What is the Law of Mass Action?

It's a principle stating that the rate of a chemical reaction is directly proportional to the product of the concentrations (active masses) of the reactants, each raised to the power of its stoichiometric coefficient.

Who proposed the Law of Mass Action?

Norwegian chemists Cato Maximilian Guldberg and Peter Waage proposed it in 1864, based on experimental studies of reversible reactions.

What does "active mass" refer to in this law?

"Active mass" is another term for the molar concentration of reactants or products, typically expressed in moles per liter.

What is the equilibrium constant (Kc)?

Kc is the ratio of the product of equilibrium concentrations of products (each raised to their stoichiometric coefficient) to the product of equilibrium concentrations of reactants (each raised to their stoichiometric coefficient). It reflects the state of a reaction at equilibrium.

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JEE Chemistry

Law of Mass Action

Law of mass action

This law helps us understand the behaviour of solutions during dynamic equilibrium by describing how the reaction rate relates to the concentrations of the reactants.

1.0Introduction

The Law of Mass Action states that the rate of a chemical reaction is directly proportional to the product of the active masses (concentrations) of the reactants, each raised to the power of its stoichiometric coefficient in the balanced chemical equation.

Based on experimental studies of various reversible reactions, Norwegian chemists Cato Maximilian Guldberg and Peter Waage proposed in 1864 that the following equilibrium equation relates the concentrations in an equilibrium mixture:

$K_{c} = \frac{[ C ] [ D ]}{[ A ] [ B ]}$

Kc is the equilibrium constant, and the expression on the right side is called the equilibrium constant expression. This equation is also called the law of mass action because, in the early days of chemistry, concentration was referred to as “active mass.”

The law of mass action is a theory stating that the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants in the chemical equation. The reaction rate for a substance is proportional to its active mass, which is the concentration of the substance (in moles per litre).

It explains that the rate of a chemical reaction is directly proportional to the product of the reactants' concentrations (active masses), each raised to the power of its stoichiometric coefficient at a given temperature.

Active Mass:

The term "active mass" refers to the molar concentration of reactants or products.
The Law of Mass Action states that the higher the concentration of reactants, the more molecules are available to react and form products.
For gases, it can be expressed in terms of partial pressure.

Rate of Reaction:

For a reaction:
aA+bB→cC+dD
The rate of the forward reaction is:

$r_{f} = k_{f} [A]^{a} [B]^{b}$
The rate of the reverse reaction is:
$r_{r} = k_{r} [C]^{c} [D]^{d}$

Equilibrium Constant (K):

At equilibrium, the rates of the forward and reverse reactions are equal:
$r_{f} = r_{r}$
The equilibrium constant K is given by:
$K = \frac{k _{f}}{k _{r}} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}} $

To better understand the law of mass action , consider the reaction between gaseous H₂ and I₂, conducted in a sealed vessel at 731 K:

H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) $⇌$ 2HI(g)H2(g)+I2(g)⇌2HI(g)

Initial Conditions:
1 mol of H₂ reacts with 1 mol of I₂ to form 2 mol of HI.

Six experiments were conducted under varying initial conditions:

In experiments 1, 2, 3, and 4, only gaseous H₂ and I₂ were introduced into a sealed vessel, and equilibrium was observed when the intensity of the purple colour remained constant.
In experiments 5 and 6, equilibrium was achieved from the opposite direction by starting with HI only.

Observations:

From experiments 1 to 4, it was noted that:

Moles of H₂ reacted=Moles of I₂ reacted= 12 × Moles of HI formed

$Moles of H₂ reacted = Moles of I₂ reacted = \frac{1}{2} \times Moles of HI formed$ . Moles of H₂ reacted=Moles of I₂ reacted=21×Moles of HI formed.

Experiments 5 and 6 showed that at equilibrium:

$[H_{2}]_{e q} = [I_{2}]_{e q}$

Based on these facts, a relationship between the concentrations of reactants and products can be derived. A simple expression to consider is:

$K_{c} = \frac{[ H I ] _{e q}^{2}}{[ H _{2} ] _{e q} \cdot [ I _{2} ] _{e q}}$

where Kc represents the equilibrium constant.

The equilibrium constant expression highlights the relationship between the concentrations of reactants and products in a chemical reaction at equilibrium. The concentrations' exponents are the stoichiometric coefficients from the balanced chemical equation.

For the reaction:

$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$

the equilibrium constant (K_C)is written as:

$K_{C} = K_{c} = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]}$

Here, the subscript c indicates that the equilibrium constant is expressed in terms of molar concentrations $(mol/L m o l / L)$ .

2.0Key Points on the Equilibrium Constant

Equilibrium Concentrations:
The concentrations used in the expression are equilibrium values. The subscript “eq” is often omitted, as it is implied.
Equilibrium Law:
At a given temperature, the equilibrium constant is the ratio of the product of concentrations of the products (each raised to their stoichiometric coefficient) to the product of concentrations of the reactants (each raised to their stoichiometric coefficient).
General Form:
For a reaction:
aA + bB $⇌$ cC + dD,
the equilibrium constant is:

$K_{c} = \frac{[ C ] ^{c} [ D ] ^{d}}{[ A ] ^{a} [ B ] ^{b}}$ For Reversed Reactions:
If the reaction is reversed, the equilibrium constant becomes the reciprocal:

$Kc' = 1 Kc . K_{c}^{'} = \frac{1}{K _{c}} . Kc' = Kc 1$ .

Effect of Stoichiometric Changes:
If the stoichiometric coefficients in the equation are multiplied by a factor nnn, the equilibrium constant is raised to the power nnn:

$Kc'' = (Kc) n . K_{c}^{''} = (K_{c})^{n} . Kc'' = (Kc ) n$ .

Example:
For the reaction:

$\frac{1}{2} H_{2} (g) + \frac{1}{2} I_{2} (g) ⇌ H I (g)$ ,
the equilibrium constant becomes:
$K_{c}^{''} = K_{c} ..$

Reaction	Equilibrium Constant
H₂(g)+I₂(g)⇌2HI(g)	Kc=[H2][I2]/[HI]2
Reverse reaction	K_c′=1Kc
Reaction with coefficients multiplied by n	Kc′′=(Kc)n

3.0Homogeneous Equilibria

In homogeneous systems, all reactants and products exist in the same phase. Examples include:

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g) (g a seo u s p ha se)$
$CH_{3} COOC_{2} H_{5} (a q) + H_{2} O (l) ⇌ CH_{3} COOH (a q) + C_{2} H_{5} OH (a q)$

Equilibrium Constant in Gaseous Systems

For reactions involving gases, equilibrium constants can be expressed in terms of:

Concentration ( $K_{c}$ ): $Kc = [p ro d u c t s] coe ff i c i e n t s [re a c t an t s] coe ff i c i e n t s K_{c} = \frac{[ products ] ^{coefficients}}{[ reactants ] ^{coefficients}} Kc = [re a c t an t s] coe ff i c i e n t s [p ro d u c t s] coe ff i c i e n t s .$
Partial Pressure ( $K_{p}$ ): $K_{p} = \frac{( p _{products} ) ^{coefficients}}{( p _{reactants} ) ^{coefficients}}$

The relationship between K_Cand K_P:

$K_{p} = K_{c} (RT)^{Δ n}, Δ n = (moles of gaseous products) - (moles of gaseous reactants)$ .

Δn=(moles of gaseous products)−(moles of gaseous reactants).

Example Reactions

$H_{2} (g) + I_{2} (g) ⇌ 2 H I (g) H 2 (g) + I 2 (g) ⇌ 2 H I (g)$ :

$K_{p} = K_{c}$ , as $Δ n$ = 0.

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)
Relationship $K_{p} = K_{c} (RT)^{- 2}$ , since Δn=−2.

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