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Law of Mass Action, Chemical Equilibrium Assignment Help, Tutor Help
Rate of Chemical Reaction, Law of Mass Action and Law of Chemical Equilibrium
Law of mass action is stated by Guldberg and Waage. It states about the influence of the
concentration of the reactants on the rate of reaction.
Law of mass action states that the rate at which substance reacts is proportional to its
active mass and the rate of chemical reaction is proportional to the product of the active
masses of the reactants.
As said in the previous session, active mass is the number of moles per liter. It is represented
by placing the chemical formula of the substance in square brackets. For example HCl is represented
as [HCl].
Explanation of law of mass action:
Consider a sample reaction of
A + B ↔ X +Y
As per the law of mass action the rate of forward reaction α [A] [B]
= Kf [A] [B]
Where Kf is proportionality constant and is termed as rate constant for forward reaction.
Similarly, rate of backward reaction α [X] [Y]
=Kb [X] [Y]
Were Kb is the proportionality constant and is termed as rate constant for backward reaction.
Explanation of law of mass action:
Consider a sample reaction of
A + B ↔ X +Y
As per the law of mass action the rate of forward reaction α [A] [B]
= Kf [A] [B]
Where Kf is proportionality constant and is termed as rate constant for forward reaction.
Similarly, rate of backward reaction α [X] [Y]
=Kb [X] [Y]
Were Kb is the proportionality constant and is termed as rate constant for backward reaction.
Law of Chemical Equilibrium:
When the above stated law of mass action is applied to a reaction in equilibrium, the
result is termed as the law of chemical equilibrium.
For example the reaction, aA + bB ↔ xX + yY
According to law of mass action
The rate of forward reaction α [A]a [B]b
= Kf [A]a [B]b
Similarly, the rate of backward reaction α [X]x [Y]y
=Kb [X]x [Y]y
Similarly, the rates of both the reactions are same at the state of equilibrium.
Kf [A]a [B]b =Kb [X]x [Y]y
According to law of mass action
The rate of forward reaction α [A]a [B]b
= Kf [A]a [B]b
Similarly, the rate of backward reaction α [X]x [Y]y
=Kb [X]x [Y]y
Similarly, the rates of both the reactions are same at the state of equilibrium.
Kf [A]a [B]b =Kb [X]x [Y]y
kf [X]x [Y]y
---- = -----------
Kb [A]a [B]b
---- = -----------
Kb [A]a [B]b
Kf and Kb are constant at constant temperature and the ratio of Kf / Kb is also constant at
constant temperature. It is represented by K and is termed as equilibrium constant. The above
reaction is given as
[X]x [Y]y
K = -----------
[A]a [B]b
and the expression is termed as law of chemical equilibrium.
K = -----------
[A]a [B]b
The law of chemical equilibrium states the product of molar concentration of the products
raised to the power equal to its co-efficient, divided by the product of the molar concentration
of the reactants raised to its co-efficient, is constant at constant temperature and is termed as
equilibrium constant.
Characteristics of equilibrium constant
Calculation of Molar concentration
- Its value remains constant at a given temperature irrespective of the direction of approach.
- The value of the equilibrium constant remains constant at given temperature and pressure irrespective of the concentration of the reactants and products.
- The value of equilibrium constant depends on the nature and temperature of the reaction but it remains unaffected in the presence or absence of catalyst.
- It gives information about the reaction proceeding in a particular direction at a given temperature.
From the above discussion, it can be seen that it is essential to determine the molar
concentration of the reactants and products in order to determine the equilibrium constant.
Here is an example for the calculation of molar concentration.
Consider the reaction
H2 + I2 ↔ 2HI
Let a and b be the initial moles of the two reactants.
Let x be the number of moles of the reactants that have reacted at the point of equilibrium.
Now, the above reaction can be given as follows:
Here is an example for the calculation of molar concentration.
Consider the reaction
H2 + I2 ↔ 2HI
Let a and b be the initial moles of the two reactants.
Let x be the number of moles of the reactants that have reacted at the point of equilibrium.
Now, the above reaction can be given as follows:
Initial moles
a
b 0
H2 + I2 ↔ 2HI
At equilibrium (a-x) (b-x) 2x
H2 + I2 ↔ 2HI
At equilibrium (a-x) (b-x) 2x
Now, let us see how to calculate the molar concentration of a dissociation reaction.
Initial moles
x
0 0
2NH3 ↔ N2 + 3H2
At equilibrium (a-x) ½ x 3/2 x
2NH3 ↔ N2 + 3H2
At equilibrium (a-x) ½ x 3/2 x
Where x is the degree of dissociation.
Solved problems
- Law of mass action is stated by --------------
- The rate at which a substance reacts is proportional to
- Active mass of the substance
- Product of number of moles of the reactants
- Total number of moles of the reactants and products
- None
- Rate of forward and backward reaction are given by the constants
- Kb and Kf
- Kf and Kb
- K and Kc
- Any of the above
- Which of the following is true
- The rate of chemical reaction remains constant at constant temperature
- the value of the chemical equilibrium constant remains constant irrespective of the presence of the catalyst
- Molar concentration of a substance can be obtained by dividing the number of moles by total volume in liters.
- All
- Information about the reaction proceeding in a particular direction is given by ----------------
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