Degree of Dissociation
Chemical Equilibrium
5
JEE Qs
8%
Hard
75
min
Master the setup of ICE tables involving alpha to correctly derive equilibrium concentrations/partial pressures, which is fundamental for Kp and Kc calculations.
🧮 Key Formulas
✅ Key Points for JEE
- 1Degree of dissociation (alpha) is the fraction of reactant that has dissociated at equilibrium, ranging from 0 to 1.
- 2It is crucial for calculating equilibrium concentrations or partial pressures for systems where a single reactant dissociates into multiple products.
- 3Le Chatelier's Principle dictates how alpha changes with temperature, pressure (for gaseous reactions), and concentration.
- 4For gaseous reactions, changes in alpha directly affect the total number of moles, which in turn influences total pressure and individual partial pressures.
- 5Always set up an ICE (Initial, Change, Equilibrium) table using initial moles/concentrations and alpha to derive expressions for equilibrium quantities before substituting into Kp/Kc expressions.
⚠️ Common Mistakes
- ✕Confusing alpha (a fraction) with the actual number of moles dissociated.
- ✕Incorrectly applying stoichiometry to determine equilibrium moles of products, especially when the dissociating species forms multiple products or has a coefficient greater than one.
- ✕Ignoring the change in total moles (and thus total pressure for constant volume) when dissociation occurs in gaseous systems, leading to errors in Kp calculations.
- ✕Assuming alpha is constant; it varies with conditions like temperature and pressure, unlike the equilibrium constant (K).
📝 Practice Questions
See allQ58.For the reaction, H2( g) + I2( g) ⇌2HI( g) Attainment of equillibrium is predicted correctly by : 2025 (24 Jan Shift 2) JEE Main Previous Year Paper (1) (2) (3) (4) −−
Q51.Consider the equilibrium CO( g) + 3H2( g) ⇌CH4( g) + H2O( g) If the pressure applied over the system increases by two fold at constant temperature then (A) Concentration of reactants and products increases. (B) Equilibrium will shift in forward direction. (C) Equilibrium constant increases since concentration of products increases. (D) Equilibrium constant remains unchanged as concentration of reactants and products remain same. Choose the correct answer from the options given below : (1) (A), (B) and (C) only (2) (A) and (B) only (3) (A), (B) and (D) only (4) (B) and (C) only
Q65.A vessel at 1000 K contains CO2 with a pressure of 0.5 atm . Some of CO2 is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm , then Kp is : (1) 1.8 atm (2) 0.3 atm (3) 3 atm (4) 0.18 atm
Q55.Consider the reaction X2Y( g) = X2( g) + 12 Y2( g) The equation representing correct relationship between the degree of dissociation (x) of X2Y(g) with its equilibrium constant Kp is ______ . Assume x to be very very small. (1) (2) x = 3√2Kpp x = 3√2Kp2p (3) (4) x = 3√Kpp x = 3√Kp2p 2025 (23 Jan Shift 2) JEE Main Previous Year Paper
Q53.At temperature T, compound AB2( g) dissociates as AB2( g) ⇌AB(g) + 12 B2( g) having degree of dissociation x (small compared to unity). The correct expression for x in terms of Kp and p is 2025 (29 Jan Shift 1) JEE Main Previous Year Paper (1) 4√2Kpp (2) 3√2Kpp (3) 3√2 pK2p (4) √Kp
Q37.At −20∘C and 1 atm pressure, a cylinder is filled with equal number of H2, I2 and HI molecules for the reaction H2( g) + I2( g) ⇌2HI(g), the Kp for the process is x × 10−1 . x = _____ [Given : R = 0.082 L atm K−1 mol−1 ] (1) 0.01 (2) 10 (3) 2 (4) 1
NCERT Chapters
- Class 11 Chemistry Ch 7: Equilibrium