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Chapter: 11th 12th std standard Class Organic Inorganic Physical Chemistry Higher secondary school College Notes

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Law of chemical equilibrium and equilibrium constant with example

Law of Mass action Two Norwegian Chemists, Guldberg and Waage, studied experimentally a large number of equilibrium reactions. In 1864, they postulated a generalisation called the Law of Mass action. It states that: "the rate of a chemical reaction is proportional to the active masses of the reactants". By the term `active mass', it is meant the molar concentration i.e., number of moles per litre.

Law of chemical equilibrium and equilibrium constant

Law of Mass action

 

Two Norwegian Chemists, Guldberg and Waage, studied experimentally a large number of equilibrium reactions. In 1864, they postulated a generalisation called the Law of Mass action. It states that:

 

"the rate of a chemical reaction is proportional to the active masses of the reactants". By the term `active mass', it is meant the molar concentration i.e., number of moles per litre.

Law of Mass Action based on the Molecular Collision theory

We assume that a chemical reaction occurs as the result of the collisions between the reacting molecules. Although some of these collisions are ineffective, the chemical change produced is proportional to the number of collisions actually taking place. Thus at a fixed temperature the rate of a reaction is determined by the number of collisions between the reactant molecules present in unit volume and hence its concentration, which is generally referred as the active mass.

 

Equilibrium constant and equilibrium law

Let us consider a general reaction

A + B  - - Kf --- > < ---Kr--- C + D

and let [A], [B], [C] and [D] represent the molar concentrations of A,B,C and D at the equilibrium point. According to the Law of Mass action,

Rate of forward reaction a[A][B] = Kf[A][B]

Rate of forward reaction a[C][D] = Kr[C][D]

where kf and kr are rate constants for the forward and reverse reactions.

At equilibrium, rate of forward reaction = rate of reverse reaction.

Therefore,

kf[A] [B]   = kr [C] [D]

kf / kr  = [C] [D] / [A] [B]

At any specific temperature kf/kr is a constant since both kf and kr are constants. The ratio kf/kr is called equilibrium constant and is represented by the symbol Kc. The subscript `c' indicates that the value is in terms of concentration of reactants and products. The equation (1) may be written as

Kf (Equilibrium constant) = [C] [D] (Products concentration) / [A] [B] (Reactants concentrations)

This equation is known as the equilibrium constant expression or equilibrium law. Hence [C], [D] [A] and [B] values are the equilibrium concentrations and are equal to equilibrium concentrations.

 

Equilibrium Constant Expression for a Reaction in General Terms

 

The general reaction may be written as

aA + bB -- >           < -- cC + dD.

where a,b,c and d are numerical quotients of the substance A,B,C and D respectively. The equilibrium constant expression is

Kc = [C]c [D]d / [A]a [B]b

where Kc is the Equilibrium constant. The general definition of the equilibrium constant may thus be stated as :

 

The product of the equilibrium concentrations of the products divided by the product of the equilibrium concentrations of the reactants, with each concentration term raised to a power equal to the coefficient of the substance in the balanced equation.

For Example

 

(a) Consider the equilibrium constant expression for the reaction

N2(g) + 3H2(g) -- >   < -- 2NH3 (g)

1.                 The equation is already balanced. The numerical quotient of H2 is 3 and NH3 is 2.

2.                 The concentration of the `product' NH3 is [NH3]2.

3.                 The product of concentrations of the reactants is [N2] [H2]3.

4.                 Therefore, the equilibrium constant expression is

Kc = [NH3]2 / [N2] [H2]3

 

(b) Consider the equilibrium constant expression for the reaction

N2O5 (g)-- >        < -- NO2(g) + O2 (g)

1.     The equation as written is not balanced. Balancing yields

2N2O5 -- >      < -- 4NO2 + O2.

2.     The coefficient of the product NO2 is 4 and of the reactant N2O5 is 2.

3.     The product of the concentrations of products is [NO2]4 [O2].

4.     The concentration of the reactant is [N2O5]2

5.     The equilibrium constant expression can be written as

Kc = [NO2]4 [O2] / [N2O5]2

(c) Consider the equilibrium constant expression of the reaction.

 

CH4(g) + H2O(g)      CO(g) + 3H2(g)

 

1.     Write the product of concentrations of `products' divided by the product (multiplication) of concentrations of `reactants'.

 

2.     The concentration of H2 is to be raised by its coefficient in the balanced equation. Thus, the equilibrium constant expression is :

 

Kc = [CO] [H2]3  / [CH4] [H2O]

 

When all the reactants and products are gases, we can formulate the equilibrium constant expression in terms of partial pressures exactly similar to equation (1).

 

The partial pressure of a gas in the equilibrium mixture is directly proportional to its molar concentration at a given temperature.

K p  = pc pd / PjAPkB

             

The gaseous equilibrium reaction of SO2 can be written as follows

2S02 (g) + O2 (g) --- > < -- 2SO3 (g)

Kp = (PSO3)2 / (PSO2)2. PO2 atm-1

As an example, Kp value can be calculated from the following data:

 

The total pressure in the reaction flask is 1 atm and the partial pressures of oxygen, SO2 and SO3 at equilibrium are 0.1 atm, 0.57 atm, 0.33 atm respectively.



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