# Law of Mass Action

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#### Statement of Law of Mass Action:

• In 1864, Norwegian chemists Cato Guldberg and Peter Wage put forward the law of mass action
• It states that  “The rate of a chemical reaction is directly proportional to the product of active masses of reactants, at given temperature at that instant”.
• Where active mass is molar concentration per unit volume of that substance.  It is denoted by enclosing the symbol or the formula of that substance in the square bracket and expressed in mol dm-3.

#### Explanation of Law of Mass Action:

• Consider a hypothetical reaction A + B →  Products.
• According to the law of mass action the rate of reaction R ∝ [A] [B].
• For any general reaction aA + bB → Products.
• According to the law of mass action the rate  of reaction R ∝ [A]a [B]b

#### Equilibrium mixture:

• The mixture of reactants and products formed at the chemical equilibrium state is called equilibrium mixture.

#### Equilibrium Concentrations:

• The concentrations of the reactants and products in a chemical equilibrium state for a reversible reaction are called equilibrium concentrations.

### The Expression for Equilibrium Constant in Terms of Concentrations:

Consider a hypothetical reversible reaction A + B  ⇌  C + D

According to the law of mass action the rate of the forward reaction

Rf ∝ [A] [B]

∴  R =  Kf [A] [B]

Where Kf  = rate constant for the forward reaction

According to the law of mass action the rate of the backward reaction

Rb ∝ [C] [D]

∴  Rb =  Kb [C] [D]

Where Kb = rate constant for the backward reaction

Now, for a chemical equilibrium,

R= Rb

K[A] [B] = Kb [C] [D]

Where KC is known as equilibrium constant and the equation is called “mass law equation”.

Consider a general reversible reaction

aA + bB + cC + …   ⇌  lL  + mM  + nN + ….,    Then,

• The equilibrium constant ‘ KC‘ is defined as the ratio of the product of the equilibrium concentrations of the products to that of the reactants with each concentration term raised to the power equal to the stoichiometric coefficients of the substance in the balanced chemical equation.
• Concentration is expressed in terms of mol dm-3 or mol L-1 or M.
• The equilibrium constant is denoted by Kc.
• More is the numerical value of ‘K’ then more is the concentration of products in comparison with reactants & vis-a-vis.

#### Concentration Quotient of a Chemical Reaction:

• At a given temperature, the ratio of the product of the equilibrium concentrations of the products to that of the reactants with each concentration term raised to the power equal to the stoichiometric coefficients of the substance in the balanced chemical equation is called concentration ratio or concentration quotient.
• It is denoted by QC.
• At equilibrium, the concentration ratio is equal to the equilibrium constant Kc.
• Its significance is that it helps in the prediction of the direction in which the net reaction is proceeding at given concentrations or partial pressures of reactants and products.
• The generalization is
• If QC > KC: The reaction is taking place in a backward direction i.e. in the direction of reactants.
• If QC < KC: The reaction is taking place in a forward direction i.e. in the direction of products.
• If QC = KC: The reaction is in an equilibrium state and hence no net reaction is taking place.

#### Law of Chemical Equilibrium:

• At a given temperature, the ratio of the product of the equilibrium concentrations of the products to that of the reactants with each concentration term raised to the power equal to the stoichiometric coefficients of the substance in the balanced chemical equation has a constant value.

### The Expression for Equilibrium Constant in Terms of Partial Pressure:

#### Partial Pressure:

• It is the pressure exerted by a gas in a mixture of gases if it alone occupies the entire volume of the mixture of gases.
• The partial pressure of a gaseous component is proportional to mole fraction.
• The partial pressure of a gas is calculated using following formula

Where,

n = Number of moles of gaseous component

N = Total moles of a gaseous system

P = Total pressure of the gaseous system

#### The Expression for Equilibrium Constant in Terms of Partial Pressure:

• In the gaseous system, the concentrations in concentration quotients are replaced by partial pressure, because for given temperature the partial pressure of the gas is directly proportional to its concentration.

Consider a hypothetical reversible reaction

aA(g) + bB(g) ⇌  cC(g) + dD(g)

Then the equilibrium constant in terms of partial pressures is given by

#### Steps Involved in Writing Equilibrium Constant Expression of a Reaction:

• Write the balanced chemical equation for the reaction.
• Write the products of equilibrium concentrations of the products in the numerator. Omit pure solids, pure liquids and the solvents in dilute solutions.
• Write the products of equilibrium concentrations of the reactants in the denominator. Omit pure solids, pure liquids and the solvents in dilute solutions.
• Raise each concentration term to the power equal to stoichiometric coefficients of the species in the equation.

#### Homogeneous Equilibrium:

• The equilibrium in which all the substances involved exist in a single homogeneous phase is called homogeneous equilibrium.
• Examples:

N2(g) +  3 H2(g)  ⇌     2NH3(g)

H2(g) +  I2(g) ⇌     2HI(g)

2SO2(g) +  O2(g)  ⇌   2SO3(g)

NH3(aq) + H2O(l) ⇌     2NH4+(aq) +  OH-(aq)

2N2O(g) ⇌ 2N2(g)  +   O2(g)

CH3COOC2H5(aq) + H2O(l) ⇌ CH3COOH(aq) +   C2H5OH(aq)

#### Heterogeneous Equilibrium:

• The equilibrium in which the substance involved are present in different phases is called heterogeneous equilibrium.
• Examples:

CaCO3(s) ⇌  CaO(s) + . CO2(g)

CaCO3(s)+H2O(l)+CO2(g) ⇌ Ca2+(aq)+ 2HCO3(aq)

(NH4)2CO3(s) ⇌ 2NH3(g) + CO2(g)+  H2O(g)

2Mg(s) + O2(g)  ⇌    2MgO(s)

#### Note: Pure Liquids and Solids are Ignored While Writing the Equilibrium Constant Expression:

• Now, for pure solids and pure liquids, the molar mass M and density ρ are constant. Hence the concentration remains constant.
• When pure solids are involved in equilibrium, its concentration remains constant no matter how much of it is present. Therefore by convention, the concentrations of all solids involved in equilibrium are taken as unity.
• When pure liquids are involved in equilibrium, its concentration remains constant no matter how much of it is present. Therefore by convention, the concentrations of all liquids involved in equilibrium are taken as unity.
• For equilibria in the aqueous medium, the concentration of solvent (water) will not change appreciably because it is present in large excess. Hence by convention, the concentration of solvent (water) is taken as unity.
• Hence in general, pure liquids, pure solids, and solvents can be ignored while writing the equilibrium constant expression.

#### Writing the Expression for Kc and Kp:

• BaCO3(s)  ⇌  BaO(s) + CO2(g)

• 4NH3(g) + 5O2(g) ⇌   4NO(g)+ 6H2O(g)

• NH3(g) + HCl(g)  ⇌  NH4Cl(s)

• 3Cl2(g) + 2NO2(g)  ⇌  2NO2Cl3(g)

• Fe3+(aq) + SCN(aq) ⇌ FeNCS2+(aq)

• CH4(g) + 2H2S(g)  ⇌  CS2(g)+ 4H2(g)

• MgCO3(s)  ⇌  MgO(s)  + CO2(g)

• AgBr(s) ⇌ Ag+(aq)+ Br(aq)

Kc = [Ag+][Br ]

• CH3COCH3(l) ⇌CH3COCH3(g)

Kc = [CH3COCH3(g)]

• CH4(g) + 2O2(g) ⇌ CO2(g)+ 2H2O(g)

• Al(s) + 3H+(aq) ⇌ Al3+(aq)+ 3/2H2(g)

• HPO42-(aq) + H2O(l) ⇌ H3O+(aq)+ PO43-(aq)

• Ag2O(s) + 2HNO3(aq) ⇌  2AgNO3(aq)+ H2O(l)

• Ni(s) + 4CO(g) ⇌  Ni(CO)4(g)

• CuO(s) + H2(g) ⇌  Cu(s)+ H2O(g)

• N2(g) + 3H2(g) ⇌  2NH3(g)

• Fe3+(aq) + 3OH(aq) ⇌  Fe(OH)3(s)

• 2N2O(g) ⇌ 2N2(g) + O2(g)

• C(s)+  CO2(g) ⇌  2CO(g)

### Relation Between Kc and Kp:

• Consider a hypothetical reversible reaction involving homogeneous gaseous phase

aA(g) + bB(g) ⇌  cC(g) + dD(g)

The equilibrium constant in terms of the partial pressure is given by

This is the relation between Kc and Kp.

• Where Δn = (Sum of the exponents in the numerator of concentration quotient) – (Sum of the exponents in the denominator of concentration quotient)

#### Steps Involved in Finding Relation Between Kc and Kp :

• Write the balanced chemical equation for the reaction.
• Find the change in the number of moles of gaseous species by following formula.

• Now, use the relation

### Examples to Find Relation between Between Kc and Kp :

#### Example – 4:

• For the reaction N2(g) + 3H2(g) ⇌  2NH3(g), the value of the equilibrium constant Kp is 3.6 x 10-2 at 500 K. Calculate the value of Kc for the reaction at the same temperature. Take R = 0.0831 bar lit mol-1 K-1.

#### Example- 5:

• For the reaction 2SO2(g)) + O2(g) ⇌ 2SO3(g), the value of the equilibrium constant Kp is 2.0 X 1010 bar-1 at 450 K. Calculate the value of Kc for the reaction at the same temperature. Take R = 0.0831 bar lit mol-1 K-1.

#### Example – 6:

• For the reaction 2NOCl(g) ⇌ 2NO(g)+ Cl2(g), the value of the equilibrium constant Kp is1.8 X 10-2 at 500 K. Calculate the value of Kc for the reaction at the same temperature. Take R = 0.0831 bar lit mol-1 K-1.

#### Example – 7:

• For the reaction CaCO3(s)  ⇌ CaO(s)+ CO2(g), the value of equilibrium constant Kp is 167 at 1073 K. Calculate the value of Kc for the reaction at the same temperature. Take R = 0.0831 bar lit mol-1 K-1.

#### Example – 8:

• Find the ratio of Kp/Kc for the reaction, CO(g) + 1/2O2(g)  ⇌  2CO2(g), at 500 K. Take R = 0.0831 bar lit mol-1 K-1.

#### Example – 9:

• Find the ratio of Kp/K for the reaction, N2(g) + O2(g)  ⇌   2NO(g)

#### Example – 10:

• The equilibrium constant Kp for the reaction H2(g)  + I2(g)  ⇌  2HI(g) is 130 at 510 K. Calculate Kc for following reactions a) 2HI(g)  ⇌ H2(g)  + I2(g),  b) HI(g)  ⇌ 1/2H2(g)  + 1/2I2(g), c) 1/2H2(g)  + 1/2I2(g)  ⇌  HI(g).

∴  Kc = 130

• For the reaction 2HI(g)  ⇌ H2(g)  + I2(g),

• For the reaction, HI(g)  ⇌ 1/2H2(g)  + 1/2I2(g)

• For reaction,  1/2H2(g)  + 1/2I2(g)  ⇌  HI(g).

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