Law of Chemical Equilibrium & Mass Action Made Easy | CHEMCA

Law of Chemical Equilibrium & Mass Action Made Easy

Published by Abhishek Sengar | CHEMCA India

Every single calculation you do in the Chemical Equilibrium and Ionic Equilibrium chapters relies on one fundamental mathematical relationship.

This relationship is known as the Law of Chemical Equilibrium (traditionally derived from the Law of Mass Action). If you understand how to write this expression flawlessly, you have essentially mastered 50% of the entire unit! Let's decode the official statement and see how to write it.

Video Tutorial: The Law of Equilibrium

Watch Abhishek Sengar sir from CHEMCA break down the exact statement of the law and demonstrate how to write the K_c expression for any general balanced equation.

Decoding the Statement

Consider a general, balanced reversible chemical reaction:

aA + bB ⇌ cC + dD

The Law of Chemical Equilibrium states that at a given, constant temperature, the ratio of the product of the equilibrium concentrations of the products to the product of the equilibrium concentrations of the reactants, with each concentration term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation, is a constant value.

Let's break that massive sentence down into simple, actionable steps:

Step 1: Products on Top. The concentrations of the right side ([C] and [D]) always go in the numerator.
Step 2: Reactants on Bottom. The concentrations of the left side ([A] and [B]) always go in the denominator.
Step 3: Coefficients become Powers. Look at the numbers balancing the equation (a, b, c, d). These numbers become exponents!

K_c = ([C]^c [D]^d) / ([A]^a [B]^b)

Fig: Visually tracking how coefficients transform into exponents in the equilibrium expression.

Crucial Note:
The square brackets [ ] specifically denote Molar Concentration (moles per liter, or Molarity). The values placed inside these brackets MUST be the concentrations at equilibrium, not the initial concentrations!

Practice Questions for JEE & NEET

Let's test your understanding of how K_c is calculated and what factors actually affect it!

Question 1 (The Temperature Trap): A student is analyzing the equilibrium N_2(g) + 3H_2(g) ⇌ 2NH_3(g). If the student suddenly adds more N₂ gas to the vessel at a constant temperature, what will happen to the numerical value of the Equilibrium Constant (K_c)?

Answer: The value of K_c remains exactly the same!

Reasoning:

This is one of the most famous trap questions in physical chemistry. The Equilibrium Constant (K_c) is a true constant. It only depends on Temperature.

Adding more N₂ changes the *Reaction Quotient* (Q_c) temporarily, causing the reaction to shift forward (Le Chatelier's Principle) to consume the extra N₂. But once the dust settles and a new equilibrium is reached, the final ratio of the concentrations will mathematically reduce back to the exact same K_c value as before!

Question 2 (Unit Calculation): Unlike a standard mathematical constant like Pi (π), the equilibrium constant often has units. Calculate the SI units of K_c for the Haber Process: N_2(g) + 3H_2(g) ⇌ 2NH_3(g).

Answer: (mol/L)^-2 or L² mol^-2

Reasoning:

The unit of K_c depends on the change in the number of gaseous moles (Δn_g).
Formula: Unit of K_c = (mol/L)^Δn_g.

Δn_g = (Moles of Gaseous Products) - (Moles of Gaseous Reactants)
Δn_g = 2 - (1 + 3) = 2 - 4 = -2.

Therefore, the unit is (mol/L)^-2, which can be algebraically flipped to L² / mol².

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