JEE Chemistry Equilibrium MCQs Set 30

Chemistry Objective Questions and Answers: Equilibrium

Explore reliable objective questions for Equilibrium tailored for JEE learners. Utilizing these Chemistry multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Download Equilibrium MCQs with Answers

View or download the dedicated Equilibrium MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Chemistry evaluations.

Multiple Choice Questions

Question 1: Krypton is added to the equilibrium mixture of a gaseous reaction at constant volume. The effect on the position of equilibrium is:
(a) forward favoured if \( \Delta n \) is negative
(b) backward favoured if \( \Delta n = 0 \)
(c) forward favoured if \( \Delta n \) is positive
(d) no effect, whatever the value of \( \Delta n \)
Show Answer & Explanation

Answer: (d) no effect, whatever the value of \( \Delta n \)

Explanation:
1. At constant volume the concentrations (partial pressures) of the reacting gases are unchanged.
2. So there is no effect for any \( \Delta n \).

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Question 2: \( a \) mol of \( \ce{PCl5} \) dissociates: \( \ce{PCl5 <=> PCl3 + Cl2} \). At equilibrium the mole fraction of \( \ce{PCl3} \) is 0.25 and the total pressure is 2.0 atm. The partial pressure of \( \ce{Cl2} \) is:
(a) 0.25 atm
(b) 0.5 atm
(c) 2.5 atm
(d) 1.0 atm
Show Answer & Explanation

Answer: (b) 0.5 atm

Explanation:
1. \( \ce{PCl3} \) and \( \ce{Cl2} \) form in equal amounts, so the mole fraction of \( \ce{Cl2} \) is also 0.25.
2. \( p_{Cl_2} = 0.25\times2.0 = 0.5 \) atm.

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Question 3: The value of \( \Delta G^\circ \) for the phosphorylation of glucose in glycolysis is +15 kJ/mol. The value of \( K_c \) at 300 K is nearly:
(a) \( e^{-3} \)
(b) \( 10^{-6} \)
(c) \( e^{-6} \)
(d) \( e^{6} \)
Show Answer & Explanation

Answer: (c) \( e^{-6} \)

Explanation:
1. \( \ln K_c = -\Delta G^\circ/RT = -15000/(8.314\times300) = -6.0 \).
2. \( K_c = e^{-6} \).

Teacher's Note:
1. The van't Hoff relation shows that K increases with temperature for an endothermic reaction and decreases for an exothermic one.
2. Use the standard free energy relation to connect K with energy changes.

Question 4: For a reversible process in a state of equilibrium, which relation is correct?
(a) \( \Delta G = -2.30\,RT\log K \)
(b) \( \Delta G^\circ = 2.30\,RT\log K \)
(c) \( \Delta G = 2.30\,RT\log K \)
(d) \( \Delta G^\circ = -2.30\,RT\log K \)
Show Answer & Explanation

Answer: (d) \( \Delta G^\circ = -2.30\,RT\log K \)

Explanation:
1. At equilibrium \( \Delta G = 0 \) and \( \Delta G^\circ = -RT\ln K = -2.30\,RT\log K \).

Teacher's Note:
1. The van't Hoff relation shows that K increases with temperature for an endothermic reaction and decreases for an exothermic one.
2. Use the standard free energy relation to connect K with energy changes.

Question 5: The equilibrium \( \ce{NH4Cl(s) <=> NH4+(aq) + Cl-(aq)} \) in a saturated solution has \( \Delta H = +3.5 \) kcal/mol. A change that shifts it to the right is:
(a) addition of \( \ce{NH4Cl} \) crystals
(b) a decrease in temperature
(c) an increase in temperature
(d) addition of \( \ce{NH4OH} \) solution
Show Answer & Explanation

Answer: (c) an increase in temperature

Explanation:
1. The dissolution is endothermic, so a higher temperature shifts it to the right.
2. Adding more solid does not change the equilibrium.

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Question 6: For \( \ce{PCl5(g) <=> PCl3(g) + Cl2(g)} \) at constant temperature the BACKWARD reaction is favoured by:
(a) introducing \( \ce{Cl2} \) at constant volume
(b) introducing \( \ce{PCl5} \) at constant volume
(c) increasing the volume of the container
(d) introducing an inert gas at constant pressure
Show Answer & Explanation

Answer: (a) introducing \( \ce{Cl2} \) at constant volume

Explanation:
1. Added product \( \ce{Cl2} \) shifts the equilibrium backward.
2. The other three changes shift it forward.

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Question 7: The equilibrium constant for \( \ce{P4(g) + 6Cl2(g) <=> 4PCl3(g)} \) is 0.125. The value for \( \ce{4PCl3(g) <=> P4(g) + 6Cl2(g)} \) is:
(a) 8
(b) 6
(c) 0.25
(d) 0.125
Show Answer & Explanation

Answer: (a) 8

Explanation:
1. The second reaction is the reverse of the first, so \( K' = 1/0.125 = 8 \).

Teacher's Note:
1. Reverse the reaction and K becomes 1/K; multiply the equation by n and K is raised to the power n; add equations and multiply the K values.
2. Combine the given equilibria so that the unwanted species cancel.

Question 8: Pure ammonia is placed in a vessel at a temperature where its dissociation is appreciable. At equilibrium:
(a) \( K_p \) does not change significantly with pressure
(b) the concentration of ammonia does not change with pressure
(c) the concentration of hydrogen is less than that of nitrogen
(d) the degree of dissociation does not change with pressure
Show Answer & Explanation

Answer: (a) \( K_p \) does not change significantly with pressure

Explanation:
1. \( K_p \) depends only on temperature.
2. The degree of dissociation decreases with pressure for \( \ce{2NH3 <=> N2 + 3H2} \), and \( [\ce{H2}] = 3[\ce{N2}] \).

Teacher's Note:
1. Write the moles of each species at equilibrium in terms of the degree of dissociation, then find the total moles.
2. Partial pressure equals mole fraction times total pressure.

Question 9: 1 mol of ethyl alcohol was treated with 1 mol of acetic acid at 25 C. Two-thirds of the acid changes into ester at equilibrium. The equilibrium constant is:
(a) 4
(b) 1
(c) 3
(d) 2
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. Ester = water = 2/3, alcohol = acid = 1/3 (volume cancels).
2. \( K = (2/3)^2/(1/3)^2 = 4 \).

Teacher's Note:
1. Write the equilibrium expression from the balanced equation and leave out pure solids and liquids.
2. Always substitute equilibrium concentrations, never initial ones.

Question 10: For \( \ce{C2H4 + H2 <=> C2H6} \), the equilibrium constant has units of:
(a) dm\(^3\) mol\(^{-1}\)
(b) mol dm\(^{-3}\)
(c) dm\(^6\) mol\(^{-2}\)
(d) mol\(^2\) dm\(^{-3}\)
Show Answer & Explanation

Answer: (a) dm\(^3\) mol\(^{-1}\)

Explanation:
1. \( K_c = [\ce{C2H6}]/([\ce{C2H4}][\ce{H2}]) \), with units (mol/dm\(^3\))\(^{1-2}\).
2. The unit is dm\(^3\) mol\(^{-1}\).

Teacher's Note:
1. Write the equilibrium expression from the balanced equation and leave out pure solids and liquids.
2. Always substitute equilibrium concentrations, never initial ones.

Question 11: The equilibrium constant expression for \( \ce{A + 2B <=> 2C + D} \) is:
(a) \( \dfrac{[2C][D]}{[A][2B]} \)
(b) \( \dfrac{[C][D]}{[A][B]} \)
(c) \( \dfrac{[C]^2[D]}{[A][2B]} \)
(d) \( \dfrac{[C]^2[D]}{[A][B]^2} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{[C]^2[D]}{[A][B]^2} \)

Explanation:
1. Each concentration is raised to the power of its stoichiometric coefficient.

Teacher's Note:
1. Write the equilibrium expression from the balanced equation and leave out pure solids and liquids.
2. Always substitute equilibrium concentrations, never initial ones.

Question 12: For \( \ce{A + B <=> C} \) the specific rates of the forward and reverse reactions are \( 1.25\times10^{-3} \) and \( 2.75\times10^{-5} \). The equilibrium constant is:
(a) 0.022
(b) 45.45
(c) 2.20
(d) 0.4545
Show Answer & Explanation

Answer: (b) 45.45

Explanation:
1. \( K = k_f/k_r = 1.25\times10^{-3}/2.75\times10^{-5} = 45.45 \).

Teacher's Note:
1. Write the equilibrium expression from the balanced equation and leave out pure solids and liquids.
2. Always substitute equilibrium concentrations, never initial ones.

Question 13: The equilibrium constant for \( \ce{H2 + Cl2 <=> 2HCl} \) is given by:
(a) \( K = [H_2][Cl_2]/(2[HCl]) \)
(b) \( K = 2[HCl]/([H_2][Cl_2]) \)
(c) \( K = [HCl]^2/([H_2][Cl_2]) \)
(d) \( K = [H_2][Cl_2]/[HCl]^2 \)
Show Answer & Explanation

Answer: (c) \( K = [HCl]^2/([H_2][Cl_2]) \)

Explanation:
1. Products over reactants, each raised to its stoichiometric coefficient.

Teacher's Note:
1. Write the equilibrium expression from the balanced equation and leave out pure solids and liquids.
2. Always substitute equilibrium concentrations, never initial ones.

Question 14: For \( \ce{N2 + 3H2 <=> 2NH3} \), the units of \( K_c \) and \( K_p \) are respectively:
(a) mol\(^{-1}\) L\(^{-1}\) and bar\(^{-1}\)
(b) mol\(^{-2}\) L\(^2\) and bar\(^{-2}\)
(c) mol\(^{-1}\) L and bar\(^{-2}\)
(d) mol\(^{-2}\) L\(^2\) and bar\(^{-1}\)
Show Answer & Explanation

Answer: (b) mol\(^{-2}\) L\(^2\) and bar\(^{-2}\)

Explanation:
1. \( \Delta n = -2 \), so \( K_c \) has (mol/L)\(^{-2}\), i.e. mol\(^{-2}\) L\(^2\).
2. \( K_p \) has bar\(^{-2}\).

Teacher's Note:
1. Write the equilibrium expression from the balanced equation and leave out pure solids and liquids.
2. Always substitute equilibrium concentrations, never initial ones.

Question 15: The equilibrium constant for \( \ce{N2 + 3H2 <=> 2NH3} \) is 70 at a certain temperature. The equilibrium constant for \( \ce{NH3 <=> 1/2 N2 + 3/2 H2} \) at the same temperature is nearly:
(a) \( 1.4\times10^{-2} \)
(b) \( 2.9\times10^{-2} \)
(c) \( 1.2\times10^{-1} \)
(d) \( 2.0\times10^{-4} \)
Show Answer & Explanation

Answer: (c) \( 1.2\times10^{-1} \)

Explanation:
1. The required reaction is the reverse of half the given one.
2. \( K' = (1/70)^{1/2} = 0.12 \).

Teacher's Note:
1. Reverse the reaction and K becomes 1/K; multiply the equation by n and K is raised to the power n; add equations and multiply the K values.
2. Combine the given equilibria so that the unwanted species cancel.

Question 16: For \( \ce{4NH3(g) + 7O2(g) <=> 4NO2(g) + 6H2O(g)} \), \( K_p \) is related to \( K_c \) by:
(a) \( K_p = K_c \)
(b) \( K_p = K_c(RT)^3 \)
(c) \( K_p = K_c(RT) \)
(d) \( K_p = K_c(RT)^{-1} \)
Show Answer & Explanation

Answer: (d) \( K_p = K_c(RT)^{-1} \)

Explanation:
1. \( \Delta n = (4+6) - (4+7) = -1 \).
2. \( K_p = K_c(RT)^{-1} \).

Teacher's Note:
1. Use Kp = Kc(RT)^(change in gas moles) and take R in units that match the pressure and concentration units.
2. Count only gaseous species when finding the change in moles.

Question 17: When \( K_c > 1 \) for a chemical reaction:
(a) equilibrium is achieved slowly
(b) equilibrium is achieved rapidly
(c) product concentrations are much greater than reactant concentrations at equilibrium
(d) reactant concentrations are much greater than product concentrations
Show Answer & Explanation

Answer: (c) product concentrations are much greater than reactant concentrations at equilibrium

Explanation:
1. A large K means products dominate at equilibrium.
2. K says nothing about how fast equilibrium is reached.

Teacher's Note:
1. Write the equilibrium expression from the balanced equation and leave out pure solids and liquids.
2. Always substitute equilibrium concentrations, never initial ones.

Question 18: In which reaction will an increase in the volume of the reaction system favour the formation of products?
(a) \( \ce{3O2(g) <=> 2O3(g)} \)
(b) \( \ce{C(s) + H2O(g) <=> CO(g) + H2(g)} \)
(c) \( \ce{H2(g) + I2(g) <=> 2HI(g)} \)
(d) \( \ce{4NH3(g) + 5O2(g) <=> 4NO(g) + 6H2O(l)} \)
Show Answer & Explanation

Answer: (b) \( \ce{C(s) + H2O(g) <=> CO(g) + H2(g)} \)

Explanation:
1. A volume increase favours the side with more gas moles.
2. The first reaction has 1 mole of gas on the left and 2 on the right.

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Question 19: For \( \ce{4NH3(g) + 3O2(g) <=> 2N2(g) + 6H2O(g)} \), \( \Delta H = -1268 \) kJ. Which change shifts the reaction to the right?
(a) decreasing the volume of the container
(b) increasing the temperature
(c) adding a catalyst
(d) removing the gaseous water by absorbing it
Show Answer & Explanation

Answer: (d) removing the gaseous water by absorbing it

Explanation:
1. Removing a product shifts the equilibrium to the right.
2. A higher temperature (exothermic) and a smaller volume (7 moles to 8 moles) shift it left; a catalyst does not shift it.

Teacher's Note:
1. Le Chatelier's principle: the system shifts to oppose the change applied.
2. An inert gas shifts the equilibrium only when it changes the partial pressures, that is at constant total pressure with a change in gas moles.

Question 20: The equilibrium constants for \( \ce{NO(g) + 1/2 O2(g) <=> NO2(g)} \) (\( K_1 \)) and \( \ce{2NO2(g) <=> 2NO(g) + O2(g)} \) (\( K_2 \)) are related as:
(a) \( K_1 = \sqrt{K_2} \)
(b) \( K_1 = 1/\sqrt{K_2} \)
(c) \( K_1 = 2K_2 \)
(d) \( K_1 = 1/K_2 \)
Show Answer & Explanation

Answer: (b) \( K_1 = 1/\sqrt{K_2} \)

Explanation:
1. The second reaction is the reverse of twice the first.
2. \( K_2 = 1/K_1^2 \), so \( K_1 = 1/\sqrt{K_2} \).

Teacher's Note:
1. Reverse the reaction and K becomes 1/K; multiply the equation by n and K is raised to the power n; add equations and multiply the K values.
2. Combine the given equilibria so that the unwanted species cancel.

Question 21: The equilibrium constants of \( \ce{2SO3 <=> 2SO2 + O2} \) and \( \ce{SO2 + 1/2 O2 <=> SO3} \) are \( K_1 \) and \( K_2 \). The relation between them is:
(a) \( K_2 = 1/\sqrt{K_1} \)
(b) \( K_2 = \sqrt{K_1} \)
(c) \( K_2 = 1/K_1 \)
(d) \( K_2 = K_1 \)
Show Answer & Explanation

Answer: (a) \( K_2 = 1/\sqrt{K_1} \)

Explanation:
1. The second reaction is the reverse of half the first.
2. \( K_2 = K_1^{-1/2} \).

Teacher's Note:
1. Reverse the reaction and K becomes 1/K; multiply the equation by n and K is raised to the power n; add equations and multiply the K values.
2. Combine the given equilibria so that the unwanted species cancel.

Question 22: The formation of ammonia can be written as (a) \( \ce{1/2 N2 + 3/2 H2 <=> NH3} \) (\( K_1 \)) and (b) \( \ce{1/3 N2 + H2 <=> 2/3 NH3} \) (\( K_2 \)). Which relations are correct? (A) \( K_1 = K_2^2 \) (B) \( K_1^3 = K_2^2 \) (C) \( K_1^{2/3} = K_2 \) (D) \( K_1 = K_2^{3/2} \)
(a) A and B
(b) B only
(c) A only
(d) C and D
Show Answer & Explanation

Answer: (d) C and D

Explanation:
1. Equation (b) = (2/3) \(\times\) equation (a), so \( K_2 = K_1^{2/3} \).
2. Hence \( K_1 = K_2^{3/2} \), and \( K_1^2 = K_2^3 \) (not \( K_1^3 = K_2^2 \)).

Teacher's Note:
1. Reverse the reaction and K becomes 1/K; multiply the equation by n and K is raised to the power n; add equations and multiply the K values.
2. Combine the given equilibria so that the unwanted species cancel.

Question 23: At 445 C, \( K_c = 0.020 \) for \( \ce{2HI <=> H2 + I2} \). A mixture has [HI] = 2.0 M, \( [\ce{H2}] = 0.50 \) M and \( [\ce{I2}] = 0.10 \) M. The correct statement is:
(a) \( Q_c < K_c \); more HI will be produced
(b) \( Q_c = K_c \); the system is at equilibrium
(c) \( Q_c > K_c \); more \( \ce{H2} \) and \( \ce{I2} \) will be produced
(d) \( Q_c < K_c \); more \( \ce{H2} \) and \( \ce{I2} \) will be produced
Show Answer & Explanation

Answer: (d) \( Q_c < K_c \); more \( \ce{H2} \) and \( \ce{I2} \) will be produced

Explanation:
1. \( Q_c = (0.50\times0.10)/2.0^2 = 0.0125 \).
2. \( Q_c < K_c = 0.020 \), so the reaction goes forward and produces more \( \ce{H2} \) and \( \ce{I2} \).

Teacher's Note:
1. Compare the reaction quotient Q with K: Q < K means the net reaction goes forward and Q > K means it goes backward.
2. Q has the same form as K but uses the concentrations present at that moment.

Question 24: In which reaction is \( K_p > K_c \)?
(a) \( \ce{H2 + I2 <=> 2HI} \)
(b) \( \ce{PCl3 + Cl2 <=> PCl5} \)
(c) \( \ce{2SO3 <=> 2SO2 + O2} \)
(d) \( \ce{N2 + 3H2 <=> 2NH3} \)
Show Answer & Explanation

Answer: (c) \( \ce{2SO3 <=> 2SO2 + O2} \)

Explanation:
1. \( K_p > K_c \) when \( \Delta n > 0 \).
2. \( \ce{2SO3 -> 2SO2 + O2} \) has \( \Delta n = +1 \).

Teacher's Note:
1. Use Kp = Kc(RT)^(change in gas moles) and take R in units that match the pressure and concentration units.
2. Count only gaseous species when finding the change in moles.

Question 25: The \( K_p/K_c \) ratio for \( \ce{4NH3(g) + 5O2(g) <=> 4NO(g) + 6H2O(g)} \) at 127 C is (R = 0.08314 L bar/mol K):
(a) 0.0301
(b) 1.0001
(c) 0.0831
(d) 33.26
Show Answer & Explanation

Answer: (d) 33.26

Explanation:
1. \( \Delta n = 10 - 9 = +1 \).
2. \( K_p/K_c = RT = 0.08314\times400 = 33.26 \).

Teacher's Note:
1. Use Kp = Kc(RT)^(change in gas moles) and take R in units that match the pressure and concentration units.
2. Count only gaseous species when finding the change in moles.

Free study material for Chemistry

Equilibrium Objective Questions & Solutions for JEE Chemistry

Chapter MCQs with Answers for JEE Chemistry

Explore reliable practice questions for Equilibrium tailored for JEE Chemistry learners. Use these multiple-choice formats to evaluate preparedness and strengthen problem-solving skills.

Expert Practice Material for JEE Chemistry

Built using the official NCERT book for JEE, these Chemistry objective sets provide reliable academic guidance. Pair your practice with our recommended NCERT solutions to master optimal problem-solving approaches.

Complete Your Chapter Revision

Follow up your worksheet practice by attempting the interactive online Chemistry MCQ test for this chapter to evaluate your execution speed. All platform resources are free to access.

FAQs

Where can I access latest JEE Chemistry Equilibrium MCQs Set 30?

You can get most exhaustive JEE Chemistry Equilibrium MCQs Set 30 for free on StudiesToday.com. These MCQs for JEE Chemistry are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Chemistry JEE material?

Yes, our JEE Chemistry Equilibrium MCQs Set 30 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Chemistry MCQs help in scoring full marks in JEE exams?

By solving our JEE Chemistry Equilibrium MCQs Set 30, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.

Do you provide answers and explanations for JEE Chemistry Equilibrium MCQs Set 30?

Yes, Chemistry MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Chemistry JEE MCQs online?

Yes, you can also access online interactive tests for JEE Chemistry Equilibrium MCQs Set 30 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.