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Factors Affecting the Position of Equilibrium

 

The following are factors that may affect the position of equilibrium:

1. Temperature

A change in the temperature of a system already in equilibrium could either shift the equilibrium to the right (favoring the forward reaction) or to the left (favoring the backward reaction).

This depends on whether the forward reaction is exothermic or endothermic. For exothermic forward reaction (∆H = - ve):

Increase in the temperature of a reversible reaction in equilibrium whose forward reaction is exothermic will lead to the backward reaction being favored.

Example, for the reaction:

H2(g) + I2(g) reversible reaction arrow 2HI(g)   ∆H = -ve

The forward reaction is exothermic, while the backward reaction will be endothermic. At equilibrium, there is a constant output of heat taking place which maintains the temperature of the system. Increase in temperature will cause the equilibrium position to move backwards, i.e. producing more H2 and I2.

Explanation: naturally, to produce HI and to attain equilibrium, heat is given off by the system (exothermic). Now, if the system is given heat from an external source, the heat available become excess and the equilibrium position is disturbed.

According to Le Chatelier’s principle, the system will respond in order to annul this effect. In response, the system drives to the direction which requires heat to occur (i.e. the backward reaction), so that the excess heat now in the system can be absorbed.

A decrease in the temperature of the system on the other hand will offset the equilibrium, to favor the forward reaction (i.e. more HI is produced).

Explanation: as has been said, at equilibrium, the system maintains a certain temperature. If the system is deprived of heat by way of reducing the temperature, the equilibrium position will be disturbed and the system responds by causing more of H2 and I2 to react, to produce more heat to make up for the lost one. Subsequently, more HI is produced.

For endothermic forward reaction (∆H =+ve):

Increase in temperature - as was mentioned above, an increase in the temperature of the system by giving heat to it will disturb the equilibrium position. And in response, the system will move towards the direction that would absorb the excess heat.

Therefore, for a reversible reaction whose forward reaction is endothermic (absorbs heat to occur), the forward reaction will be favored.

Example, for the reaction:

H2O(l) reversible reaction arrow H2O(g)   ∆H = +ve

Increase in temperature will favor the production of more steam, H2O(g) rather than water, H2O(l) in the reacting vessel.

Decrease in temperature - this will cause the prevailing temperature of the system at equilibrium to reduce. The system has to annul this effect by favoring the reaction which produces heat (in order to make - up for the lost heat).

Hence the backward reaction (which is exothermic) is favored, i.e. more water is formed than steam.

Note:

  • The enthalpy change given for a reversible reaction is the enthalpy change for the forward reaction.

  • When forward reaction is exothermic (∆H = -ve), then the backward reaction is endothermic (∆H = +ve)

  • The balance of the above two enthalpies gives the prevailing enthalpy of the system at equilibrium, and which maintains its temperature.

  • For exothermic forward reactions - increase in temperature favors backward reaction, while decrease in temperature favors forward reaction.

  • For endothermic forward reaction - increase in temperature favors forward reaction, while decrease in temperature favors backward reaction.

2. Pressure

Gaseous, rather than solids and liquids reactions are much affected by change in pressure. This is because the volumes of solids and liquids are very little affected even by large changes in pressure. To predict the effect of pressure on a gaseous reaction in equilibrium, we may consider both Boyle’s and Avogadro’s laws.

Boyle’s law states that the volume of a fixed mass of a gas changes inversely with pressure at constant temperature. Hence, if the pressure of the system at equilibrium is increased, provided every other factor that may affect the system is constant, the equilibrium is disturbed and it moves towards a state of lower volume, in order to cancel the effect.

In the same way, if pressure is reduced, a state of higher volume is favored. Avogadro’s law states that equal volumes of all gases contain the same number of molecules at the same temperature and pressure.

Therefore, in a stoichiometry, the ratio of the coefficients of all species is taken to be equivalent to the ratio of their combining volumes. By combining these two laws, we can successfully predict the direction of equilibrium resulting from a change of pressure.

Example, for these reactions below:

N2O4(g) reversible reaction arrow 2NO2(g)    

From the stoichiometry of the reaction, the ratio of moles of N2O4 to 2NO2 is 1 to 2. Hence, by Avogadro’s law, this can be expressed as 1 vol to 2 vols. i.e.

N2O4(g) reversible reaction arrow 2NO2(g)    
1 vol.         2vols.    

By increasing pressure - a state of lower volumes is favored (by Boyle’s law). Hence, the equilibrium moves towards the production of more N2O4 (i.e. backward reaction is favored).

By decreasing pressure - a state of higher volumes is favored. I.e. the forward reaction is favored. The same explanation applies to all similar cases: Example,

(a) 2O3(g) reversible reaction arrow 3O2(g)    
     2 vol.         3 vols.    

Increase in pressure - more of ozone, O3 is produced. That is, the backward reaction is favored. Decrease in pressure - more of oxygen, O2 is produced. That is, the forward reaction is favored.

(b) N2(g) + 3H2(g) reversible reaction arrow 2NH3(g)    
     1 vol.   3 vols         2 vols.    
 Total    4 vols.            2 vols.    

Increase in pressure - more of NH3 (occupies less volume) is produced. That is, forward reaction is favored.

Decrease in pressure - more of N2 and H2 (occupy higher volume) are produced. That is, backward reaction is favored.

(c) H2(g) + I2(g)  reversible reaction arrow    2HI(g)    
     1 vol.   1 vol         2 vols.    
 Total  2 vols.               2 vols.    

Notice that there is no difference between total volume on both sides, therefore, pressure change will not affect the equilibrium position of the system. That is, the equilibrium constant and position remain unchanged.

3. Concentration

At equilibrium, the concentrations of materials or species on both sides of the reaction are equal. For the hypothetical reaction:

A + B  reversible reaction arrow    C + D    

 At equilibrium, [A] + [B] = [C] + [D]

Now, if there is a change in the concentration of any of the materials, the equilibrium position will be disturbed, and in accordance with Le Chatelier’s principle, the system will respond in order to cancel the effect.

Therefore, for the hypothetical reaction above, increase in the concentration of any of A or B or both will cause the equilibrium to move forward, i.e. more C and D will be produced together. This is because, to remove the excess A or B, or both, the system has to cause more reactions between A and B, and this leads to the production of more C and D together.

A and B become reduced. Decrease in the concentration of A or B or both will cause the reverse reaction. More of C and D are caused to react in order to replace the lost A or B or both, that is, the backward reaction is favored. C and D become reduced.

In the same way, increase in the concentration of either C or D or both will favor backward reaction, that is, more A and B are produced together. C and D become reduced. And decrease in concentration of C or D or both will favor the forward reaction, that is, the system causes more of A and B to react in order to replace the lost C or D or both. A and B become reduced.

4. Change in Volume of The System

Gaseous reactions are mostly affected by change in volume. An increase in the volume of the vessel containing a reversible gaseous reaction at equilibrium, while keeping the temperature and concentration of the species involved constant, will cause the equilibrium position to shift to the production of more of the product(s) of higher total volume.

This is because, pressure is reduced, hence more volume become available to be occupied. A decrease in the volume of the reacting vessel will favor the side of lower total volume, because pressure is increased, hence less volume is available to be occupied.

Example, for the reaction:

2NO2(g)  reversible reaction arrow   N2O4(g)    
2 vols.           1 vol.    

Increasing the volume of the system (that is, the reacting vessel) - more of NO2 is formed (it occupies higher volume), the backward reaction is favored.

Decreasing the volume of the system - more of N2O4 is formed because it occupies the lower volume - forward reaction is favored. However, an equilibrium with no difference in total volume on either side will not be affected by a change in the volume of the system.

5. Common Ion Effect

In accordance with Le Chatelier’s principle, addition of the products of a reaction to a system at equilibrium will cause the equilibrium to shift in the direction of the reactants.

Thus, addition of a strong acid to a solution of a weak acid will suppress the ionization of the latter because the H3O+ is a common dissociate. Example, for the ionization of ethanoic acid in solution.

CH3COOH(aq)  reversible reaction arrow   CH3COO-(aq) + H+(aq)    

Addition of HCl or any other strong acid to the solution will cause the equilibrium to shift to the left, favoring backward reaction.

That is, the ionization of ethanoic acid is reduced and more unionized CH3COOH is produced.

Another example is the addition of a chloride, example, NaCl to a solution of AgCl.

AgCl(s)  reversible reaction arrow   Ag+(aq) + Cl-(aq)    

The backward reaction is favored, more precipitation of AgCl are formed.


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