JEE Chemistry Chemical Equilibrium MCQs Set 06

Multiple Choice Questions (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.

Practice Chemical Equilibrium MCQs for JEE Chemistry

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: For the reaction \( \ce{Cu(s) + 2Ag+(aq) <=> Cu^{2+}(aq) + 2Ag(s)} \), the equilibrium constant is:
(a) \( K_c = [Cu^\( \ce{2+} \)][Ag]^2 / ([Cu][Ag^+]^2) \)
(b) \( K_c = [Cu^\( \ce{2+} \)][Ag]^2 / [Cu][Ag^+] \)
(c) \( K_c = [Cu^\( \ce{2+} \)] / [Ag^+]^2 \)
(d) \( K_c = [Ag^+]^2 / [Cu^\( \ce{2+} \)] \)
Show Answer & Explanation

Answer: (c) \( K_c = [Cu^\( \ce{2+} \)] / [Ag^+]^2 \)

Explanation:
1. Pure solids have unit activity and are left out of the expression: Kc = \( [Cu2+]/[Ag+]^2 \).

Teacher's Note:
1. Leave out solids and pure liquids.

Question 2: Kp for the reaction \( \ce{CaCO3(s) <=> CaO(s) + CO2(g)} \) is correctly expressed as:
(a) \( K_p = p_\( \ce{CaO} \)\,p_\( \ce{CO_2} \)/p_\( \ce{CaCO_3} \) \)
(b) \( K_p = p_\( \ce{CaCO_3} \)/(p_\( \ce{CaO} \)\,p_\( \ce{CO_2} \)) \)
(c) \( K_p = p_\( \ce{CO_2} \) \)
(d) \( K_p = p_\( \ce{CaCO_3} \)/p_\( \ce{CO_2} \) \)
Show Answer & Explanation

Answer: (c) \( K_p = p_\( \ce{CO_2} \) \)

Explanation:
1. Solids do not appear in Kp, so only the gas contributes: Kp = p(CO2).

Teacher's Note:
1. Only gases appear in Kp.

Question 3: The equilibrium constant for A <=> B is K. For the reaction mA <=> mB it is:
(a) mK
(b) K/m
(c) Km
(d) \( K^m \)
Show Answer & Explanation

Answer: (d) \( K^m \)

Explanation:
1. Multiplying the coefficients of an equation by m raises K to the power m.

Teacher's Note:
1. K changes as a power when the equation is multiplied.

Question 4: The equilibrium constants of a reaction at 298 K and 308 K are \( 1.0 x 10^{-2} \) and \( 2.0 x 10^{-2} \) respectively. The reaction is:
(a) Exothermic
(b) Endothermic
(c) Either exothermic or endothermic
(d) Cannot be predicted
Show Answer & Explanation

Answer: (b) Endothermic

Explanation:
1. K increases with temperature, so by van't Hoff equation the reaction is endothermic.

Teacher's Note:
1. Rising K with T means endothermic.

Question 5: The equilibrium constant of \( \ce{2AB <=> A2 + B2} \) is 36. For \( \ce{AB <=> 1/2 A2 + 1/2 B2} \) it is:
(a) 49
(b) 24
(c) 6
(d) 2
Show Answer & Explanation

Answer: (c) 6

Explanation:
1. Dividing the equation by 2 takes the square root: \( K' = \sqrt{36} = 6 \).

Teacher's Note:
1. Halve the equation, take the square root.

Question 6: For the homogeneous reaction \( \ce{4NH3(g) + 5O2(g) <=> 4NO(g) + 6H2O(g)} \), Kc has the unit of:
(a) \( (Conc)^{-1} \)
(b) Conc
(c) \( (Conc)^{10} \)
(d) It is dimensionless
Show Answer & Explanation

Answer: (b) Conc

Explanation:
1. Kc = \( [NO]^4[H2O]^6/([NH3]^4[O2]^5) \), so the unit is \( conc^{10-9} \) = conc.

Teacher's Note:
1. Unit = conc^(change in moles).

Question 7: For \( \ce{N2(g) + 2H2(g) <=> N2H4(g)} \), Kc = \( 7.4 x 10^{-26} \). The equilibrium mixture contains:
(a) Mostly products
(b) Mostly reactants
(c) Cannot be predicted
(d) Sometimes reactants and sometimes products
Show Answer & Explanation

Answer: (b) Mostly reactants

Explanation:
1. A very small K means the product concentration is very small compared with the reactant concentrations.

Teacher's Note:
1. K much less than 1 favours reactants.

Question 8: In which reaction is the concentration of the reactant equal to that of the product at equilibrium?
(a) A <=> B; K = 0.01
(b) R <=> P; K = 1
(c) X <=> Y; K = 10
(d) L <=> J; K = 0.025
Show Answer & Explanation

Answer: (b) R <=> P; K = 1

Explanation:
1. For a 1:1 reaction, K = [P]/[R] = 1 means equal concentrations.

Teacher's Note:
1. K = 1 means equal concentrations here.

Question 9: At 500 K, K for cis-C2H2Cl2 <=> trans-C2H2Cl2 is 0.6. For trans <=> cis at the same temperature K is:
(a) 0.6
(b) 1.66
(c) 1
(d) 16.6
Show Answer & Explanation

Answer: (b) 1.66

Explanation:
1. Reversing a reaction inverts K: 1/0.6 = 1.66.

Teacher's Note:
1. K(reverse) = 1/K.

Question 10: The active mass of 5.6 L of O2 at STP is:
(a) 5.6/22.4
(b) 8/5.6
(c) 32/5.6
(d) 0.25/5.6
Show Answer & Explanation

Answer: (d) 0.25/5.6

Explanation:
1. Moles = 5.6/22.4 = 0.25 mol in a volume of 5.6 L, so active mass = 0.25/5.6 mol/L.

Teacher's Note:
1. Active mass is molar concentration.

Question 11: \( \ce{AB2(g) <=> AB(g) + B(g)} \). The initial pressure of AB2 is 600 mm Hg and the total pressure at equilibrium is 800 mm Hg. Kp is:
(a) 50
(b) 100
(c) 166.8
(d) 400
Show Answer & Explanation

Answer: (b) 100

Explanation:
1. Pressure rise = x = 200 mm. p(AB2) = 400, p(AB) = p(B) = 200.
2. Kp = 200 x 200/400 = 100 mm Hg.

Teacher's Note:
1. Total pressure rise equals x.

Question 12: For which reaction is Kp = Kc?
(a) \( \ce{N2 + 3H2 <=> 2NH3} \)
(b) \( \ce{2SO2 + O2 <=> 2SO3} \)
(c) \( \ce{PCl3 + Cl2 <=> PCl5} \)
(d) \( \ce{N2 + O2 <=> 2NO} \)
Show Answer & Explanation

Answer: (d) \( \ce{N2 + O2 <=> 2NO} \)

Explanation:
1. Kp = Kc(RT)^(change in moles), so they are equal only when the change in moles of gas is zero.

Teacher's Note:
1. Check delta n.

Question 13: For \( \ce{H2(g) + I2(g) <=> 2HI(g)} \) at 721 K, Kc = 50. The value of Kp is:
(a) 0.02
(b) 0.2
(c) 50
(d) 50/RT
Show Answer & Explanation

Answer: (c) 50

Explanation:
1. The change in moles of gas is zero, so Kp = Kc = 50.

Teacher's Note:
1. No change in moles gives Kp = Kc.

Question 14: For \( \ce{S8(g) <=> 4S2(g)} \) in a 1 L vessel, Kc = \( 6.30 x 10^{-6} \). With 2 mol S8 and 0.2 mol S2 present, which statements are correct? (a) The reaction quotient is \( 8 x 10^{-4} \). (b) The reaction proceeds backward. (c) The reaction proceeds forward.
(a) a and b
(b) b and c
(c) a and c
(d) a, b and c
Show Answer & Explanation

Answer: (a) a and b

Explanation:
1. \( Q = (0.2)^4/2 = 8 x 10^{-4} \), which is greater than Kc, so the reaction proceeds in the backward direction.

Teacher's Note:
1. Compare Q with K.

Question 15: For \( \ce{N2O4(g) <=> 2NO2(g)} \), Kp = \( 4x^2 P/(1 - x^2) \) at degree of dissociation x and total pressure P. At constant temperature:
(a) Kp increases with increase of P
(b) Kp increases with increase of x
(c) Kp increases with decrease of x
(d) Kp remains constant with change in P or x
Show Answer & Explanation

Answer: (d) Kp remains constant with change in P or x

Explanation:
1. Kp depends only on temperature. Changing P changes x such that Kp stays constant.

Teacher's Note:
1. K depends only on temperature.

Question 16: For \( \ce{C(s) + CO2(g) <=> 2CO(g)} \), the partial pressures of CO2 and CO are 2.0 and 4.0 atm at equilibrium. Kp is:
(a) 0.5
(b) 5.0
(c) 30.0
(d) 8.0
Show Answer & Explanation

Answer: (d) 8.0

Explanation:
1. Kp = \( p(CO)^2/p(CO2) = 16/2 = 8 \).

Teacher's Note:
1. Square the product pressure.

Question 17: 2 mol N2 and 6 mol H2 are mixed in a 1 L vessel. If 50% of N2 is converted to NH3 at equilibrium, Kc for \( \ce{N2 + 3H2 <=> 2NH3} \) is:
(a) 4/27
(b) 27/4
(c) 2/27
(d) 20
Show Answer & Explanation

Answer: (a) 4/27

Explanation:
1. Equilibrium: N2 = 1, H2 = 3, NH3 = 2 mol/L.
2. Kc = \( 2^2/(1 x 3^3) = 4/27 \).

Teacher's Note:
1. Use equilibrium concentrations.

Question 18: \( \ce{NH2COONH4(s) <=> 2NH3(g) + CO2(g)} \). If the total equilibrium pressure is 3 atm, Kp is:
(a) 4
(b) 20
(c) 25
(d) 15
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. p(NH3) = 2 atm and p(CO2) = 1 atm, so Kp = \( 2^2 x 1 = 4 atm^3 \).

Teacher's Note:
1. NH3 : CO2 = 2 : 1.

Question 19: For \( \ce{A + B <=> C + D} \), the initial concentrations of A and B are equal and at equilibrium [D] = 2[A]. The equilibrium constant is:
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. Let a be the initial concentration and x the amount converted. x = 2(a - x), so x = 2a/3 and [A] = a/3.
2. \( K = x^2/(a - x)^2 = (2a/3)^2/(a/3)^2 = 4 \).

Teacher's Note:
1. Let x be the change.

Question 20: For \( \ce{CO(g) + Cl2(g) <=> COCl2(g)} \), Kp/Kc is equal to:
(a) 1/RT
(b) RT
(c) \( \sqrt\( \ce{RT} \) \)
(d) \( (RT)^2 \)
Show Answer & Explanation

Answer: (a) 1/RT

Explanation:
1. Change in moles of gas = 1 - 2 = -1, so Kp/Kc = \( (RT)^{-1} \).

Teacher's Note:
1. Kp = \( Kc(RT)^{\delta n} \).

Question 21: The equilibrium constant for \( \ce{N2 + O2 <=> 2NO} \) at temperature T is \( 4 x 10^{-4} \). For \( \ce{NO <=> 1/2 N2 + 1/2 O2} \) at the same temperature Kc is:
(a) \( 25 \times 10^2 \)
(b) 50
(c) \( 4 \times 10^\( \ce{-4} \) \)
(d) 10.00
Show Answer & Explanation

Answer: (b) 50

Explanation:
1. Reverse gives \( 1/(4 x 10^{-4}) = 2500 \), and halving the equation gives the square root: 50.

Teacher's Note:
1. Reverse, then take the square root.

Question 22: For \( \ce{AB(g) <=> A(g) + B(g)} \), the pressure at which one-third of AB is dissociated is numerically equal to:
(a) 8 Kp
(b) 16 Kp
(c) 4 Kp
(d) 9 Kp
Show Answer & Explanation

Answer: (a) 8 Kp

Explanation:
1. Kp = \( \alpha^2 P/(1 - \alpha^2) = (1/9)P/(8/9) = P/8 \), so P = 8 Kp.

Teacher's Note:
1. Use alpha = 1/3.

Question 23: Haber's process \( \ce{N2 + 3H2 <=> 2NH3} \), delta H = -22.4 kcal. The effect of increasing temperature on the equilibrium is:
(a) Shifts to the right
(b) Unaffected
(c) Shifts to the left
(d) Shifts first right then left
Show Answer & Explanation

Answer: (c) Shifts to the left

Explanation:
1. The forward reaction is exothermic, so a higher temperature favours the backward reaction.

Teacher's Note:
1. Heat is treated as a product.

Question 24: Inert gas is added at constant volume to \( \ce{PCl5(g) <=> PCl3(g) + Cl2(g)} \) and \( \ce{N2O4(g) <=> 2NO2(g)} \). The effect is:
(a) Dissociation increases in both
(b) Dissociation decreases in both
(c) Dissociation of PCl5 increases and N2O4 decreases
(d) No effect on either equilibrium
Show Answer & Explanation

Answer: (d) No effect on either equilibrium

Explanation:
1. At constant volume, the concentrations of the reacting gases do not change, so Q remains equal to K.

Teacher's Note:
1. Inert gas at constant volume has no effect.

Question 25: For \( \ce{N2(g) + O2(g) <=> 2NO(g)} \), delta H = +180 kJ/mol. Which statement is NOT true?
(a) Change in pressure at constant volume does not affect the equilibrium
(b) The change in moles is zero
(c) Formation of NO increases at higher temperature
(d) Formation of NO decreases at higher temperature
Show Answer & Explanation

Answer: (d) Formation of NO decreases at higher temperature

Explanation:
1. The reaction is endothermic, so a higher temperature favours the forward reaction (more NO).

Teacher's Note:
1. Endothermic reactions are favoured by heat.

Question 26: In \( \ce{4H2O(g) + 3Fe(s) <=> Fe3O4(s) + 4H2(g)} \), the yield of H2 can be increased by:
(a) Increasing the pressure
(b) Passing more steam
(c) Increasing the mass of iron
(d) Decreasing the pressure
Show Answer & Explanation

Answer: (b) Passing more steam

Explanation:
1. The number of gas moles does not change, so pressure has no effect. Solids do not change the equilibrium. More steam shifts it forward.

Teacher's Note:
1. Add reactant to shift forward.

Question 27: For a reaction with Kp less than Kc, an increase of pressure favours:
(a) The backward reaction
(b) No reaction
(c) The forward reaction
(d) Both equally
Show Answer & Explanation

Answer: (c) The forward reaction

Explanation:
1. Kp < Kc means delta n is negative, so the forward reaction reduces the moles of gas and is favoured by high pressure.

Teacher's Note:
1. High pressure favours fewer gas moles.

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Chemical Equilibrium Objective Questions & Solutions for JEE Chemistry

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