Law of mass Action
This low is stated as below:
“The rate at which a substance reacts is directly proportional to its molar concentration and hence the rate of a chemical reaction is proportional to the product of the active masses of the reactant species at a given temperature”
Consider a reaction:  A+ B products
Rate of reaction α [A] [B] where [A] = molar conc. Of reactant A.
[B]= molar conc. Of reactant B.
Also consider a general reaction: aA + bB + cC             products
Rate of reaction α [A]a [B]b [C]c  = k [A]a[B]b[C]c
Here k= rate constant of the reaction at a given temperature.
Thus, the law of mass action is defined as:
“The rate of the reaction is directly proportional to the products of the active mass of the reactant spices, every raises to the power equal to its coefficients as represented by the balanced chemical eqn at a given temp.”
Le-Chatelier’s  Principle
According by this principle: “If a system in equilibrium is subjected to a change in conc. , pressure and temp. , the equilibrium shifts in the direction that tends to undo the effect of the changes”:
This principle helps in predicting the effect of change in concentration, pressure or temperature at the state of equilibrium.
(a)    Effect of change of concentration.  For a reaction in equilibrium at a particular temperature,                                           if the conc. Of one of the reactants is increased, the equilibrium will shifts towards the product side after partially using the reactant.
Consider the following reaction at equilibrium.
H2 (g) + I2(g) D 2HI (g);       Kc = [HI]/ [H2] [I2]
If at the state of equilibrium, the conc. Of H2 or I2 is increased, then more of HI will be formed so as to keep Kc constant.  Also if the molar conc. Of HI is increased at equilibrium, the equilibrium shifts to the left, thereby increasing the conc. of H2 and I2 so as to keep Kc constant.
(b)Effect of Pressure. If the system in equilibrium consists of gases, then the state of equilibrium is distributed by the change of pressure.
(I)Consider that number of gaseous moles on the reactant side is more than the number of gaseous moles on the product side in an equilibrium reaction. Then the increase in pressure favours the formation of products. For example:
                      N2 (g) + 3H2 (g)  D  2NH3 (g)
For such reactions, increase in pressure shifts the equilibrium towards right and the formation of ammonia is favoured.
(ii)Consider that number of gaseous moles on the reactant side is less than the number of gaseous moles on the product side in an equilibrium reaction. Then, the increase in pressure shifts the equilibrium in the backward direction.
For example:          N2O4 (g)    D   2NO2 (g)
For such reactions, the increase in pressure does not favour the formation of product.
(iii)Consider that the number of gaseous moles on the reactant side is equal to those on the product side. Then, the change in pressure does not change the state of equilibrium.  Examples of reactions are:        H2 (g) + I2 (g)  D  2HI (g)
                                                   N2 (g) + O2 (g)  D 2NO (g)

(c)Effect of Change of Temperature.  According to le-chatelier’s principle.
(i)                  If the temperature of the system in equilibrium is increased. Then equilibrium shifts in a direction in which heat is absorbed. It may be noted that if a reaction is exothermic in the forward direction, it is equally endothermic in the backward direction.
              An increase in temperature favours the endothermic reaction whereas decrease in temperature favours an exothermic reaction.
Example:  Consider the formation of ammonia:
                                N2 (g) + 3H2 (g) D 2NH3 (g)           ∆H=-92kJ
 It is exothermic in the forward direction. Clearly, it is endothermic in the backward direction.   Thus, the formation of NH3is favoured by decrease in temperature.
   (d)Effect of catalyst. The addition of a catalyst increases the rates of the opposing reaction to the same extent. This hastens the approach of equilibrium.
                Thus, a catalyst has no effect on the state of equilibrium.
    (e)Effect of adding an inert gas
          (I) If the inert gas is added to the equilibrium system at constant volume, then the total pressure will increase. The molar concentrations of the reactant and the product species will not change. Hence, the state of equilibrium will not change.
(ii)If the inert gas is added at constant pressure, there will be net increase in volume. Due to this, the molar concentration of reactant and product species will decrease.
    




Visit www.topposts.in for more and latest news