Degree of Dissociation

Introduction

Degree of dissociation describes how much of a starting species breaks into products by the time equilibrium is reached. It is a fraction between zero and one for a simple dissociation starting from that species alone. Stating the initial basis and balanced stoichiometry prevents a percentage from being mistaken for an equilibrium constant.

Core explanation

For A ⇌ B + C, begin with n₀ moles A and no B or C. If fraction α dissociates, A consumed is αn₀, leaving n A = n₀(1 − α). Products are n B = n C = αn₀, assuming coefficients one and no other sources. The total gas moles, if all are gases, become n total = n₀(1 + α). This total changes because one A produces two gas particles.

The same bookkeeping works with concentration at fixed volume: [A]eq = C₀(1 − α), [B]eq = [C]eq = C₀α. In an ideal concentration expression, Kc = [B][C]/[A] = C₀α²/(1 − α). This equation shows that α depends on both K and the initial concentration. It is not a constant property independent of conditions.

For A ⇌ 2B, product amount is 2αn₀, not αn₀. The initial A consumed remains αn₀. The stoichiometric coefficient controls product formation, while α refers to the fraction of original A units that reacted. With starting products already present, a single α definition needs careful adaptation because net reverse change is possible.

Degree of dissociation is often reported as a percentage: 100α%. A 20% dissociation means α = 0.20 and 80% of the initially specified A remains under the simple model. It does not mean the equilibrium mixture is 20% product by total mole fraction; because total moles can change, the product mole fraction may differ.

Gas partial-pressure problems can combine α with total pressure. Calculate equilibrium mole fractions from the resulting mole amounts, then p i = y iPtotal. Avoid assuming each partial pressure equals its mole amount times a fixed pressure without dividing by the changing total gas amount.

Step-by-step reasoning

1. Write the dissociation equation and initial amount n₀ or C₀. 2. Set amount consumed = α times that initial amount. 3. Use coefficients to generate product amounts. 4. Form K or mole fractions from equilibrium amounts and solve for α if needed.

Visual explanation

Draw ten A units initially. If α = 0.30, cross out three A units and replace each with its balanced set of products, leaving seven A units.

Real-world analogy

If 30 of 100 packages are opened, the opening fraction is 0.30. Counting the items released requires knowing how many items each package contains, analogous to reaction coefficients.

Real-world example

The equilibrium dissociation of a gaseous compound can be summarized by α from measured composition. Engineers can use that fraction to estimate how much original feed remains.

Why?

Why does α depend on starting concentration in A ⇌ B + C? Dissociation increases particle count, and the equilibrium quotient contains the baseline concentration as well as α.

Common misconception

“α is the mole fraction of product.” It is the fraction of starting A that reacted; product mole fractions require the full final mole inventory.

Worked example

Initially 2.00 mol N₂O₄ undergoes N₂O₄ ⇌ 2NO₂ with α = 0.25. Consumed N₂O₄ is 0.50 mol, leaving 1.50 mol. NO₂ formed is 2(0.50) = 1.00 mol. Total gas amount is 2.50 mol, so y(NO₂) = 1.00/2.50 = 0.40, not 0.25.

Quick check

1. If α = 0.10 for A ⇌ B + C starting with 1.0 mol A, how much A remains? Answer: 0.90 mol A under the simple model.

Exam focus

Define the species whose initial amount is multiplied by α. Apply balanced coefficients to products and recompute total moles before gas mole fractions.

Advanced insight

The equilibrium extent ξ can be written ξ = αn₀ for a one-reactant initial basis. This links percentage-dissociation language to the general stoichiometric extent used in reaction-network calculations.

Summary

Degree of dissociation is the reacted fraction of a specified starting species. Balanced coefficients convert that fraction into all equilibrium amounts, and K may then determine α.

Practice questions

1. What fraction remains if α = 0.35? Answer: 1 − 0.35 = 0.65 of the starting species remains. 2. For A ⇌ 2B, how much B forms from 1 mol A with α = 0.20? Answer: 0.40 mol B, because 0.20 mol A reacts and produces twice as much B. 3. Is percent dissociation necessarily the same as product mole percent? Answer: No. The total number of moles may change during dissociation.