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# Equilibrium Constant

We know that a chemical reaction involves the conversion of reactant molecules to products. A chemical reaction can be represented with the help of chemical equation. Chemical equation is a symbolic representation of a chemical reaction which indicates the molecular formulas of reactants and products which is separated by the single or double headed arrow. The single headed arrow indicates that the reaction is irreversible in nature whereas the double headed arrow indicates that the reaction is reversible in nature. The irreversible reaction moves in one direction only from reactant to product such as thermal decomposition of potassium chlorate results the formation of potassium chloride with oxygen gas. But oxygen gas and KCl cannot revert into potassium chlorate again.

The reversible reactions can move in both directions. In other words, in a reversible reaction reactant can convert into product and product can again convert again to reactant molecule. As the reaction proceeds in the forward direction, the concentration of a product increases but as the rate of backward reaction increases, it converts the reactant molecules to products again. During reversible reactions, the rate of forward reaction becomes equal to rate of backward reaction. This stage of reversible reaction is known as equilibrium and can be represented in terms of equilibrium constant. Let’s discuss the role of equilibrium constant in reversible reaction.

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## Equilibrium Constant Definition

The definition for equilibrium constant is given below."At a given temperature, the product of concentrations of the reaction product each raised to the respective stoichiometric coefficients in the balanced chemical equation divided by the product  of concentrations of the reactants raised to their individual stoichiometric coefficients has a constant value"For gaseous reactions, the concentration terms are replaced by partial pressures because at a given temperature the partial pressure of gases is proportional to its concentration.
Equilibrium constant expression for the gaseous reaction:

aA +bB cC + dD

is given by

Pc, PD, PA and PB are partial pressures of A, B, C and D.

### Characteristics of Equilibrium Constant

1. It should be noted that the reciprocal of Kc is also a constant and it will be the equilibrium constant for the backward reaction.
2. Equilibrium constant varies with temperature. But at a particular temperature, its value remains constant. Its value increases with temperature.
3. It is independent of the initial concentration.

### Relationship between Kp and Kc

The ideal gas equation is written as

PV = nRT
Or, P = (n / V) RT

P is the partial pressure, n is the amount of gas, v is the volume in m3 and T is the temperature in Kelvin.

If the concentration is in mol /L and P is in atmosphere.

Then

P = C RT

PA = CA RT, PB = CB RT, PC = CC RT and PD = CD RT.

So,

Kp = Kc RT ∆n

Where, ∆n = number of moles of products – number of moles of reactants in the balance chemical equation.

## Equilibrium Constant Units

The units of equilibrium constant depend upon the number of the reactants and products involved in the reaction. However, solids, pure liquids and solvent present in excess are not considered while writing the units.
1. If the number of moles of the product is the same as the number of moles of reactants, k has no units.Example

H2(g) + I2(g) 2HI(g)

Kc for this reaction would be written as:

Kc = [HI]2 / [H]2 [I]2 = (mol / L)2 / (mol / L) (mol / L) no units.

2. If the number of moles of the product is not the same, then the K will have some units.For the reaction

PCl5(g) PCl3 + Cl2(g)

Kc = equilibrium constant equation =

So, the unit of this equilibrium constant is mol.L-1.

## Equilibrium Constant Expression

If two gases A and B are capable of equilibrium, then we have

Arbitrary amounts of A(g) and B(g) are introduced into a closed container at 298 K. Because the amounts are arbitrary, the quotient of the partial pressures Q = PB / PA in the container will not be the equilibrium constant Keq.

The chemical potentials GA and GB will probably not be in the standard state and instead they will differ

Here, the 1.0 in the denominators signify that the partial pressures are relative to the standard state of 1.0 bar. The difference between the chemical potentials of the reactant state and product state is

This is simply put as

As the reaction progresses ΔrG is not zero and "G" of the reacting system is not constant with time "t".

When a chemical reaction reaches completion, the pressure quotient Q = PB / PA is arrived at a value such that

and hence we have

The free energy change of the system has arrived at a Gibbs potential energy minimum. Under these conditions, we have ΔrG = 0 so that

the expression now changes to

## Conditions that Affect the Equilibrium

The conditions that affect the equilibrium are
• According to Le Chatelier’s principle, if a system at equilibrium is disturbed by changing the variables such as pressure, temperature or concentration, then the system will tend to adjust itself so as to minimize the effect of that change as far as possible.
• This principle of Le Chatelier’s highlights the behavior of a system at equilibrium, if it is subjected to changes in parameters like pressure, temperature, or concentration.

## Calculating Equilibrium Constant

The ways to calculate the equilibrium constant are
• The determination of equilibrium constant requires the equilibrium concentrations or partial pressures of the reactants and the products.
• The equilibrium concentrations are determined without affecting the equilibrium state.
• Measurements of physical quantities like density, refractive index, light absorption, electrical conductivity etc. may be employed for evaluating the equilibrium concentrations and in such cases, the equilibrium state is not affected.

## Chemical Equilibrium

There are two types of chemical equilibria; namely homogeneous equilibria in which all the reactants and products are the same phase and heterogeneous equilibria, where the reactants and the products are present in two or more phases.

### Chemical Equilibria in Homogeneous Systems

1. Homogeneous reactions in gaseous systems

The gaseous reactions are broadly classified into two categories :
1. Gaseous reactions involving no change in the number of moles.
2. Gaseous reactions in which the number of moles of reactants differ from the number of moles of products.
2. Homogeneous equilibrium in liquid system
1. The principle of Law of Mass action is also applicable to reactions in liquid systems.
2. The equilibrium constant is found to be independent of the volume of the system.

### Chemical Equilibria in Heterogeneous Systems

In heterogeneous systems there are two more phases.
1. In heterogeneous equilibrium, the equilibrium constant involves only the activities or partial pressures of the gaseous constituents and does not include any terms for the concentrations of either pure solids or liquids.
2. By applying the law of mass action, and assuming one of the reactants is solid, we could get condensed equilibrium constant equal to the product of partial pressures of the products formed.

## Law of Mass action

The law of mass action states that at constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of reacting species, with each concentration term raised to the power equal to the numerical coefficient of the species in the chemical equilibrium.
So, for the reaction

aA + bB Products

The law of mass action can be depicted as:
Rate = k [A]a [B]b

Where k = rate constant.

## Law of Chemical Equilibrium

The law of equilibrium can be obtained by applying the law of mass action to a reversible reaction at equilibrium.
Let us consider the reaction:

A + B C + D

Applying the law of mass for the forward reaction and backward reactions

Rate of forward reaction ∞ [A] [B] = kf [A] [B]

Kf is the rate constant for the forward reaction. [A] and [B] are the molar concentrations of reactant A and B.The rate of backward reaction, similarly, would be:

Rate of backward ∞ [C] [D] = Kb [C] [D]

Kb is the rate constant of the backward reaction.
Therefore, kf [A] [B] = Kb [C] [D]
Or

Since Kf and kb are constants, therefore, the ratio kf / kb is also a constant and represented by Kc. And the above expression is the equilibrium constant expression.

The constant Kc is called the equilibrium constant for a reversible reaction. For all chemical reactions in general, the expression for the equilibrium constant is interrelated to stoichiometric coefficients of the chemical equation for the reaction.

## Factors Affecting Equilibrium

Some of the factors which affect the equilibrium constant value are:

1. Concentration

If the concentration of one ingredient is changed, the equilibrium is lost.

2. Temperature

Change in temperature shifts the equilibrium to the favorable side.

3. Catalyst involved in the reaction

Presence of a catalyst may affect either side of the reaction, thereby destroying the equilibrium.

## Equilibrium Constant Problems

Below you could see problems

### Solved Examples

Question 1: 1. Phosphorus pentachloride decomposes according to the reaction PCl5 → PCl3 + Cl2 3 moles of PCl3 is placed in 1.0 liters closed vessel and allowed to reach equilibrium at 523 K. Equilibrium constant Kc is 1.8. What is the composition of the mixture of equilibrium?
Solution:

Let us write the equilibrium constant expression

Kc = $\frac{[PCl3] [Cl2]}{[PCl5]}$

The initial concentration of PCl5 is [PCl5] Initial = 3 mol. L-1

Let us construct the Equilibrium constant table

 Compounds Initial Concentration Final Concentration PCl5 3.0 mol. L-1 3.0 mol. -x L-1 PCl3 - X Cl2 - X

Substituting the values from the table

Kc = $\frac{[PCl3] [Cl2]}{[PCl5]}$
= x * x / (3.0 – x) = 1.8

x calculated from this equation is 1.59.

So, equilibrium concentrations of all the compounds would be:

PCl5 = 3.0 – 1.59 = 1.41 mol/l
PCl3 = 1.59 mol/l
Cl2 = 1.59 mol /l

Question 2: A mixture of H2 and I2 is allowed to react at 448 degree Celsius. When the equilibrium is established, the concentrations of participants are [H2] = 0.46 mol/l, [I2] = 0.39 mol/l and [HI] = 3.0 mol/l. Calculate the Kc value for the reaction.
Solution:

Writing the equilibrium constant expression for the reaction:

Kc = $\frac{[HI]^ 2}{ [H2][I2]}$

Lets substitute the values as:

Kc = ${[3.0]_2}{[0.46][0.39]}$= 50.

So, the equilibrium constant value for this reaction is 50. There are no units, since concentration terms on the top and the bottom cancel out each other.

Question 3: At 700K, the equilibrium constant Kc is 3.1 x 10-7 mol/l. Calculate the Kp for the reaction. 2SO3 (g) --> 2SO2 (g) + O2 (g)
Solution:
For the reaction: ∆n = (2+1) = 3.

Kp = Kc (RT) ∆n

R = gas constant = 0.082 L. atm/K. mol
T = 700 Kelvin.

Substituting the values for delta n, R and temperature, the value of Kp would be equal to:

Kp = 1.78 x 10-5 atm.

 More topics in Equilibrium Constants Chemical Reactions Reversible Reaction Dynamic Equilibrium Reaction Order Zero Order Reaction First Order Reaction Second Order Reaction Steady State Approximation
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