompanied by a decrease in volume.
• Chemical reactions accompanied by no change in volume.

By Le-Chatelier’s principle, at a constant temperature, increase in pressure will favour a reaction which is accompanied by a decrease in volume and decrease in pressure will favour a reaction which is accompanied by the increase in volume.

Chemical reactions accompanied by an increase in volume:

Consider following reaction.

PCl_5(g)  ⇌ PCl_3(g)   +    Cl_2(g)

1 Vol             1 Vol           1 Vol

1 Vol             2 Vol

In this reaction, 1 volume of reactants gives 2 volumes of products.  Thus in this reaction volume is increased.

Chemical reactions involving gases and accompanied by an increase in volume are favoured by a reduction in pressure. Thus by decreasing the pressure at equilibrium, equilibrium is shifted towards the right.

Chemical reactions accompanied by a decrease in volume:

Consider following reaction.

N_2(g)  +   3H_2(g)   ⇌   2 NH_3(g)

1 Vol         3 Vol              2 Vol

4 Vol                2 Vol

In this reaction, 4 volumes of reactants give 2 volumes of products.  Thus in this reaction volume is decreased.

Chemical reactions involving gases and accompanied by a decrease in the volume are favoured by an increase in pressure. Thus by increasing the pressure at equilibrium, equilibrium is shifted towards the right.

Chemical reactions accompanied by no change in volume:

Consider following reaction.

H_2(g)    +   I_2(g) ⇌   2 HI_(g)

1 Vol          1 Vol          2 Vol

2 Vol              2 Vol

In this reaction, 2 volumes of reactants give 2 volumes of products.  Thus in this reaction volume is not changed.

Chemical reactions involving gases and accompanied by no change in volume are not affected by the change in pressure.

Effect of Temperature on the Chemical Equilibrium:

If the temperature of the exothermic chemical reaction is increased, then the concentration of products reduces and thus the equilibrium is shifted towards left.  Hence the reduction in temperature favours exothermic reaction at equilibrium. and increase in temperature favours endothermic reaction.

It is to be noted that in a reversible reaction if one reaction is exothermic then another reaction is endothermic. Thus the effect of change of temperature on the two reactions is different.

By Le-Chatelier’s principle. for an exothermic reaction at equilibrium lowering of temperature will favour the forward reaction And for an endothermic reaction, an increase in temperature will favour the forward reaction.

Effect of the Catalyst on the Chemical Equilibrium:

The catalyst is a substance which increases or decreases the rate of a reaction without taking part in the chemical reaction. In a reversible reaction at equilibrium, catalyst affects the rate of both forward reaction and backward reaction by the same extent.    Hence catalyst at equilibrium does not affect chemical equilibrium.

Effect of Inert Gas Addition:

If the volume is kept constant and an inert gas such as argon is added which does not take part in the reaction, the equilibrium remains undisturbed. It is due to the fact that the addition of an inert gas at constant volume does not change the partial pressure or the molar concentration of substances involved in the reaction. The change will take place if and only if the added gas is a reactant or product involved in the reaction.

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Characteristics of Equilibrium Constant:

• It has a definite value for every chemical reaction at a particular temperature.
• It is independent of the initial concentrations of the reacting species.
• It changes with the change in the temperature.
• It depends on the nature of the reaction.
• It is independent of the change of pressure, volume and concentrations of the reactants and products.
• It is not affected by the introduction of the catalyst.
• The expression for it may contain the concentrations of gases or molecules and ions in solution but not of pure solids or pure liquids.
• The expression for it and its magnitude depends on the stoichiometric form of the balanced chemical equation.
• When the equation for equilibrium is multiplied by a factor, then the equilibrium constant must be raised to the power equal to the factor.

If K_c is equilibrium constant for reaction aA + bB ⇌  cC + dD

Then its value for reaction naA + nbB ⇌  ncC + ndD is given by

K’_c = (K_c)ⁿ

• When the addition of two equilibria leads to another equilibrium then the product of their equilibrium constants gives the value of K_C of the resultant equilibrium.

K_(resultant) = K_{(Reaction 1)}  x K_{(Reaction 2)}

• If K₁, k₂, K₃, …. are equilibrium constant for recation₁, reaction₂, reaction₃, ……. Then the value of K_C for reaction, a x recation₁+ b x reaction₂, c x reaction₃, ……. is given by

K_C = (K₁)^a(K₂)^b(K₃)^c……..

• For a reversible reaction, the value of K_C for the backward reaction is inverse of the equilibrium constant for the forward reaction

K_(backward) = 1 / K_(forward)

• If it is expressed in terms of concentration, it has different units for different reactions.

K_C is Dimensionless Unitless Quantity:

Depending upon the stoichiometric coefficients of a chemical reaction, the equilibrium constant should have a unit. Its unit should be

(mol dm^-3)^{(∑n products – ∑n reactants)}

Note that 1 dm³ = 1 L. Hence concentration can be expressed as (mol L^-1)

These days we specify equilibrium constant in terms of dimensionless quantities by specifying the standard state of reactants and products. The standard state pressure for pure gas is 1 bar and the partial pressures of the gases are measured with respect to this standard state.

If a gas has a partial pressure of 1.5 bar, then in terms of standard state its pressure would be equal to 1.5 bar/1bar = 1.5, a dimensionless number. Similarly, for concentrations, the standard state is 1 M ( 1mol dm^-3). If the concentration of the substance is 2.0 M. Then in terms of standard state it will be expressed as 2.0 M/1 M = 2. Thus using partial pressures and concentrations K_P and K_C obtained are dimensionless. Hence equilibrium constant is considered as dimensionless, unitless quantity.

K_C of a General Reaction and its Multiple:

Consider a hypothetical reversible reaction

aA + bB ⇌  cC + dD

The equilibrium constant of the reaction given by

Consider a multiple of above reaction

naA + nbB ⇌  ncC + ndD

The equilibrium constant of the reaction given by

Thus when the equation for an equilibrium is multiplied by a factor, then the equilibrium constant must be raised to the power equal to the factor.

K_C for the addition of Two Chemical Equilibria:

Consider following equilibria

1) N_2(g) +   O_2(g)  ⇌     2NO_(g)

The equilibrium constant for the reaction is

2)   2NO_(g) +   O_2(g)  ⇌     2NO_2(g) 

The equilibrium constant for the reaction is

3)  N_2(g) +  2O_2(g)   ⇌   2NO_2(g)

The equilibrium constant for the reaction is

Multiplying equation (1) by (2) we get

Thus When the addition of two equilibria leads to another equilibrium then the product of their equilibrium constants gives the equilibrium constant of the resultant equilibrium.

Temperature Dependence of Equilibrium Constant:

At chemical equilibrium, the rate of the forward reaction is equal to the rate of backward reaction. When the temperature is increased, in general, the rate of both the forward reaction and the backward reaction increases. As the energy of activation of the forward reaction and the backward reaction are different, the extent of the increase of the forward reaction and the backward reaction is different. Thus the value of K_f (rate constant for the forward reaction) and K_b (rate constant for the backward reaction) changes to a different extent. Thus the ratio K_f / K_b changes. i.e. the value of equilibrium constant changes with the change in temperature.

The value of the equilibrium constant for endothermic reaction increases with an increase in temperature, while the value of the equilibrium constant for exothermic reaction decreases with an increase in temperature. The temperature dependence of equilibrium constant can be written mathematically as

Uses of K_C:

It helps in the prediction of the extent of reaction:

The magnitude of the K_C tells us about the extent in which the reactants are converted into the products before the equilibrium is attained. Larger values of K indicates that the extent of reactants converting into products is greater. The generalization is

• If K_C > 10³ Products predominates the reactants. i.e. the concentration of products is very high compared to that of reactants t equilibrium and the reaction proceeds nearly to completion
• If K_C < 10^-3 Reactants predominates the products. i.e. the concentration of products is very less compared to that of reactants t equilibrium and the reaction hardly proceeds.
• 10^-3 < K_C < 10^-3 , appreciable concentrations of both the reactants and products are present at equilibrium

It gives us an idea of relative stabilities of reactants and products:

If the K_C is large products are more stable than the reactants. Whereas, If the value of the equilibrium constant is small reactants are more stable than the products.  The generalisation is

If K_C > 10³ Products are stable than reactants.

If K_C < 10^-3 Reactants are stable than products.

It helps in the prediction of the direction of a net reaction.

The value of K_C helps in the prediction of the direction in which the net reaction is proceeding at given concentrations or partial pressures of reactants and products. If Q_c is the concentration quotient and K_c be the equilibrium constant of a chemical reaction.

For reaction aA + bB ⇌  cC + dD

When computing Q_C the concentration at that instant are used

When computing K_C the concentration at equilibrium are used. The generalization is

• If Q_C > K_C , the reaction is taking place in a backward direction i.e. in the direction of reactants
• If Q_C <  K_C , the reaction is taking place in a forward direction i.e. in the direction of products.
• If Q_C =  K_C The reaction is in the equilibrium state and hence no net reaction is taking place.

It helps in the calculation of equilibrium constants and equilibrium pressures. 

If the equilibrium concentrations of various reactants and products are known for a reaction, the equilibrium constant can be calculated. On the other hand, if the equilibrium constant is known, the equilibrium concentrations can be calculated.

Steps Involved in the Calculation of K_C and Equilibrium Pressures:

1. Write the chemical equation for the equilibrium
2. Write expression for K_C or K_P for the reaction.
3. Express all unknown concentrations or partial pressures in terms of a single variable x.
4. Substitute equilibrium concentrations or partial pressures in terms of x in the expression for K_C or K_P.
5. Solve the equation for x
6. Substitute the value obtained for x in the expression in step 3 to calculate equilibrium concentrations or equilibrium partial pressures

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In this article, we shall stuudy to write expression for equilibrium constant.

Steps Involved in Writing Expression for Equilibrium Constant of a Reaction:

• Write the balanced chemical equation for the reaction.
• Write the products of equilibrium concentrations of the products in the numerator. Omit pure solids, pure liquids and the solvents in dilute solutions.
• Write the products of equilibrium concentrations of the reactants in the denominator. Omit pure solids, pure liquids and the solvents in dilute solutions.
• Raise each concentration term to the power equal to stoichiometric coefficients of the species in the equation.

Homogeneous Equilibrium:

The equilibrium in which all the substances involved exist in a single homogeneous phase is called homogeneous equilibrium.

Examples:

N_2(g) +  3 H_2(g)  ⇌     2NH_3(g)

H_2(g) +  I_2(g) ⇌     2HI_(g)

2SO_2(g) +  O_2(g)  ⇌   2SO_3(g)

NH_3(aq) + H₂O_(l) ⇌     2NH₄+_(aq) +  OH-_(aq)

2N₂O_(g) ⇌ 2N_2(g)  +   O_2(g)

CH₃COOC₂H_5(aq) + H2O_(l) ⇌ CH₃COOH_(aq) +   C2H5OH_(aq)

Heterogeneous Equilibrium:

The equilibrium in which the substance involved are present in different phases is called heterogeneous equilibrium.

Examples:

CaCO_3(s) ⇌  CaO_(s) + . CO_2(g)

CaCO_3(s)+H₂O(l)+CO₂(g) ⇌ Ca²⁺_(aq)+ 2HCO₃^–_(aq)

(NH₄)₂CO_3(s) ⇌ 2NH_3(g) + CO_2(g)+  H₂O_(g)

2Mg_(s) + O2_(g)  ⇌    2MgO_(s)

Note: Pure liquids and solids are ignored while writing the equilibrium constant expression.

Now, for pure solids and pure liquids, the molar mass M and density ρ are constant. Hence the concentration remains constant.

• When pure solids are involved in equilibrium, its concentration remains constant no matter how much of it is present. Therefore by convention, the concentrations of all solids involved in equilibrium are taken as unity.
• When pure liquids are involved in equilibrium, its concentration remains constant no matter how much of it is present. Therefore by convention, the concentrations of all liquids involved in equilibrium are taken as unity.
• For equilibria in the aqueous medium, the concentration of solvent (water) will not change appreciably because it is present in large excess. Hence by convention, the concentration of solvent (water) is taken as unity.
• Hence in general, pure liquids, pure solids, and solvents can be ignored while writing the equilibrium constant expression.

Writing the Expression for K_c and K_p:

BaCO_3(s)  ⇌  BaO_(s) + CO_2(g)

4NH_3(g) + 5O_2(g) ⇌   4NO_(g)+ 6H₂O_(g)

NH_3(g) + HCl_(g)  ⇌  NH₄Cl_(s)

3Cl_2(g) + 2NO_2(g)  ⇌  2NO₂Cl_3(g)

Fe³⁺_(aq) + SCN^–_(aq) ⇌ FeNCS²⁺_(aq)

CH_4(g) + 2H₂S_(g)  ⇌  CS_2(g)+ 4H_2(g)

MgCO_3(s)  ⇌  MgO_(s)  + CO_2(g)

AgBr_(s) ⇌ Ag⁺_(aq)+ Br^–_(aq)

K_c = [Ag⁺][Br ^–]

CH₃COCH_3(l) ⇌CH₃COCH_3(g)

K_c = [CH₃COCH_3(g)]

CH_4(g) + 2O_2(g) ⇌ CO_2(g)+ 2H₂O_(g)

Al_(s) + 3H⁺_(aq) ⇌ Al³⁺_(aq)+ 3/2H_2(g)

HPO₄^2-_(aq) + H₂O_(l) ⇌ H₃O⁺_(aq)+ PO₄^3-_(aq)

Ag₂O_(s) + 2HNO_3(aq) ⇌  2AgNO_3(aq)+ H₂O_(l)

Ni_(s) + 4CO_(g) ⇌  Ni(CO)_4(g)

CuO_(s) + H_2(g) ⇌  Cu_(s)+ H₂O_(g)

N_2(g) + 3H_2(g) ⇌  2NH_3(g)

Fe³⁺_(aq) + 3OH^–_(aq) ⇌  Fe(OH)_3(s)

2N₂O_(g) ⇌ 2N_2(g) + O_2(g)

C_(s)+  CO_2(g) ⇌  2CO_(g)

Consider a hypothetical reversible reaction involving homogeneous gaseous phase

aA(g) + bB(g) ⇌  cC(g) + dD(g)

The equilibrium constant in terms of the partial pressure is given by

This is the relation between K_c and K_p.

Where Δn = (Sum of the exponents in the numerator of concentration quotient) – (Sum of the exponents in the denominator of concentration quotient)

Steps Involved in Finding Relation Between K_c and K_p :

• Write the balanced chemical equation for the reaction.
• Find the change in the number of moles of gaseous species by the following formula.

• Now, use the relation

Examples to Find Relation between Between K_c and K_p :

Example – 01:

Example – 02:

Example – 03:

Example – 04: 

For the reaction N_2(g) + 3H_2(g) ⇌  2NH_3(g), the value of the equilibrium constant K_p is 3.6 x 10^-2 at 500 K. Calculate the value of K_c for the reaction at the same temperature. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 05:

For the reaction 2SO_2(g)) + O_2(g) ⇌ 2SO_3(g), the value of the equilibrium constant K_p is 2.0 X 10¹⁰ bar^-1 at 450 K. Calculate the value of K_c for the reaction at the same temperature. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 06:

For the reaction 2NOCl_(g) ⇌ 2NO_(g)+ Cl_2(g), the value of the equilibrium constant K_p is1.8 X 10^-2 at 500 K. Calculate the value of K_c for the reaction at the same temperature. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 07:

For the reaction CaCO_3(s)  ⇌ CaO_(s)+ CO_2(g), the value of equilibrium constant K_p is 167 at 1073 K. Calculate the value of K_c for the reaction at the same temperature. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 08:

Find the ratio of K_p/K_c for the reaction, CO(g) + 1/2O_2(g)  ⇌  2CO_2(g), at 500 K. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 09:

Find the ratio of K_p/K_c  for the reaction, N_2(g) + O_2(g) ⇌   2NO_(g)

Example – 10:

The equilibrium constant K_p for the reaction H_2(g)  + I_2(g)  ⇌  2HI_(g) is 130 at 510 K. Calculate K_c for following reactions a) 2HI_(g)  ⇌ H_2(g)  + I_2(g),  b) HI_(g)  ⇌ 1/2H_2(g)  + 1/2I_2(g), c) 1/2H_2(g)  + 1/2I_2(g)  ⇌  HI_(g).

∴  K_c = 130

For the reaction 2HI_(g)  ⇌ H_2(g)  + I_2(g),

For the reaction, HI_(g)  ⇌ 1/2H_2(g)  + 1/2I_2(g)

For reaction, 1/2H_2(g)  + 1/2I_2(g)  ⇌  HI_(g).

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In this article, we shall study the law of mass action and its application to chemical equlibrium.

Rate of Chemical Reaction:

The change in the concentration of the reactants (or products) per unit time is called the rate of reaction Mathematically it can be expressed as, The rate of Reaction = Change in Concentration of Products or Reactants/time in which change is taking place Hence the rate of chemical reaction is the change in concentration of the reactants in unit time. It’s S.I. unit is mol dm^-3 s^-1.

Active Mass:

Active mass is the molar concentration per unit volume of that substance. It is denoted by enclosing the symbol or formula of that substance in the square bracket. For solutions, it is expressed in “ moles dm^-3”. For gases, it is expressed in “ mol dm^-3” or pressure in the atmosphere (atm) or bar or pascal (Pa).

Factors Affecting the Rate of Chemical Reaction:

Concentration:

If the concentration of reactants increases then the rate of reaction increases.

Pressure:

Change in pressure plays an important role in gaseous reactions. There can be three types of gaseous reactions:

Reaction with the increase  in volume:

The reduction in pressure for a gaseous reaction accompanied by an increase in volume increases the rate of reaction.

e.g.  PCI_5(g)    ⇌    PCl_3(g)   +     Cl_2(g)

Reaction with the decrease in volume:

The increase in pressure for a gaseous reaction accompanied by a decrease in volume increases the rate of reaction.

e.g. N_2(g) +  3 H_2(g)   ⇌    2NH_3(g)

Reaction with no change in volume:

The gaseous reaction of equilibrium, involving no change in volume is independent of pressure.

e.g.  H_2(g) + Cl_2(g)  ⇌  2HCl_(g)

Temperature:

In general rate of reaction increases with increase in temperature.

Catalyst:

In a chemical reaction, catalyst increases the rate of reaction.

Light:

The reactions which take place in the presence of light are called photochemical reactions. Such reactions are influenced by light.

Size of particles:

If solid reactants are used in powdered form instead of granular form, the rate of reaction increases due to an increase in the surface area available for reaction

Law of Mass Action:

In 1864, Norwegian chemists Cato Guldberg and Peter Wage put forward the law of mass action

Law of mass action states that  “The rate of a chemical reaction is directly proportional to the product of active masses of reactants, at given temperature at that instant”.

Where active mass is molar concentration per unit volume of that substance.  It is denoted by enclosing the symbol or the formula of that substance in the square bracket and expressed in mol dm^-3.

Explanation of Law of Mass Action:

Consider a hypothetical reaction

A + B →  Products.

According to the law of mass action the rate of reaction

R ∝ [A] [B].

For any general reaction

aA + bB → Products.

According to the law of mass action the rate  of reaction

R ∝ [A]^a [B]^b

Equilibrium mixture:

The mixture of reactants and products formed at the chemical equilibrium state is called equilibrium mixture.

Equilibrium Concentrations:

The concentrations of the reactants and products in a chemical equilibrium state for a reversible reaction are called equilibrium concentrations.

The Expression for Equilibrium Constant in Terms of Concentrations:

Consider a hypothetical reversible reaction

A + B  ⇌  C + D

According to the law of mass action the rate of the forward reaction

R_f ∝ [A] [B]

∴  R_f  =  K_f [A] [B]

Where K_f  = rate constant for the forward reaction

According to the law of mass action the rate of the backward reaction

R_b ∝ [C] [D]

∴  R_b =  K_b [C] [D]

Where K_b = rate constant for the backward reaction

Now, for a chemical equilibrium,

R_f = R_b

K_f [A] [B] = K_b [C] [D]

Where K_C is known as equilibrium constant and the equation is called “mass law equation”.

Consider a general reversible reaction

aA + bB + cC + …   ⇌  lL  + mM  + nN + ….,    Then,

The equilibrium constant ‘ K_C‘ is defined as the ratio of the product of the equilibrium concentrations of the products to that of the reactants with each concentration term raised to the power equal to the stoichiometric coefficients of the substance in the balanced chemical equation.

Concentration is expressed in terms of mol dm^-3 or mol L^-1 or M. The equilibrium constant is denoted by Kc.

More is the numerical value of ‘K’ then more is the concentration of products in comparison with reactants & vis-a-vis.

Concentration Quotient of a Chemical Reaction:

At a given temperature, the ratio of the product of the equilibrium concentrations of the products to that of the reactants with each concentration term raised to the power equal to the stoichiometric coefficients of the substance in the balanced chemical equation is called concentration ratio or concentration quotient. It is denoted by Q_C.

At equilibrium, the concentration ratio is equal to the equilibrium constant Kc.

Its significance is that it helps in the prediction of the direction in which the net reaction is proceeding at given concentrations or partial pressures of reactants and products.

The generalization is

• If Q_C > K_C: The reaction is taking place in a backward direction i.e. in the direction of reactants.
• If Q_C < K_C: The reaction is taking place in a forward direction i.e. in the direction of products.
• If Q_C = K_C: The reaction is in an equilibrium state and hence no net reaction is taking place.

Law of Chemical Equilibrium:

At a given temperature, the ratio of the product of the equilibrium concentrations of the products to that of the reactants with each concentration term raised to the power equal to the stoichiometric coefficients of the substance in the balanced chemical equation has a constant value.

The Expression for Equilibrium Constant in Terms of Partial Pressure:

Partial Pressure:

It is the pressure exerted by a gas in a mixture of gases if it alone occupies the entire volume of the mixture of gases. The partial pressure of a gaseous component is proportional to the mole fraction. The partial pressure of a gas is calculated using the following formula

Where, n = Number of moles of gaseous component

N = Total moles of a gaseous system

P = Total pressure of the gaseous system

The Expression for Equilibrium Constant:

In the gaseous system, the concentrations in concentration quotients are replaced by partial pressure, because for given temperature the partial pressure of the gas is directly proportional to its concentration.

Consider a hypothetical reversible reaction

aA_(g) + bB_(g) ⇌  cC(g) + dD_(g)

Then the equilibrium constant in terms of partial pressures is given by

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Gibb’s Energy or Free Energy:

Gibb’s energy or free energy is a thermodynamic function which helps us in the development of a criterion of spontaneity or feasibility of a process. It refers to the capacity of the system to do useful work.

Gibb’s energy is defined as the amount of energy available from a system at a given set of conditions that can be put into useful work. It is also called free energy.

Mathematically,   G = U + PV – TS

Where      G = Gibb’s energy

U = Internal energy of the system

P = Pressure of the system

V = Volume of the system

T = Absolute temperature of the system

S = Entropy of the system

The absolute value of Gibb’s energy cannot be calculated. But the change in it can be calculated as

ΔG = ΔU + Δ(PV) – Δ(TS)

For constant pressure and constant temperature process

ΔG = ΔU + PΔV – TΔS

But ΔU + PΔV = ΔH

∴ ΔG = ΔH – TΔS

This relation is known as Gibb’s Helmholtz equation.

Where,      ΔG = Change in Gibb’s energy

ΔH = Change in enthalpy of the system

T = Absolute temperature of the system

ΔS = Change in entropy of the system

The spontaneity of a Reaction w.r.t. Gibb’s Energy and Total Entropy:

Let ΔS_Total be the total enthalpy of the system and ΔG_T,P is Gibb’s energy at constant temperature and pressure.

• If ΔS_Total > 0 (positive) or ΔG_T,P < 0 (negative), the process will be spontaneous and proceeds in a backward direction. It nears to completion. (K > 1)
• If ΔS_Total < 0 (negative) or ΔG_T,P > 0 (positive), the process will be non-spontaneous. and favoured in backward direction. (K < 1)
• If ΔS_Total = 0 or ΔG_T,P = 0, the process will be in equilibrium state. (K = 1)

Relation Between Gibb’s Energy and Chemical Equilibrium:

For spontaneous process Gibb’s energy is negative. For a reversible reaction, there is a decrease in Gibb’s energy during the course of reaction whether we start from reactants or products. Let us consider a hypothetical reaction

A ⇌ B

A graph is drawn by taking Gibb’s energy on the y-axis and the change in the composition of the reacting mixture with time on the x-axis. The graph is as follows.

The minima in the curve correspond to the composition of the reaction mixture at the equilibrium state at which Gibb’s energy is minimum.

From graph following points should be noted.

• In reaching the equilibrium state whether from A or from B, the ΔG is negative.
• At equilibrium state, there is no change in Gibb’s energy. i.e. ΔG at equilibrium = 0.
• If minima of the curve lie very close to the products, then it means that the equilibrium composition strongly favours the products and hence K >> 1. i.e. reaction will proceed to completion.
• If minima of the curve lie very close to the reactants, then it means that the equilibrium composition strongly favours the reactants and hence K < 1. i.e. the reaction hardly proceeds.

Relation Between Standard Gibb’s Energy change and Equilibrium Constant:

Let us consider a hypothetical reaction

A + B  ⇌  C + D

The relation between Gibb’s energy change (ΔG)and standard Gibb’s energy change ΔG^o is given by

ΔG = ΔG^o  + RT ln Q

Where      R = Universal gas constant

T = Absolute temperature of the system

Q = Concentration coefficient

At equilibrium ΔG = 0 and Q = K_c

∴   0 = ΔG^o  + RT ln K_c

∴    ΔG^o  = – RT ln K_c

∴    ΔG^o  = – 2.303 RT log ₁₀ K_c

This is the relation between the standard Gibb’s energy change and equilibrium constant. In exponential form, the expression can be written as

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In this article, we shall study the use of the Nernst equation to find e.m.f. of cell and electrodes.

Convention Followed While Calculation of Cell Potential (e.m.f.):

• In the symbolic representation of the cell, the right-hand side electrode is the cathode (positive electrode) and the left-hand side is the anode (negative electrode).
• All standard potentials are reduction potentials that are they refer to a reduction reaction.
• The cathode has a higher standard potential than the anode.
• For spontaneous reaction to take place the cell potential should be positive.

Illustrations for Use of Nernst Equation:

When Reactions are given:

Example – 1: 

Cr_(s) + 3Fe³⁺ _(aq) → Cr³⁺_(aq) + 3Fe²⁺ _(aq)

The cell formation is

Cr_(s)| Cr³⁺_(aq)|| Fe²⁺_(aq),Fe ³⁺_(aq)| Pt

The half cell reactions are

Cr_(s) → Cr³⁺(aq)+ 3e^–  (Oxidation)

Fe³⁺_(aq)+ 3e^– → Fe²⁺_(aq)(Reduction)

Hence n = 3

Nernst equation is

Example – 2:  

Al³⁺(aq) + 3e-  → Al_(s) ,

Here n = 3,  Nernst equation is

When Type of Electrode is Given:

Redox Electrode:

Example – 1:  

Pt | Sn²⁺, Sn⁴⁺

The Reduction reaction is

Sn⁴⁺_(aq)+ 2e^– → Sn²⁺_(aq)(Reduction),

Hence n = 2, Nernst equation is

Example – 2:

Pt | Fe²⁺, Fe³⁺

The Reduction reactions are

Fe³⁺_(aq) + 1e^– →  Fe ²⁺_(aq)(Reduction)

Hence n = 1, Nernst equation is

Metal Metal Ion Electrode:

Example – 1: 

Zn_(s)| Zn⁺⁺_(aq)

Reduction reaction for it is

Zn_(s)  →  Zn⁺⁺_(aq)     +   2e^–

Here n = 2,  Nernst equation is

Example – 2:

Al_(s)| Al³⁺_(aq)

The reduction reaction is

Al³⁺(aq) + 3e-  → Al_(s)

Here n = 3,  Nernst equation is

Metal Sparingly Soluble Salt Electrode:

Example – 1:

Cl^– _(aq) | AgCl_(s)| Ag

The Reduction reaction is

AgCl_(s)+ e^– → Cl^– _(aq) + Ag_(s) (Reduction)

Hence n = 1, Nernst equation is

Gas Electrode:

Example – 1:

Cl^– _(aq) |   Cl_2(g), (1 atm)| Pt

The Reduction reaction is

½ Cl_2(g) + e ^–  →   Cl^– _(aq)  (Reduction)

Hence n = 1, Nernst equation is

Important Terms:

Half-Cell:

An electrode in contact with an electrolyte containing its own ions is called a half cell. e.g. In Daniel cell, the zinc rod dipped in zinc sulphate solution is called zinc half cell.

Half-Cell Reaction:

The reaction taking place in a half cell or reaction taking place at each electrode is called half-cell reaction. e.g. In Daniel cell in zinc half cell oxidation takes place. Therefore the half-cell reaction is

Zn_(s)  →  Zn⁺⁺_(aq)     +   2e^–

Cell:

A combination of two half-cells such that oxidation takes place at one half cell and reduction takes place at other half-cell is called the cell. e.g. A Daniel cell is formed by the combination of zinc half cell and copper half cell. Oxidation takes place at zinc half cell and the reduction takes place at the copper half cell.

Single Electrode Potential:

The difference of potential between the electrode and its salt solution around it at equilibrium is called a single electrode potential. Electrode potential depends upon

• Nature of the element/ metal,
• Concentration or activity of ions in solution
• Temperature and
• Pressure in case of gas.

Standard Electrode Potential (E°):

The difference of potential between the electrode and its salt solution around it containing ion concentration at a unit activity at 298 K is called standard electrode potential.

Oxidation  Electrode Potential (E_ox):

The difference of potential between the electrode and its salt solution around it at equilibrium and at constant temperature due to oxidation is called oxidation potential.

Standard Oxidation Potential (E°_ox):

The difference of potential between the electrode and its salt solution around it containing ion concentration at a unit activity at 298 K  due to oxidation is called standard oxidation potential (S.O.P.).

Reduction Electrode Potential (E°_red):

The difference of potential between the electrode and its salt solution around it at equilibrium and at constant temperature due to reduction is called reduction potential.

Standard Reduction Potential (E°_red):

The difference of potential between the electrode and its salt solution around it containing ion concentration at a unit activity at 298 K  due to reduction is called standard reduction potential (S.R.P.).

Standard e.m.f. of Cell:

The algebraic sum of the standard oxidation potential of one electrode (anode) and the standard reduction potential of another electrode (cathode) is called the standard e.m.f. of a cell.

Note:

The oxidation potential of electrode is equal to the reduction potential of the electrode with the opposite sign

Gibb’s Energy Change:

In thermodynamics, the Gibbs free energy is a thermodynamic potential that measures the maximum or reversible work that may be performed by a thermodynamic system at a constant temperature and pressure (isothermal, isobaric).

As the cell reaction in an electrochemical cell progresses, electrons move through a wire connecting the two electrodes until the equilibrium point of the cell reaction is reached, at which point the flow of electrons ceases. Just cell performs the work. In electrochemistry, the maximum amount of electrical work a galvanic cell can do at constant temperature and pressure is Gibb’s free energy.

The amount maximum work a galvanic cell can do is given as

Electrical work = Amount of charge (nF) × Cell potential (E_cell)

Electrical work = n F E_cell

The reversible electrical work done in a galvanic cell reaction is equal to the decrease in its Gibb’s energy

Thus,     Electrical work = – ΔG

∴ – ΔG = n F E_cell

∴ ΔG = –  n F E_cell

The standard Gibb’s energy change is given by

ΔG° = –  nFE°_cell

Gibb’s energy is an extensive property, which depends on the amount of substance. But the electrical potential is an intensive property which does not depend on the amount of substance. Thus E°_cell remains constant. Thus if ΔG° changes there is the corresponding change in the number of electrons. It can be explained as follows

ΔG° = –  n F E°_cell 

If the stoichiometric equation of redox reaction is multiplied by 2, then the standard Gibb’s energy ΔG° gets doubled and the number of electrons ‘n’ also gets doubled.

From (1) and (2) we can see that the e.m.f. of cell in both cases is the same. It shows that electrical potential is an intensive property that does not depend on the amount of substance.

Relation between Standard Cell Potential and Equilibrium Constant:

The Gibb’s free energy of a galvanic cell is given by

G° = –  n F E°_cell

By thermodynamical and equilibrium concept

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