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ChemistryMediumClass 11

Equilibrium Constant — Kc, Kp, Kx and relations

Chemical Equilibrium

5

JEE Qs

8%

Hard

75

min

Always correctly identify the phases of all species and ensure only gaseous species contribute to Δn, while excluding pure solids/liquids from K expressions.

🧮 Key Formulas

For aA(g) + bB(g) <=> cC(g) + dD(g):
Kc = ([C]^c * [D]^d) / ([A]^a * [B]^b)
Kp = (P_C^c * P_D^d) / (P_A^a * P_B^b)
Kp = Kc * (RT)^Δn
Δn = (sum of stoichiometric coefficients of gaseous products) - (sum of stoichiometric coefficients of gaseous reactants)
P_i = x_i * P_total (where P_i is partial pressure, x_i is mole fraction, P_total is total pressure)
Kx = (x_C^c * x_D^d) / (x_A^a * x_B^b)
Kp = Kx * (P_total)^Δn

✅ Key Points for JEE

  • 1Equilibrium constant (Kc, Kp, Kx) is a constant value for a given reaction at a specific temperature; it is independent of initial concentrations/pressures, catalyst presence, or path.
  • 2For heterogeneous equilibria (reactions involving solids or pure liquids), their concentrations/partial pressures are considered constant and are excluded from the equilibrium constant expression.
  • 3Δn (change in number of moles) for the Kp-Kc relation must be calculated ONLY considering gaseous reactants and products. The sign of Δn is crucial.
  • 4The value of the gas constant 'R' used in Kp = Kc(RT)^Δn must be consistent with the units of pressure (e.g., R=0.0821 L atm mol⁻¹ K⁻¹ for pressure in atm, or R=8.314 J mol⁻¹ K⁻¹ for pressure in Pa).
  • 5Equilibrium constants for reverse reactions are the inverse (1/K) of the forward reaction; for reactions multiplied by a factor 'n', K becomes K^n.

⚠️ Common Mistakes

  • Incorrectly including concentrations or partial pressures of pure solids or pure liquids in the equilibrium constant expressions.
  • Errors in calculating Δn, such as including non-gaseous species or getting the sign wrong (products minus reactants).
  • Using an inconsistent value of the gas constant 'R' (e.g., using 8.314 J mol⁻¹ K⁻¹ when pressure is in atmospheres).
  • Confusing the roles of stoichiometric coefficients (powers) and initial/equilibrium concentrations in setting up the K expression.

📝 Practice Questions

See all

Q58.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) −−

2025·Graph basedMedium

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

2025·MCQMedium

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

2025·MCQMedium

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

2025·MCQMedium

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

2025·MCQMedium

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

2024·MCQMedium

NCERT Chapters

  • Class 11 Chemistry Ch 7: Equilibrium