JEE Chemistry Chemical Equilibrium MCQs Set 05

Download JEE MCQs for JEE Chemistry: Chemical Equilibrium

Access targeted multiple-choice questions for Chemical Equilibrium designed to align with the latest JEE academic syllabus for JEE Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Chapter-wise Objective Questions: Chemical Equilibrium

Access the complete set of multiple-choice questions for Chemical Equilibrium below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: The ratio of active masses (molar concentrations in the same volume) of 22 g CO2, 3 g H2 and 7 g N2 in a gaseous mixture is:
(a) 22 : 3 : 7
(b) 0.5 : 3 : 7
(c) 1 : 3 : 1
(d) 1 : 3 : 0.5
Show Answer & Explanation

Answer: (d) 1 : 3 : 0.5

Explanation:
1. Moles: CO2 = 22/44 = 0.5; H2 = 3/2 = 1.5; N2 = 7/28 = 0.25.
2. Ratio = 0.5 : 1.5 : 0.25 = 1 : 3 : 0.5.
3. Answer: (d)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 2: Select the correct statement:
(a) Equilibrium constant changes with addition of a catalyst
(b) A catalyst increases the rate of the forward reaction only
(c) The ratio of the mixture at equilibrium is not changed by a catalyst
(d) Catalysts are active only in solution
Show Answer & Explanation

Answer: (c) The ratio of the mixture at equilibrium is not changed by a catalyst

Explanation:
1. A catalyst speeds up the forward and backward reactions equally and only shortens the time to reach equilibrium.
2. The equilibrium composition and K are unchanged.
3. Answer: (c)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 3: For Ag+ + 2NH3 ⇌ Ag(NH3)2+ at 298 K, the molar concentrations of Ag+, Ag(NH3)2+ and NH3 are 10−1, 10−1 and 103 respectively. The value of Kc at 298 K for this equilibrium is:
(a) 10−6
(b) 106
(c) 2 × 10−3
(d) 2 × 106
Show Answer & Explanation

Answer: (a) 10−6

Explanation:
1. Kc = [Ag(NH3)2+]/([Ag+][NH3]2).
2. = 10−1/(10−1 × (103)2) = 10−6.
3. Answer: (a)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 4: For the reaction C(s) + CO2(g) ⇌ 2CO(g), the partial pressures of CO and CO2 are 2.0 and 4.0 atm respectively at equilibrium. The Kp for the reaction is:
(a) 0.5
(b) 4.0
(c) 8.0
(d) 1
Show Answer & Explanation

Answer: (d) 1

Explanation:
1. Solids are omitted: Kp = pCO2/pCO2 = 2.02/4.0 = 1.
2. Answer: (d)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 5: For which reaction is Kp = Kc?
(a) 2NOCl(g) ⇌ 2NO(g) + Cl2(g)
(b) N2(g) + 3H2(g) ⇌ 2NH3(g)
(c) H2(g) + Cl2(g) ⇌ 2HCl(g)
(d) 2SO2(g) + O2(g) ⇌ 2SO3(g)
Show Answer & Explanation

Answer: (c) H2(g) + Cl2(g) ⇌ 2HCl(g)

Explanation:
1. Kp = Kc(RT)Δn, so they are equal when Δn = 0.
2. Only H2 + Cl2 ⇌ 2HCl has Δn = 2 − 2 = 0.
3. Answer: (c)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 6: For the reaction CuSO4·5H2O(s) ⇌ CuSO4·3H2O(s) + 2H2O(g), which is a correct representation?
(a) Kp = pH2O2
(b) Kc = [H2O]2
(c) Kp = Kc(RT)2
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. Only the gas appears in the expressions: Kp = pH2O2 and Kc = [H2O]2.
2. Δn = 2, so Kp = Kc(RT)2. All three are correct.
3. Answer: (d)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 7: Consider two gaseous equilibria involving SO2 at 299 K: SO2(g) + ½O2(g) ⇌ SO3(g), K1; and 4SO3(g) ⇌ 4SO2(g) + 2O2(g), K2. The equilibrium constants are related by:
(a) K2 = 1/(K1)4
(b) K2 = K14
(c) K2 = (1/K1)1/4
(d) K2 = 1/K1
Show Answer & Explanation

Answer: (a) K2 = 1/(K1)4

Explanation:
1. The second reaction is the first reversed (K → 1/K1) and multiplied by 4 (K raised to the power 4).
2. K2 = (1/K1)4.
3. Answer: (a)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 8: Which oxide of nitrogen is most stable? (1) 2NO2 ⇌ N2 + 2O2, K = 6.7 × 1016 mol L−1; (2) 2NO ⇌ N2 + O2, K = 2.2 × 1030; (3) 2N2O5 ⇌ 2N2 + 5O2, K = 1.2 × 1034 mol L−5; (4) 2N2O ⇌ 2N2 + O2, K = 3.5 × 1033 mol L−1.
(a) NO2
(b) NO
(c) N2O5
(d) N2O
Show Answer & Explanation

Answer: (a) NO2

Explanation:
1. K for decomposition measures the tendency to decompose.
2. The smallest K (6.7 × 1016, NO2) means the least decomposition, so NO2 is the most stable of the four.
3. Answer: (a)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 9: Which one of the following statements about the equilibrium constant is correct?
(a) The equilibrium constant of a reaction changes with temperature
(b) The equilibrium constant of a reaction depends upon the concentration of reactants with which we start
(c) The equilibrium constant of 3Fe(s) + 4H2O(g) ⇌ Fe3O4(s) + 4H2(g) is the same whether it is carried out in an open vessel or a closed vessel
(d) The equilibrium constant of a reaction becomes double if the reaction is multiplied by 2 throughout
Show Answer & Explanation

Answer: (a) The equilibrium constant of a reaction changes with temperature

Explanation:
1. K depends only on temperature (and the way the equation is written).
2. (b) is false, (c) is false because the reaction needs a closed vessel to reach equilibrium, and (d) is false: K becomes K2.
3. Answer: (a)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 10: For the reaction N2 + 3H2 ⇌ 2NH3, the value of Kc does not depend upon: (a) initial concentration of the reactants, (b) pressure, (c) temperature, (d) catalyst.
(a) Only c
(b) a, b, c
(c) a, b, d
(d) a, b, c, d
Show Answer & Explanation

Answer: (c) a, b, d

Explanation:
1. Kc depends only on temperature.
2. So it does not depend on the initial concentrations, the pressure or the catalyst: a, b and d.
3. Answer: (c)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 11: The equilibrium constant for A + B ⇌ C + D is K when the initial concentrations of A and B are each 0.1 M. Under similar conditions in another experiment, the initial concentrations of A and B are taken as 2 and 3 mol L−1 respectively. The equilibrium constant will be:
(a) K/2
(b) K
(c) K2
(d) 1/K
Show Answer & Explanation

Answer: (b) K

Explanation:
1. K depends only on temperature, not on the initial concentrations, so it remains K.
2. Answer: (b)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 12: The equilibrium constant (Kp) for PCl5(g) ⇌ PCl3(g) + Cl2(g) is 16. If the volume of the container is reduced to one half of its original volume, the value of Kp for the reaction at the same temperature will be:
(a) 32
(b) 64
(c) 16
(d) 4
Show Answer & Explanation

Answer: (c) 16

Explanation:
1. Kp depends only on temperature; changing the volume does not alter it.
2. Answer: (c)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 13: If some He gas is introduced into the equilibrium PCl5 ⇌ PCl3 + Cl2 at constant pressure and temperature, the equilibrium constant of the reaction:
(a) Increases
(b) Decreases
(c) Is unchanged
(d) Nothing can be said
Show Answer & Explanation

Answer: (c) Is unchanged

Explanation:
1. K depends only on temperature, which is constant.
2. (The extent of dissociation would increase, but K itself does not change.)
3. Answer: (c)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 14: The dissociation of CO2 can be expressed as 2CO2 ⇌ 2CO + O2. If 2 mol of CO2 is taken initially and 40% of the CO2 is dissociated completely, the total number of moles at equilibrium is:
(a) 2.4
(b) 2.0
(c) 1.2
(d) 5
Show Answer & Explanation

Answer: (a) 2.4

Explanation:
1. CO2 dissociated = 0.8 mol, leaving 1.2 mol.
2. CO formed = 0.8 mol; O2 formed = 0.4 mol.
3. Total = 1.2 + 0.8 + 0.4 = 2.4 mol.
4. Answer: (a)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 15: In the reaction 2P(g) + Q(g) ⇌ 3R(g) + S(g), 2 mol each of P and Q are taken in a 1 L flask. At equilibrium which is true?
(a) [P] < [Q]
(b) [P] = [Q]
(c) [Q] = [R]
(d) None of these
Show Answer & Explanation

Answer: (a) [P] < [Q]

Explanation:
1. P is consumed twice as fast as Q (2 : 1), so starting with equal amounts [P] ends below [Q].
2. Answer: (a)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 16: In the equilibrium PCl5(g) ⇌ PCl3(g) + Cl2(g), the mole fraction of PCl5 is 0.4 and that of Cl2 is 0.3. The mole fraction of PCl3 is:
(a) 0.3
(b) 0.7
(c) 0.4
(d) 0.6
Show Answer & Explanation

Answer: (a) 0.3

Explanation:
1. Mole fractions add up to 1: 1 − 0.4 − 0.3 = 0.3.
2. Answer: (a)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 17: 'a' mol of PCl5 undergoes thermal dissociation as PCl5 ⇌ PCl3 + Cl2. The mole fraction of PCl3 at equilibrium is 0.25 and the total pressure is 2.0 atm. The partial pressure of Cl2 at equilibrium is:
(a) 2.5
(b) 1.0
(c) 0.5
(d) None of these
Show Answer & Explanation

Answer: (c) 0.5

Explanation:
1. PCl3 and Cl2 are formed in equal amounts, so the mole fraction of Cl2 is also 0.25.
2. Partial pressure = 0.25 × 2.0 = 0.5 atm.
3. Answer: (c)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 18: In a 0.25 L tube, dissociation of 4 mol of NO takes place. If its degree of dissociation is 10%, the value of Kp for the reaction 2NO ⇌ N2 + O2 is:
(a) 1/(18)2
(b) 1/(8)2
(c) 1/16
(d) 1/32
Show Answer & Explanation

Answer: (a) 1/(18)2

Explanation:
1. NO left = 3.6 mol; N2 = O2 = 0.2 mol (Δn = 0, so volume cancels).
2. Kp = Kc = (0.2 × 0.2)/(3.6)2 = 1/324 = 1/(18)2.
3. Answer: (a)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 19: Evaluate Kp for the reaction H2 + I2 ⇌ 2HI. If 2 moles each of H2 and I2 are taken initially, at equilibrium moles of HI are 2.
(a) 2.5
(b) 4
(c) 0.25
(d) 1.0
Show Answer & Explanation

Answer: (b) 4

Explanation:
1. HI = 2 mol uses 1 mol each of H2 and I2, leaving 1 mol each.
2. Kp = Kc = (2)2/(1 × 1) = 4.
3. Answer: (b)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 20: 4 moles of A are mixed with 4 moles of B. When 2 mol of C are formed at equilibrium according to A + B ⇌ C + D, the equilibrium constant is:
(a) 4
(b) 1
(c) √2
(d) √4
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. C = D = 2 mol, A = B = 4 − 2 = 2 mol.
2. K = (2 × 2)/(2 × 2) = 1.
3. Answer: (b)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 21: Two moles of ammonia are introduced into an evacuated 500 mL vessel at high temperature. The decomposition reaction is 2NH3(g) ⇌ N2(g) + 3H2(g). At equilibrium NH3 becomes 1 mole. K would be:
(a) 0.42
(b) 6.75
(c) 1.7
(d) 1.5
Show Answer & Explanation

Answer: (b) 6.75

Explanation:
1. NH3 left = 1 mol, so N2 = 0.5 mol and H2 = 1.5 mol.
2. Concentrations (÷ 0.5 L): [NH3] = 2, [N2] = 1, [H2] = 3.
3. K = (1)(3)3/(2)2 = 6.75.
4. Answer: (b)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 22: 1.50 mol each of hydrogen and iodine were placed in a sealed 10 L container maintained at 717 K. At equilibrium 1.25 mol each of hydrogen and iodine were left behind. The equilibrium constant Kc for H2(g) + I2(g) ⇌ 2HI(g) at 717 K is:
(a) 0.4
(b) 0.16
(c) 25
(d) 50
Show Answer & Explanation

Answer: (b) 0.16

Explanation:
1. HI formed = 2 × 0.25 = 0.5 mol.
2. Kc = (0.5)2/(1.25 × 1.25) = 0.25/1.5625 = 0.16.
3. Answer: (b)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 23: For the equilibrium cis-2-pentene ⇌ trans-2-pentene, the standard free energy change at 400 K is −3.67 kJ/mol. If excess trans-2-pentene is added to the system, then:
(a) Additional trans-2-pentene will form
(b) Excess cis-2-pentene will form
(c) Equilibrium will proceed in the forward direction
(d) Equilibrium will remain unaffected
Show Answer & Explanation

Answer: (b) Excess cis-2-pentene will form

Explanation:
1. Adding product (trans) increases Q above K, so the net reaction goes backwards, forming more cis-2-pentene.
2. Answer: (b)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 24: The equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g) shifts forward if:
(a) A catalyst is used
(b) An adsorbent is used to remove SO3 as soon as it is formed
(c) Small amounts of reactants are used
(d) None of these
Show Answer & Explanation

Answer: (b) An adsorbent is used to remove SO3 as soon as it is formed

Explanation:
1. Removing a product lowers Q below K, so the equilibrium shifts forward (Le Chatelier).
2. A catalyst does not shift equilibrium.
3. Answer: (b)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 25: In which of the following equilibrium reactions would the equilibrium shift to the right side if the total pressure is decreased?
(a) N2 + 3H2 ⇌ 2NH3
(b) H2 + I2 ⇌ 2HI
(c) N2O4 ⇌ 2NO2
(d) H2 + Cl2 ⇌ 2HCl
Show Answer & Explanation

Answer: (c) N2O4 ⇌ 2NO2

Explanation:
1. Lower pressure favours the side with more gas moles.
2. Only N2O4 ⇌ 2NO2 has more moles on the right (Δn = +1).
3. Answer: (c)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 26: For the manufacture of ammonia by N2 + 3H2 ⇌ 2NH3 + 21.9 kcal, the favourable conditions are:
(a) Low temperature, low pressure and catalyst
(b) Low temperature, high pressure and catalyst
(c) High temperature, low pressure and catalyst
(d) High temperature, high pressure and catalyst
Show Answer & Explanation

Answer: (b) Low temperature, high pressure and catalyst

Explanation:
1. Exothermic: low temperature favours the forward reaction.
2. 4 mol of gas becomes 2 mol: high pressure favours it.
3. A catalyst speeds up attainment of equilibrium.
4. Answer: (b)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 27: In the reaction 2A(g) + B(g) ⇌ C(g) + 362 kcal, which combination of pressure and temperature gives the highest yield of C at equilibrium?
(a) 1000 atm and 500°C
(b) 500 atm and 500°C
(c) 1000 atm and 50°C
(d) 500 atm and 100°C
Show Answer & Explanation

Answer: (c) 1000 atm and 50°C

Explanation:
1. The forward reaction is exothermic and reduces the number of moles (3 → 1).
2. High pressure and low temperature favour C, so 1000 atm and 50°C.
3. Answer: (c)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 28: For aA ⇌ bB + cC, ΔH = −x kcal, high pressure and low temperature are the favourable conditions for the formation of the products. Hence:
(a) a > b + c
(b) a < b + c
(c) a = b + c
(d) None of these
Show Answer & Explanation

Answer: (a) a > b + c

Explanation:
1. High pressure favours the side with fewer moles, here the products, so b + c < a, i.e. a > b + c.
2. Low temperature agrees with the reaction being exothermic.
3. Answer: (a)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 29: The reaction in which the yield of product cannot be increased by the application of high pressure is:
(a) PCl3(g) + Cl2(g) ⇌ PCl5(g)
(b) N2(g) + O2(g) ⇌ 2NO(g)
(c) N2(g) + 3H2(g) ⇌ 2NH3(g)
(d) 2SO2(g) + O2(g) ⇌ 2SO3(g)
Show Answer & Explanation

Answer: (b) N2(g) + O2(g) ⇌ 2NO(g)

Explanation:
1. N2 + O2 ⇌ 2NO has Δn = 0, so pressure has no effect on the yield.
2. The other three reduce the number of moles and are favoured by high pressure.
3. Answer: (b)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 30: In a vessel containing SO3, SO2 and O2 at equilibrium, some helium gas is introduced so that the total pressure increases while temperature and volume remain constant. According to Le Chatelier's principle, the dissociation of SO3:
(a) Increases
(b) Decreases
(c) Remains unaltered
(d) None of these
Show Answer & Explanation

Answer: (c) Remains unaltered

Explanation:
1. At constant volume the partial pressures (concentrations) of the reacting gases are unchanged by the added inert gas.
2. So the equilibrium does not shift.
3. Answer: (c)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

Question 31: For the equilibrium H2O(l) ⇌ H2O(g), what happens if pressure is applied?
(a) More water evaporates
(b) The boiling point of water is increased
(c) No effect on boiling point
(d) None of the above
Show Answer & Explanation

Answer: (b) The boiling point of water is increased

Explanation:
1. Applying pressure favours the denser phase (liquid) and a higher external pressure raises the boiling point, since vapour pressure must equal the applied pressure.
2. Answer: (b)

Teacher's Note:
1. Write the equilibrium expression first.
2. Check the answer by substituting back into K.

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Multiple Choice Questions (MCQs) for JEE Chemistry Chemical Equilibrium

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