Degree of Dissociation from Vapour Density Calculations
Vapour density measurement is a classic method tested in Class 11, JEE Main, JEE Advanced, and NEET to find the degree of dissociation of gases at equilibrium. When a gas dissociates, total moles increase, lowering average molar mass and observed vapour density. Comparing theoretical and observed vapour densities allows direct determination of dissociation fractions and equilibrium constants.
Vapour Density Principles in Dissociating Systems
Vapour density is the ratio of the mass of a gas volume to that of an equal volume of hydrogen at identical temperature and pressure. By Avogadro's law, vapour density equals molar mass divided by the molar mass of H2 (taken as 2 in standard school curricula):
Theoretical Vapour Density (D) = Mth / 2
Observed Vapour Density (d) = Mobs / 2
When a gas like PCl5 or N2O4 dissociates on heating, one mole of reactant yields multiple moles of products. Total moles increase while total mass remains constant. Consequently, average molar mass drops, making observed vapour density d strictly less than theoretical vapour density D.
Derivation of the Degree of Dissociation Formula
Consider the dissociation of one mole of gas A into n moles of gaseous products: A(g) ⇔ n B(g). Let α be the degree of dissociation. At equilibrium: moles of A left = 1 - α, moles of B formed = n α, and total equilibrium moles = 1 + (n - 1)α.
By conservation of mass: Mth = [1 + (n - 1)α] × Mobs. Therefore, Mth / Mobs = D / d = 1 + (n - 1)α. Solving for α gives the formula.
Formula for Degree of Dissociation (α):
α = (D - d) / [ (n - 1) × d ] = (Mth - Mobs) / [ (n - 1) × Mobs ]
Variables:
- α = Degree of dissociation (fraction between 0 and 1; multiply by 100 for percentage)
- D = Theoretical vapour density = Mth / 2
- d = Observed vapour density = Mobs / 2
- Mth = Theoretical molar mass of reactant (g mol-1)
- Mobs = Observed average molar mass (g mol-1)
- n = Moles of gaseous products from one mole of reactant
Note on textbook conventions: At STP, older textbooks use 22.4 L mol-1 (based on 1 atm), whereas current IUPAC standards use 22.71 L mol-1 (based on 1 bar). State the convention used if gas volumes are required.
Exam details
- Source: practice problem written for this article (not a past-paper question)
- Relevant exams: JEE Main, JEE Advanced, NEET, CUET UG, Class 11 Board Exams
- Topic: Chemical Equilibrium and States of Matter
- Difficulty: Intermediate
Worked Numerical Examples
Question: At 250 °C, phosphorus pentachloride (PCl5) dissociates into PCl3 and Cl2 gas. The observed vapour density is 62.00. Using atomic masses P = 30.97 g mol-1 and Cl = 35.45 g mol-1, calculate the degree of dissociation α and percentage dissociation of PCl5.
Solution:
Step 1: Dissociation equation: PCl5(g) ⇔ PCl3(g) + Cl2(g). 1 mole PCl5 yields 2 moles of products, so n = 2.
Step 2: Calculate theoretical molar mass and vapour density: Mth = 30.97 + (5 × 35.45) = 208.22 g mol-1. D = 208.22 / 2 = 104.11. Observed d = 62.00.
Step 3: Primary route calculation of α: α = (D - d) / [ (n - 1) × d ] = (104.11 - 62.00) / [ (2 - 1) × 62.00 ] = 42.11 / 62.00 = 0.6792. Percentage dissociation = 0.6792 × 100 = 67.92%.
Step 4: Secondary route verification: Mobs = 2 × 62.00 = 124.00 g mol-1. Total equilibrium moles = Mth / Mobs = 208.22 / 124.00 = 1.6792. Thus α = 1.6792 - 1 = 0.6792 (67.92%). Both routes give identical results.
Answer: 0.6792 (or 67.92%)
Question: A sample of dinitrogen tetroxide (N2O4) gas at 333 K and 1.00 atm total pressure has an observed vapour density of 31.00. Using atomic masses N = 14.01 g mol-1 and O = 16.00 g mol-1, calculate: (a) the degree of dissociation α, and (b) the equilibrium constant Kp in atm.
Solution:
Step 1: Dissociation equation: N2O4(g) ⇔ 2 NO2(g), so n = 2. Mth = (2 × 14.01) + (4 × 16.00) = 92.02 g mol-1. D = 46.01. Observed d = 31.00.
Step 2: Primary route: α = (D - d) / [ (n - 1) × d ] = (46.01 - 31.00) / (1 × 31.00) = 15.01 / 31.00 = 0.4842.
Step 3: Partial pressures and Kp: Total moles = 1 + α = 1.4842. p(N2O4) = [ (1 - α) / (1 + α) ] × P = (0.5158 / 1.4842) × 1.00 = 0.3475 atm. p(NO2) = [ 2α / (1 + α) ] × P = (0.9684 / 1.4842) × 1.00 = 0.6525 atm. Kp = [p(NO2)]2 / p(N2O4) = (0.6525)2 / 0.3475 = 0.4258 / 0.3475 = 1.225 atm.
Step 4: Secondary route verification: Kp = 4α2P / (1 - α2) = [ 4 × (0.4842)2 × 1.00 ] / [ 1 - (0.4842)2 ] = 0.9378 / 0.7656 = 1.225 atm. Both routes agree.
Answer: α = 0.4842; Kp = 1.225 atm
Common Mistakes Students Make
- Swapping D and d: Theoretical vapour density D comes from Mth / 2, while d is measured experimentally. In dissociation, D > d. A negative α indicates D and d were reversed.
- Misidentifying n: Factor n is moles of gaseous products from ONE mole of reactant. For 2 SO3(g) ⇔ 2 SO2(g) + O2(g), dividing by 2 gives n = 1.5, not 3.
- Including non-gaseous phases: Only gases contribute to vapour density. In reactions like NH4HS(s) ⇔ NH3(g) + H2S(g), solid NH4HS must not be included.
- Confusing fraction with percentage: Degree of dissociation α is a fraction (0 to 1). Percentage dissociation is α × 100. Provide the format requested.
Exam Relevance and Format Breakdown
The table below summarizes how degree of dissociation from vapour density is tested across competitive exams.
| Exam | Question Format | Key Concepts Tested | Preparation Strategy |
|---|---|---|---|
| JEE Main | Multiple choice and numericals | Finding α for PCl5 and N2O4, molar mass link | Use α = (D - d) / [(n - 1)d] with D = M/2 |
| JEE Advanced | Multi-correct and linked problems | Combining vapour density with Kp, non-integer n | Derive total equilibrium moles from first principles |
| NEET | Objective questions | Direct formula calculation, percentage dissociation | Master PCl5 and N2O4 numericals |
| Class 11 Boards | Derivations and numericals | Deriving the α formula, defining vapour density | State initial and equilibrium moles clearly |
Turn this concept into marks: solve previous-year CSIR-NET questions and take a timed quiz.
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