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Law of Mass Action

February 4, 2013

 Chemical equilibrium.

The experimental observations of chemical equilibrium tell us that most of the chemical reactions when carried out in closed vessels do not go to completion. Under these a conditions, a reaction starts by itself or by initiation, continues for some time at diminishing rates and ultimately appears to stop. The reactants may still be present but they do not appear to change into products any more. What happens in such case is that the products of the reaction start reacting at the same rate as the reactants. In other words, the rate of the back reaction becomes equal to the rate of the forward reaction.

Characteristic features of chemical equilibrium

Thus, in a given time as much of the products are formed as react back to give the reactants. The composition of the reaction mixture at a given temperature is the same irrespective of the initial state of the system, i.e., irrespective of the fact whether we start with the reactants or the products. The reaction in such conditions is said to be in a state of equilibrium.

The attainment of equilibrium can be recognized by noting constancy of observable properties such as pressure, concentration, density or color whichever may be suitable in a given case.

The relationship between the quantities of the reacting substances and the products formed can be worked out readily with the help of the law of mass action.

 law of Mass Action:

The law was stated by C.M.Guldberg and P.Waage in 1863 . The law gives the relation between the rate of chemical reaction and the active masses ( i.e. concentration ) of the reactants of a reaction .

Definition :  At any instant the rate of chemical reaction at a given temperature is directly to the product of the active masses of the reactants at that instant .

Active mass is taken as equivalent to molar concentration for reaction in solutions at low concentrations and for gaseous reactions at low pressures . Therefore the law of mass action is restated as

At any instant the rate of a reaction at a temperature is proportional directly to the product of themolar concentrations of the ractants at the instant .

The molar concentrations ( active masses ) are indicated in square brackets .

Application of Law of Mass Action

Let us consider a general reversible reaction

xA   +  yB     mC   +  nD

By applying the law to the forward reaction .

xA   +  yB     `|->`     products , we have

vf   `prop`   [A]x  [B]y                 where   vf  =  rate of forward reaction

vf    =  kf   [A]x   [B]y            where   kf  =  rate constant of forward reaction

By applying the law to the reverse reaction

mC   +  nD   `|->`   reactants

vb   `prop`   [C]m  [D]n        where  vb  =  rate of backward reaction

vb    =   Kb   [C]m  [D]n   where  kb  =  rate constant of backward reaction .

where [A] , [B] , [C] , [D] are equilibrium concentrations of A , B , C and D respectively .

At equilibrium   v=  vb    therefore  kf [A]x[B]y   =  kb [C]m [D]n

or      `(kf)/(kb)`    =   `([C]^m[D]^n)/([A]^x[B]^y)`

`(kf)/(kb)`    =   Kc     ;    Kc   =   concentration equilibrium constant

Kc     =    `([C]^m[D]^n)/([A]^x[B]^y)`

The equilibrium constant Kc for a reversible reaction is therefore written as

concentration equilibrium constant   Kc   =  `(a)/(b)`

where a  =  product of the equilibrium concentrations of the products

b  =  product of the equilibrium concentrations of the reactants

[ The concentration terms must be raised to the powers which are respectively the coefficients of the reactants and the products in the stoichiometric chemical equation of the reversible reaction ].

The Kc may have or may not have any units depending on the difference between ( m+n ) and ( x+y )

if  ( m+n )  =  ( x+y )   ;   [( m+n ) – ( x+y )}   =   `Delta`n = 0  then  Kc has no units

if  `Delta`n  =  positive , hen Kc has units of { mol.lit-1 }`Delta`n   and

fi  `Delta`n  =  negative  , then  Kc   has units  of  { lit . mol-1 }`Delta`n 

Some examples :

1 .       N2 (g)   +    3 H2 (g)   2 NH3 (aq)

Kc   =    `([NH3]^2)/([N2][H2]^3)`

2 .      2 SO2 (g)    +   O2 (g)   2 SO3 (g)

Kc   =   `([SO3]^2)/([SO2]^2[O2])`

3 .      CaCO3 (s)       CaO (s)    +   CO2 (g)

Kc    =    `([CaO][CO2])/([CaCO3])`
But [CaCO3] , [CaO] are taken as unity since CaCO3 and CaO are solids ,

kc    =   [CO2]

In the case of gaseous , reversible reactions , the equilibrium constant is written using equilibrium partial pressure (p) of the reactants and the products in place of the molar concentrations . For example for the reaction

xA (g)    +    yB (g)       mC (g)    +   nD (g)

The equilibrium constant is written as

Kp   =    `((Pc)^m*(Pd^n))/((Pa^x)*(Pb^y))`  

Some examples :

1 .  N2 (g)   +  3H2 (g)    2NH3 (g)

Kp   =    `((PNH3)^2)/((PN2)*(PH2)^3)`

2 .  2SO2   +   O2 (g)      2SO3 (g)

Kp    =     `((PSO3)^2)/((PSO2)^2(PO2))`

Relation between Kp and Kc

For reversible reactionswhich are gases, the equilibrium constant can be expressed as Kc and Kp by using the equilibrium molar concentrations of the equilibrium partial pressures respectively of the reactants and products . For example

xA (g)    +    yB (g)       mC (g)    +   nD (g)

Kc   =   `([C]^m[D]^n)/([A]^x[B]^y)`

Kp   =   `((Pc)^m*(Pd)^n)/((Pa)^x*(Pb)^y)`

These two  are related to each other by the equation

Kp   =    Kc * (RT)`Delta`n for ideal gaseous reactions .

where  `Delta`n = [ ( the number of molecules of gaseous products) – ( the number of molecules of gaseous reactants ) ]   in the balanced chemical equation of the reversible reaction .

R = gas constant   ;   T  =  Kelvin temperature

Some examples :

1 . N2 (g)    +   3H2 (g)    2NH3 (g)

Δn  =  { 2 – (1+3) }   =   ( 2 – 4 )   =   -2

Kp   =   Kc (RT)-2

then ,   Kp <  Kc     or   Kc  >  Kp

2 .    2SO2 (g)   +   O2 (g)       2SO3 (g)

Δ n   =    {  2 – (2+1) }   =   (2-3)   =   -1

Kp  =   Kc (RT)-1

Kp  <   Kc      or    Kc  >  Kp

Le Chatelier’s Principle

There are three main factors which alter the state of chemical equilibrium problems. These are: concentration, temperature and pressure.

The addition of a catalyst has no effect. The function of a catalyst, as already stated, is merely to hasten the approach of equilibrium. It does so by speeding up the forward as well as the backward reaction. Le Chatelier’s, a noted French chemist, studied the effect of concentration, temperature and pressure on a large number of chemical equilibrium. He summed up his conclusion in the form of a generalization known as Le Chatelier’s principle, which states as follows:

If equilibrium is subjected to a stress, the equilibrium shifts in such a way as to reduce the stress.

According to this principle if a system at equilibrium is subjected to a change of concentration, pressure or temperature, the equilibrium shifts in the direction that tends to undo the effect of the change.

Solve this problems

Example 1:

Find out:

What should be the pH of the solution when 0.100 L of 0.500 M sodium hydroxide NaOH is added to 0.150 L of 0.400 M hydrochloric acid HCl?

Example 2

Find out

To what volume be supposed to a solution have 0.050 mol of each monosaccharide (formation of disaccharide sugar from monosaccharide) be diluted in order to bring about 5% conversions to sucrose phosphate?


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