Degree of Dissociation (α) & How to Draw ICE Tables
If you want to solve numerical problems in Chemical Equilibrium, writing out the Kc formula is only the first step. To actually find the equilibrium concentrations, you must set up an ICE Table.
At the heart of the ICE table is a critical variable called the Degree of Dissociation (α). Let's decode exactly what α means and how to use it to handle complex stoichiometries seamlessly.
Video Tutorial: Constructing the ICE Table
Watch Abhishek Sengar sir from CHEMCA break down the exact definition of Alpha and construct ICE tables for various reaction stoichiometries (like 3R → 2P).
1. What is the Degree of Dissociation (α)?
The Degree of Dissociation (α) is defined as the fraction of the total initial moles of a reactant that have dissociated (broken down or reacted) to reach equilibrium.
- Range of α: The value of α always lies between 0 and 1.
- If α = 0: No reaction has occurred (0% dissociation).
- If α = 1: The reaction has gone to 100% completion (Complete dissociation).
- Note: Examiners will often give α as a percentage (e.g., "40% dissociated"). To use it in calculations, simply convert it to a decimal (α = 0.40).
If α is the fraction that dissociates, and you start with an initial concentration C, then the actual amount that reacted is exactly Cα. This is the secret to filling out the "Change" row!
2. Building the ICE Table
ICE stands for Initial, Change, and Equilibrium. Let's apply it to a basic reaction where 1 mole of Reactant gives 1 mole of Product (R ⇌ P).
| State | R | ⇌ | P |
|---|---|---|---|
| Initial (I) | C | 0 | |
| Change (C) | - Cα | + Cα | |
| Equilibrium (E) | C - Cα = C(1 - α) | Cα |
From this table, the equilibrium constant is:
Kc = [P] / [R] = (Cα) / C(1 - α) = α / (1 - α).
3. The "Stoichiometry Ratio" Trick
The standard ICE table is easy. But what happens if the reaction has different coefficients? For example: aR ⇌ bP. How do we fill out the "Change" row for the products?
The Rule: The reactant always loses -Cα. To find out how much product is formed, you multiply the reactant's loss (Cα) by the molar ratio: (Product Coefficient / Reactant Coefficient).
Fig: If 3 moles of R produce 2 moles of P, then 1 mole of R produces 2/3 moles of P. Therefore, Cα moles of R produce (2/3)Cα moles of P.
Example: 3R ⇌ 2P
| State | 3 R | ⇌ | 2 P |
|---|---|---|---|
| Initial (I) | C | 0 | |
| Change (C) | - Cα | + (2/3)Cα | |
| Equilibrium (E) | C(1 - α) | (2/3)Cα |
From here, Kc = [P]2 / [R]3.
Kc = ((2/3)Cα)2 / (C(1 - α))3.
Practice Questions for JEE & NEET
Test your understanding of α with these high-yield conceptual questions!
Question 1: In many textbooks, you will see the variable x used instead of α (e.g., the Change row is written as -x). What is the exact mathematical relationship between the amount dissociated (x) and the degree of dissociation (α)?
Answer: x = C × α
Reasoning:
This is a major source of confusion for students.
- α is a fraction or percentage (e.g., 0.20 or 20%). It tells you how much of the original pile reacted.
- x is an actual physical quantity (moles or Molarity). It is the exact number of moles that vanished.
Therefore, the actual amount that vanished (x) is equal to the initial concentration (C) multiplied by the fractional percentage that reacted (α). Both -Cα and -x are mathematically identical ways of writing the Change row!
Question 2: Consider the reaction 2A ⇌ B + 3C. If you start with an initial concentration of C for reactant A, what will be the equilibrium concentration of product C in terms of C and α?
Answer: (3/2) Cα
Reasoning:
Apply the "Stoichiometry Ratio Trick" from the lesson!
1. Reactant A will lose -Cα.
2. To find product C, multiply A's loss by the ratio of their coefficients: (Coefficient of C) / (Coefficient of A).
3. The ratio is 3 / 2.
4. Therefore, the Change for product C is + (3/2) Cα. Since it started at 0, its equilibrium concentration is (3/2) Cα.
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