Solved Objective Problems: Chemical Equilibrium | Chemistry Class 11 - NEET PDF Download

Chemical equilibrium occurs when the rates of the forward and reverse reactions become equal and the macroscopic properties (concentrations/partial pressures) of reactants and products remain constant with time. For a general reaction aA + bB ⇌ cC + dD the equilibrium constant in terms of concentration is given by the expression Kc = [C]c[D]d / ([A]a[B]b). For gas-phase reactions the pressure equilibrium constant Kp is related to Kc by the relation Kp = Kc(RT)Δn, where Δn = (c + d) - (a + b), R is the gas constant and T is the absolute temperature. According to Le Châtelier's principle, a change in concentration, pressure or temperature shifts the position of equilibrium so as to oppose the change; however, only temperature changes the numerical value of the equilibrium constant K. A catalyst speeds up both forward and reverse reactions equally and therefore does not change the value of K, it only helps the system reach equilibrium faster. The van 't Hoff relation gives the temperature dependence of K: d(ln K)/dT = ΔH°/(RT²), so for reactions with positive ΔH° (endothermic) K increases with increasing temperature, and for negative ΔH° (exothermic) K decreases with increasing temperature.

Problem 1

Problem 1 : The enthalpies of two reactions are DH₁ and DH₂ both positive with DH₂ > DH₁ . If the temperature of reacting system is increased from T₁ to T₂ predict which of the following alternative is correct ?

(A)  

(B) 

(C) 

(D) 

Ans. D 

Explanation: The temperature dependence of an equilibrium constant is given by the van 't Hoff relation. 
d(ln K)/dT = ΔH°/(RT²)
For endothermic reactions ΔH° > 0, so ln K increases with T and hence K increases as temperature increases.
For two endothermic reactions with ΔH₂ > ΔH₁, the magnitude of d(ln K)/dT is larger for the reaction with larger ΔH°.
Therefore the equilibrium constant for the reaction with ΔH₂ increases more on raising the temperature from T₁ to T₂, which corresponds to option (D).

Problem 2

Problem 2: At a certain teperature 2 moles of crbonmonoxide and 3 moles of chlo rine were allowed to reach equilibrium according to the reaction CO + Cl₂  COCl₂  in a 5 lit vessel. A t equilibrium if one mole of CO is present then equilibrium constant for the reaction is :

(A) 2  (B) 2.5  (C) 3.0  (D) 4

Ans. B
Explanation: Write the reaction with stoichiometry and initial and equilibrium mole values.
CO + Cl₂ ⇌ COCl₂
Initial moles: CO = 2, Cl₂ = 3, COCl₂ = 0
At equilibrium CO = 1 (given), so the change in CO = 2 - 1 = 1 mole reacted.
Therefore Cl₂ reacted by 1 mole and COCl₂ formed 1 mole.
Equilibrium moles: CO = 1, Cl₂ = 3 - 1 = 2, COCl₂ = 1
Convert to concentrations using vessel volume 5 L.
[CO] = 1/5 = 0.20 M
[Cl₂] = 2/5 = 0.40 M
[COCl₂] = 1/5 = 0.20 M
Equilibrium constant Kc = [COCl₂] / ([CO][Cl₂])
Kc = 0.20 / (0.20 × 0.40)
Kc = 0.20 / 0.08 = 2.5
Therefore the correct value is 2.5, option (B).

Problem 3

Problem 3 : The numerical value of the equilibrium constant or any chemical change is affected by changing the

(A) concentration of product                            (B) catalyst

(C) concentration of reacting substance   (D) temperature

Ans. D
Explanation: The numerical value of the equilibrium constant depends only on temperature.
Changing concentrations of reactants or products shifts the position of equilibrium but does not change K.
A catalyst changes the rate at which equilibrium is reached but does not change the equilibrium constant.
Therefore only temperature change affects the numerical value of K.

Problem 4

Problem 4: The reaction : 3O₂  2O₃,  DH= 69,000 calories is aided by :

(A) high temperature and low pressure  (B) high temperature and high pressure

(C) low temperature and high pressure  (D) low temperature and low pressure

Ans. (B)
Explanation:  The reaction is endothermic (ΔH > 0), so increasing temperature favours the forward reaction and promotes the formation of ozone (O₃).
The number of moles of gas decreases from 3 moles of O₂ to 2 moles of O₃. Therefore, increasing pressure shifts the equilibrium towards the product side.
Hence, both high temperature and high pressure favour the formation of ozone.

Problem 5

Problem5 : In the homogenous reaction 2X₂ + 2Y₂  2X₂Y₂, DH is negative for forward direction. The mass of X₂Y₂ increased at equilibrium by

(A) Raising the temperature (B) employing a suitable catalyst

(C) reducing the total pressure (D) raising the total pressure

Ans. D 
Explanation: Forward reaction is exothermic (ΔH negative), so lowering temperature would favour products; raising temperature would favour reactants.
Catalyst has no effect on equilibrium composition; it only speeds up attainment of equilibrium.
Count gaseous molecules: reactants total = 2 + 2 = 4 moles; products total = 2 moles.
Increasing total pressure favours the side with fewer moles of gas (products), thereby increasing the amount of X₂Y₂.
Therefore raising the total pressure increases the mass of product; option (D) is correct.

Problem 6

Problem 6: The equilibrium constant at 323^oC is 1000. What would be its value in the presence of a catalyst in the forward reaction. 

A + B   C +  D                                       38 kcal

(A) 1000 x concentration of catalyst 
(B) 1000

(C) 

(D) impossible to tell

Ans. B 
Explanation: A catalyst accelerates both the forward and reverse reactions equally and thus does not change the position of equilibrium or the numerical value of the equilibrium constant at a given temperature.
Therefore K remains 1000 in the presence of a catalyst at the same temperature.

Problem 7

Problem 7 : K_p for the reaction : CO₂(g) + H₂(g)   CO(g) + H₂O(g) is found to be 160 at a given temperature. Originally equal number of moles of H₂ and CO₂ were placed in the flask . At equilibrium, the pressure of H₂ is 1.20 atm. What is the partial pressure of CO and H₂O?

(A) 4.80 atm each (B) 9.60 atm each

(C) 2.40 atm each (D) 1.20 atm each

Ans. A
Explanation: Write the reaction and let initial partial pressures of CO₂ and H₂ both be P₀.
CO₂ + H₂ ⇌ CO + H₂O
Change at equilibrium: both CO₂ and H₂ decrease by x; both CO and H₂O increase by x.
Equilibrium partial pressures: CO₂ = P₀ - x, H₂ = P₀ - x = 1.20 atm (given), CO = x, H₂O = x.
K_p = (P_CO × P_H2O)/(P_CO2 × P_H2) = x² / ((P₀ - x)²)
Since P₀ - x = 1.20 atm, K_p = x² / (1.20 × 1.20) = x² / 1.44
Given K_p = 160, so x² = 160 × 1.44 = 230.4
x = √230.4 ≈ 15.18 atm
Thus the correct equilibrium partial pressures of CO and H₂O would be approximately 15.18 atm each for K_p = 160 and H₂ = 1.20 atm.

Note: This numerical value is not among the options listed. If K_p had been 16 instead of 160, then x² = 16 × 1.44 = 23.04, x = 4.80 atm, which corresponds to option (A). Therefore either the given K_p or the printed options appear inconsistent; the algebra above shows the correct relation and numeric result for the stated K_p.

Problem 8

Problem 8: Kc for the reaction : N₂(g)+ O₂(g)   2NO(g) is 4.0 x 10^-6 at 400 K. Kp for the above reaction is

(A) 2.4 x 10^-3
(B) 4.0 x 10^-6

(C) 4 x 10^-6 x (RT)2²
 (D) None of these

Ans. B
Explanation: For the reaction N₂ + O₂ ⇌ 2NO the change in moles of gas Δn = (2) - (1 + 1) = 0.
The relation between Kp and Kc is Kp = Kc(RT)^Δn.
Since Δn = 0, (RT)^Δn = 1, therefore Kp = Kc.
Thus Kp = 4.0 × 10^-6, option (B).

Problem 9

Problem-9: Which one of the following reactions at equlibrium , with all reactans and products in the gaseous phase, would be uneffected by an increase in pressure.

(A) N₂ +3H₂  2NH₃
(B) 2CO+ O₂  2CO₂

(C) 2H₂ + O₂  2H₂O
(D) N₂ + O₂  2NO

Ans. D 
Explanation: Pressure changes affect equilibria for gas-phase reactions when the number of moles of gas changes between reactants and products (Δn ≠ 0).
Compare numbers of moles for each option:
(A) Reactants = 1 + 3 = 4 moles, Products = 2 moles → Δn = -2 (affected by pressure)
(B) Reactants = 2 + 1 = 3 moles, Products = 2 moles → Δn = -1 (affected)
(C) Reactants = 2 + 1 = 3 moles, Products = 2 moles → Δn = -1 (affected)
(D) Reactants = 1 + 1 = 2 moles, Products = 2 moles → Δn = 0 (unaffected)

Option (D) has equal number of moles on both sides, so increase in pressure does not change the equilibrium position.

Problem 10

Problem 10: For the gas phase reaction 2NO(g) 

 N₂(g) + O₂(g) , DH = -43.5 kcal, which one of the following is true for N₂(g) + O₂(g)   2NO(g)
(A) K is independent of T
 (B) K decreases as T decreases

(C) K increases as T decreases
 (D) K varies with addition of NO

Ans. C
Explanation: The reaction 2NO → N₂ + O₂ is exothermic (ΔH < 0).
For an exothermic reaction, lowering the temperature favours the forward direction and increases the equilibrium constant.
Therefore, the equilibrium constant K increases as the temperature decreases.