CUET UG Chemistry: Physical Chemistry - Chemical Equilibrium, Le Châtelier's Principle, Kp, Kc, Kx

Must-Know (15–20 detailed bullets)

• Equilibrium constant ( K_c ) is defined as the ratio of the product of molar concentrations of products to that of reactants, each raised to their stoichiometric coefficients; for ( aA + bB \rightleftharpoons cC + dD ), ( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} ).
• ( K_p ) is the equilibrium constant in terms of partial pressures; for gaseous reactions, ( K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b} ).
• Relationship between ( K_p ) and ( K_c ): ( K_p = K_c (RT)^{\Delta n_g} ), where ( \Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} ), ( R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1} ), ( T ) in Kelvin.
• For the reaction ( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) ), ( \Delta n_g = 2 - (1 + 3) = -2 ), so ( K_p = K_c (RT)^{-2} ).
• ( K_x ) is the equilibrium constant in terms of mole fractions; it is not commonly used because it is pressure-dependent unless ( \Delta n_g = 0 ).
• Pure solids and pure liquids are not included in the expression for ( K_c ) or ( K_p ); e.g., for ( CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) ), ( K_p = P_{CO_2} ).
• If ( K > 10^3 ), products dominate at equilibrium; if ( K < 10^{-3} ), reactants dominate; if ( K \approx 1 ), neither predominates.
• Equilibrium constants are temperature-dependent only; changing pressure, concentration, or adding catalyst does not change ( K ).
• For a reverse reaction, ( K' = \frac{1}{K} ); for a reaction multiplied by a factor ( n ), ( K' = K^n ).
• Le Chatelier’s Principle: if a system at equilibrium is disturbed, it will shift to counteract the change and re-establish equilibrium.
• Increasing pressure (by decreasing volume) shifts equilibrium toward the side with fewer moles of gas; e.g., in ( N_2 + 3H_2 \rightleftharpoons 2NH_3 ), high pressure favors forward reaction.
• Increasing temperature favors endothermic direction; for endothermic reactions, ( K ) increases with temperature.
• For the reaction ( 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) ), ( \Delta H = -198\ \text{kJ mol}^{-1} ), so low temperature favors forward reaction (exothermic).
• Addition of inert gas at constant volume does not affect equilibrium because partial pressures remain unchanged.
• Addition of inert gas at constant pressure increases volume, decreasing partial pressures, and shifts equilibrium toward more moles of gas.
• A catalyst speeds up both forward and reverse reactions equally; it helps reach equilibrium faster but does not change ( K ) or equilibrium composition.
• For heterogeneous equilibrium involving gases and solids, only gaseous species appear in ( K_p ); e.g., ( NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g) ), ( K_p = P_{NH_3} \cdot P_{H_2S} ).
• When ( \Delta n_g = 0 ), ( K_p = K_c ); e.g., for ( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) ), ( K_p = K_c ).
• The reaction quotient ( Q ) has the same form as ( K ); if ( Q < K ), reaction proceeds forward; if ( Q > K ), reverse; if ( Q = K ), at equilibrium.
• For the dissociation of ( PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) ), increasing volume (decreasing pressure) shifts equilibrium to the right (more moles of gas).

Difficulty Level

Intermediate — requires conceptual clarity between ( K_p ), ( K_c ), and ( K_x ), and application of Le Chatelier’s principle to various disturbances.

Common CUET Traps (3 bullets)

• Trap: Students assume adding inert gas always shifts equilibrium. Avoid: It shifts only if added at constant pressure, not at constant volume.
• Trap: Believing catalyst changes equilibrium position. Avoid: Catalyst only reduces time to reach equilibrium; ( K ) and composition remain unchanged.
• Trap: Using concentrations for ( K_p ) or partial pressures for ( K_c ). Avoid: ( K_c ) uses mol/L, ( K_p ) uses atm or bar; convert using ( P = CRT ) if needed.

Practice MCQs (5 questions)

Q1. For the reaction ( 2NO_2(g) \rightleftharpoons N_2O_4(g) ), ( \Delta H = -57.2\ \text{kJ mol}^{-1} ). Which condition favors the formation of ( N_2O_4 )?
A. High temperature and high pressure
B. Low temperature and high pressure
C. High temperature and low pressure
D. Low temperature and low pressure
Answer: B
Explanation: Reaction is exothermic and decreases moles of gas, so low T and high P favor product.
Why others fail: Option A incorrectly assumes high T favors exothermic forward reaction.

Q2. For which reaction is ( K_p = K_c )?
A. ( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) )
B. ( PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) )
C. ( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) )
D. ( 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) )
Answer: C
Explanation: ( \Delta n_g = 0 ), so ( K_p = K_c ).
Why others fail: Options A, B, D have ( \Delta n_g \ne 0 ), so ( K_p \ne K_c ).

Q3. The equilibrium constant ( K_c ) for ( A + B \rightleftharpoons C + D ) is 100. If initial concentrations of A and B are 1 M each, what is the equilibrium concentration of C?
A. 0.9 M
B. 0.95 M
C. 0.99 M
D. 1.0 M
Answer: A
Explanation: Solving ( K_c = \frac{x^2}{(1-x)^2} = 100 ) gives ( x = 0.91 \approx 0.9\ \text{M} ).
Why others fail: Option C assumes nearly complete reaction without calculation.

Q4. For the reaction ( CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) ), which factor affects the equilibrium constant ( K_p )?
A. Addition of ( CaCO_3(s) )
B. Increase in total pressure
C. Increase in temperature
D. Addition of inert gas at constant volume
Answer: C
Explanation: Only temperature changes ( K_p ); others affect position but not ( K ).
Why others fail: Option B seems correct due to pressure change, but ( K_p = P_{CO_2} ), independent of total pressure.

Q5. At 700 K, ( K_p = 1.8 \times 10^{-5} ) for ( 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) ). What is ( K_c ) at this temperature?
A. ( 1.8 \times 10^{-5} )
B. ( 1.5 \times 10^{-6} )
C. ( 2.1 \times 10^{-4} )
D. ( 1.1 \times 10^{-6} )
Answer: B
Explanation: ( \Delta n_g = 3 - 2 = 1 ), so ( K_c = \frac{K_p}{(RT)^{\Delta n_g}} = \frac{1.8 \times 10^{-5}}{(0.0821 \times 700)^1} \approx 1.5 \times 10^{-6} ).
Why others fail: Option A ignores conversion and assumes ( K_p = K_c ).

Last?Minute Revision (15–20 one-liners)

• ( K_p = K_c (RT)^{\Delta n_g} ) — use ( R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1} ) when pressure in atm.
• Pure solids and liquids never appear in equilibrium constant expressions.
• For ( \Delta n_g = 0 ), ( K_p = K_c ) — e.g., ( H_2 + I_2 \rightleftharpoons 2HI ).
• Increasing pressure shifts equilibrium to side with fewer gas moles.
• Increasing temperature favors endothermic reaction; ( K ) increases with T for endothermic.
• Catalyst does not affect equilibrium constant or position — only rate.
• ( Q < K ): reaction proceeds forward; ( Q > K ): reverse; ( Q = K ): equilibrium.
• Inert gas at constant volume-no shift; at constant pressure-shifts to more gas moles.
• ( K > 10^3 ): products favored; ( K < 10^{-3} ): reactants favored.
• For reverse reaction, ( K' = 1/K ).
• For reaction multiplied by 2, ( K' = K^2 ).
• ( NH_4HS(s) \rightleftharpoons NH_3(g) + H_2S(g) ): ( K_p = P_{NH_3} \cdot P_{H_2S} ).
• ( PCl_5 \rightleftharpoons PCl_3 + Cl_2 ): increasing volume favors dissociation.
• ( K_x ) is dimensionless but pressure-dependent unless ( \Delta n_g = 0 ).
• Le Chatelier’s Principle applies to concentration, pressure, temperature — not catalyst.
• ( K_c ) has units unless ( \Delta n_g = 0 ); ( K_p ) may have units.
• For exothermic reactions, ( K ) decreases with increasing temperature.
• ( K ) is constant at constant temperature — verify from NCERT.
• Use ( R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} ) for energy; ( 0.0821 ) for ( K_p - K_c ) conversions.
• Mnemonic: “Low T and High P” for exothermic and fewer moles — LTHP sounds like “let hp” — helps recall favorable conditions.