JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Le-Chaterlier Principle and It's Application

Factors which Change the State of Equilibrium: Le-Chatelier's Principle.

Le-Chatelier and Braun (1884), French chemists, made certain generalizations to explain the effect of changes in concentration, temperature or pressure on the state of system in equilibrium. When a system is subjected to a change in one of these factors, the equilibrium gets disturbed and the system readjusts itself until it returns to equilibrium. The generalization is known as Le-Chatelier's principle. It may stated as:

“Change in any of the factors that determine the equilibrium conditions of a system will shift the equilibrium in such a manner to reduce or to counteract the effect of the change.”

The principle is very helpful in predicting qualitatively the effect of change in concentration, pressure or temperature on a system in equilibrium. This is applicable to all physical and chemical equilibria.

(1) Effect of change of concentration : According to Le-Chatelier's principle, “If concentration of one or all the reactant species is increased, the equilibrium shifts in the forward direction and more of the products are formed. Alternatively, if the concentration of one or all the product species is increased, the equilibrium shifts in the backward direction forming more reactants.”

Thus,

                        Increase in concentration of any of the reactants \[\underset{equilibrium\,\,to}{\mathop{\xrightarrow{Shifts\,\,the}}}\,\] Forward direction

                        Increase in concentration of any of the products \[\underset{equilibrium\,\,to}{\mathop{\xrightarrow{Shifts\,\,the}}}\,\] Backward direction

(2) Effect of change of temperature : According to Le-Chatelier's principle, “If the temperature of the system at equilibrium is increased (heat is supplied), the equilibrium will shift in the direction in which the added heat is absorbed. In other words, the equilibrium will shift in the direction of endothermic reaction with increase in temperature. Alternatively, the decrease in temperature will shift the equilibrium towards the direction in which heat is produced and, therefore, will favour exothermic reaction.”

Thus,

                                    Increase in temperature \[\underset{in\,\,the\,\,direction\,\,of}{\mathop{\xrightarrow{Shifts\,\,the\,\,equilibrium}}}\,\] Endothermic reaction

                                    Decrease in temperature \[\underset{in\,\,the\,\,direction\,\,of}{\mathop{\xrightarrow{Shifts\,\,the\,\,equilibrium}}}\,\] Exothermic reaction

(3) Effect of change of pressure : Pressure has hardly effect on the reactions carried in solids and liquids. However, it does influence the equilibrium state of the reactions that are carried in the gases. The effect of pressure depends upon the number of moles of the reactants and products involved in a particular reaction. According to Le-Chatelier's principle, “Increase in pressure shifts the equilibrium in the direction of decreasing gaseous moles. Alternatively, decrease in pressure shifts the equilibrium in the direction of increasing gaseous moles and pressure has no effect if the gaseous reactants and products have equal moles.”

Thus,

                                    Increase in pressure \[\underset{in\,\,the\,\,direction\,\,of}{\mathop{\xrightarrow{Shifts\,\,the\,\,equilibrium}}}\,\] Decreasing gaseous moles

                                    Decrease in pressure \[\underset{in\,\,the\,\,direction\,\,of}{\mathop{\xrightarrow{Shifts\,\,the\,\,equilibrium}}}\,\] Increasing gaseous moles

(4) Effect of volume change : We know that increase in pressure means decrease in volume, so the effect of change of volume will be exactly reverse to that of pressure. Thus, “decreasing the volume of a mixture of gases at equilibrium shifts the equilibrium in the direction of decreasing gaseous moles while increasing the volume shifts the equilibrium in the direction of increasing gaseous moles.”

Thus,

                        Increase in volume \[\underset{in\,\,the\,\,direction\,\,of}{\mathop{\xrightarrow{Shifts\,\,the\,\,equilibrium}}}\,\] Increasing gaseous moles

                        Decrease in volume \[\underset{in\,\,the\,\,direction\,\,of}{\mathop{\xrightarrow{Shifts\,\,the\,\,equilibrium}}}\,\] Decreasing gaseous moles

(5) Effect of catalyst : Catalyst has no effect on equilibrium. This is because, catalyst favours the rate of forward and backward reactions equally. Therefore, the ratio of the forward to reverse rates remains same and no net change occurs in the relative amount of reactants and products present at equilibrium. Thus, a catalyst does not affect the position of the equilibrium. It simply helps to achieve the equilibrium quickly. It may also be noted that a catalyst has no effect on the equilibrium composition of a reaction mixture.

Thus,    Catalyst does not shift the equilibrium in any direction

(6) Effect of addition of inert gas : The addition of an inert gas (like helium, neon, etc.) has the following effects on the equilibrium depending upon the conditions :

(i) Addition of an inert gas at constant volume : When an inert gas is added to the equilibrium system at constant volume, then the total pressure will increase. But the concentrations of the reactants and products (ratio of their moles to the volume of the container) will not change. Hence, under these conditions, there will be no effect on the equilibrium.

(ii) Addition of an inert gas at constant pressure : When an inert gas is added to the equilibrium system at constant pressure, then the volume will increase. As a result, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift in a direction in which there is increase in number of moles of gases.

Thus,

                        Addition of an inert gas \[\underset{{}}{\mathop{\xrightarrow{V\,=\,cons\tan t}}}\,\] No effect on the equilibrium.

                        Addition of an inert gas \[\underset{\begin{smallmatrix}

 Shifts\,\,the\,\,equilibrium \\

 in\,\,the\,\,direction\,\,of

\end{smallmatrix}}{\mathop{\xrightarrow{P\,=\,cons\tan t}}}\,\] Increasing gaseous moles.

Applications of Le-Chatelier's Principle.

The Le-Chateliers principle has a great significance for the chemical, physical systems and in every day life in a state of equilibrium. Let us discuss in brief a few applications.

(1) Applications to the chemical equilibrium : With the help of this principle, most favourable conditions for a particular reaction can be predicted.

            (i) Synthesis of ammonia (Haber’s process):  \[\underset{1\ vol}{\mathop{{{N}_{2}}}}\,+\underset{3\ vol}{\mathop{3{{H}_{2}}}}\,\] ? \[\underset{2\ vol}{\mathop{2N{{H}_{3}}}}\,+23kcal\] (exothermic)

            (a) High pressure \[(\Delta n<0)\] (b) Low temperature (c) Excess of \[{{N}_{2}}\] and \[{{H}_{2}}\] (d) Removal of \[N{{H}_{3}}\] favours forward reaction.

            (ii) Formation of sulphur trioxide :  \[\underset{2\ vol}{\mathop{2S{{O}_{2}}}}\,+\underset{1\ vol}{\mathop{{{O}_{2}}}}\,\] ? \[\underset{2\ vol}{\mathop{2S{{O}_{3}}}}\,+45\ kcal\]  (exothermic)

            (a) High pressure \[(\Delta n<0)\] (b) Low temperature (c) Excess of \[S{{O}_{2}}\] and \[{{O}_{2}}\], favours the reaction in forward direction.

            (iii) Synthesis of nitric oxide :  \[\underset{1\ vol}{\mathop{{{N}_{2}}}}\,+\underset{1\ vol}{\mathop{{{O}_{2}}}}\,\] ? \[\underset{2\ vol}{\mathop{2N{{O}_{{}}}}}\,-43.2\ kcal\] (endothermic )

            (a) High temperature (b) Excess of \[{{N}_{2}}\] and \[{{O}_{2}}\] (c) Since reaction takes place without change in volume i.e., \[\Delta n=0\], pressure has no effect on equilibrium.

            (iv) Formation of nitrogen dioxide :  \[\underset{2\ vol}{\mathop{2N{{O}_{{}}}}}\,+\underset{1\ vol}{\mathop{{{O}_{2}}}}\,\] ? \[\underset{2\ vol}{\mathop{2N{{O}_{2}}}}\,+27.8\ Kcal\]

            (a) High pressure (b) Low temperature (c) Excess of \[NO\] and \[{{O}_{2}}\] favours the reaction in forward direction.

            (v) Dissociation of phosphours pentachloride :  \[\underset{1\ vol}{\mathop{PC{{l}_{5}}}}\,\] ? \[\underset{1\ vol}{\mathop{PC{{l}_{3}}}}\,+\underset{1\ vol}{\mathop{C{{l}_{2}}}}\,-15\ kcal\]

            (a) Low pressure or high volume of the container, \[\Delta n>0\] (b) High temperature (c) Excess of \[PC{{l}_{5}}\].

(2) Applications to the physical equilibrium: Le-Chatelier's principle is applicable to the physical equilibrium in the following manner;

            (i) Melting of ice (Ice – water system) : \[\underset{\text{(Greater}\ \text{Volume)}}{\mathop{\text{Ice}}}\,\] ? \[\underset{\text{(Lesser}\ \text{Volume)}}{\mathop{\text{Water}}}\,-x\ kcal\]

            (In this reaction volume is decreased from 1.09 c.c. to 1.01 c.c. per gm.)

            (a) At high temperature more water is formed as it absorbs heat.       (b) At high pressure more water is formed as it is accompanied by decrease in volume.(c) At higher pressure, melting point of ice is lowered, while boiling point of water is increased.

            (ii) Melting of sulphur :\[{{S}_{(s)}}\]? \[{{S}_{(l)}}-x\ kcal\]

            (This reaction accompanies with increase in volume.)

            (a) At high temperature, more liquid sulphur is formed.         (b) At higher pressure, less sulphur will melt as melting increases volume.(c) At higher pressure, melting point of sulphur is increased.

            (iii) Boiling of water (water- water vapour system) :  \[\underset{\text{(Low}\ \text{volume)}}{\mathop{\text{Water}}}\,\]? \[\underset{\text{(Higher}\ \text{volume)}}{\mathop{\text{Water}\ \text{Vapours}}}\,-x\ kcal\]

            (It is accompanied by absorption of heat and increase in volume.)

            (a) At high temperature more vapours are formed.(b) At higher pressure, vapours will be converted to liquid as it decreases volume.(c) At higher pressure, boiling point of water is increased (principle of pressure cooker).

            (iv) Solubility of salts : If solubility of a salt is accompanied by absorption of heat, its solubility increases with rise in temperature; e.g., \[N{{H}_{4}}Cl,\ {{K}_{2}}S{{O}_{4}},\ KN{{O}_{3}}\] etc. \[KN{{O}_{3(s)}}+(aq)\xrightarrow{{}}KN{{O}_{3(aq)}}-x\ kcal\]

            On the other hand if it is accompanied by evolution of heat, solubility decreases with increase in temperature; e.g., \[CaC{{l}_{2}},Ca{{(OH)}_{2}},NaOH,\ KOH\] etc. \[Ca{{(OH)}_{2(\ s)}}+(aq)\xrightarrow{{}}Ca{{(OH)}_{2\ (aq)}}+x\ kcal\]

(3) Application in everyday life : We have studied the application of the Le-Chatelier's principle to some equilibria involved in the physical and chemical systems. In addition to these, the principle is also useful to explain certain observations which we come across in every day life. A few out of them are discussed below,

(i)  Clothes dry quicker in a windy day : When wet clothes are spread on a stand, the water evaporates and the surrounding air tends to get saturated thus hampering the process of drying. On a windy day when breeze blows, the nearby wet air is replaced by dry air which helps the process of evaporation further. Thus, clothes dry quicker when there is a breeze.

            (ii)  We sweat more on a humid day : The explanation is the same as given above. In a humid day, the air is already saturated with water vapours. This means that the water that comes out of the pores of the body as sweat does not vaporise. This will result in greater sweating in a humid day.

            (iii) Carriage of oxygen by haemoglobin in blood : The haemoglobin (Hb) in red corpuscles of our blood carries oxygen to the tissues. This involves the equilibrium, \[Hb(s)+{{O}_{2}}(g)\,\]? \[Hb{{O}_{2}}(s)\]

            The blood that is in equilibrium with the oxygen of the air in the lungs finds a situation in the tissues where the partial pressure of oxygen is low. According to Le-Chatelier's principle, the equilibrium shifts towards the left so that some of the oxyhaemoglobin changes to haemoglobin giving up the oxygen. When the blood returns to the lungs, the partial pressure of the oxygen is higher and the equilibrium favours the formation of more oxyhaemoglobin.

            (iv) Removal of carbon dioxide from the tissues by blood : Blood removes \[C{{O}_{2}}\] from the tissues. The equilibrium is,  \[C{{O}_{2}}(g)+{{H}_{2}}O(l)\]? \[{{H}_{2}}C{{O}_{3}}(aq)\]? \[{{H}^{+}}(aq)+HCO_{3}^{-}(aq)\]

            Carbon dioxide dissolves in the blood in the tissues since the partial pressure of \[C{{O}_{2}}\] is high. However in the lungs, where the partial pressure of \[C{{O}_{2}}\] is low, it is released from the blood.

            (v) Sweet substances cause tooth decay : Tooth enamel consists of an insoluble substance called hydroxyapatite, \[C{{a}_{5}}{{(P{{O}_{4}})}_{3}}OH.\] The dissolution of this substance from the teeth is called demineralization and its formation is called remineralization. Even with the healthy teeth, there is an equilibrium in the mouth as

\[C{{a}_{5}}{{(P{{O}_{4}})}_{3}}OH(s)\] \[\underset{endothermic}{\mathop{\overset{exothermic}{\mathop{{}}}\,}}\,\]  \[5C{{a}^{2+}}(aq)+3PO_{4}^{3-}(aq)+O{{H}^{-}}(aq)\]

            When sugar substances are taken, sugar is absorbed on teeth and gets fermented to give H⁺ ions. The H⁺ ions produced distrub the equilibrium by combining with OH^– to form water and with \[PO_{4}^{3-}\]to form \[HPO_{4}^{2-}\]. Removal of products cause the equilibrium to shift towards right and therefore, \[C{{a}_{5}}{{(P{{O}_{4}})}_{3}}OH\]dissolves causing tooth decay.