Le Chatelier's Principle — Effect of T, P, concentration
Chemical Equilibrium
5
JEE Qs
8%
Hard
60
min
Master the qualitative predictions of Le Chatelier's Principle, paying special attention to the *only* factor (temperature) that changes the equilibrium constant K, and the conditions under which pressure/volume changes are effective.
🧮 Key Formulas
✅ Key Points for JEE
- 1Le Chatelier's Principle predicts the qualitative direction of equilibrium shift to counteract a stress (change in T, P, concentration) and establish a new equilibrium.
- 2Changes in concentration or pressure/volume shift the equilibrium position but *do not* change the numerical value of the equilibrium constant (K).
- 3Temperature is the *only* factor that changes the value of the equilibrium constant (K). For endothermic reactions (ΔH > 0), K increases with increasing temperature; for exothermic reactions (ΔH < 0), K decreases with increasing temperature.
- 4Pressure/volume effects are significant *only* if the number of moles of gaseous products (Δn_g) is different from the number of moles of gaseous reactants. Increasing pressure (or decreasing volume) shifts the equilibrium towards the side with fewer moles of gas.
- 5Addition of an inert gas at constant volume has no effect on equilibrium (as partial pressures don't change). At constant pressure, it increases the total volume, decreasing partial pressures of reacting gases, shifting equilibrium towards the side with more moles of gas.
- 6A catalyst speeds up both forward and reverse reactions equally, thus helping to attain equilibrium faster but *does not* affect the equilibrium position or the value of K.
⚠️ Common Mistakes
- ✕Assuming that changes in concentration or pressure/volume will alter the value of the equilibrium constant (K).
- ✕Incorrectly applying the effect of pressure or volume changes to reactions where the number of gaseous moles on both sides is equal (Δn_g = 0).
- ✕Confusing the effect of temperature on K for endothermic versus exothermic reactions.
- ✕Believing that adding a catalyst shifts the equilibrium position or changes the equilibrium constant.
📝 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