Equilibrium JEE Main Questions 2025:Important Practice Questions with PYQs

Equilibrium JEE Main Questions 2025 cover topics such as Equilibria involving Physical and Chemical processes, Ionic equilibrium etc. Equilibrium chapter holds a weightage of 4 % in the exam. Some of the important questions have been provided here along with their solutions. Practice these questions to ace the exam.

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JEE Main 2025 Equilibrium Important Questions
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Equilibrium is one of the important topics of JEE Main 2025 Chemistry Syllabus as it holds 4% of the Chemistry Section. Mastering such topics along with the important Equilibrium JEE Main Questions 2025 is crucial for the aspirants of the JEE Main exam. The Equilibrium section of the JEE Main 205 Chemistry Syllabus deals with chemical equilibrium, iIonic equilibrium and solubility products and candidates and can find a minimum of 1 - 2 questions from the Equilibrium chapter in the JEE Main 2025 exam. Therefore, it is crucial that candidates habituate themselves with the important practice questions of the Equilibrium chapter for JEE Main 2025. Candidates are advised to read this article thoroughly for the JEE Main Equilibrium questions and their solutions.

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JEE Main 2025 Equilibrium Important Questions

Candidates can check the practice questions for JEE Main Equilibrium 2025 along with their solutions as provided in the table below.

Serial No.	Question	Solution
1.	What is the conjugate base of OH–? (a) O2 (b) H2O (c) O– (d) O2-	When acid gives H+ then the remaining of its part is called the conjugate base. The conjugate base of OH– is O2-. Hence option (d) is the answer.
2.	The increase of pressure on ice water system at constant temperature will lead to (a) no effect on that equilibrium (b) a decrease in the entropy of the system (c) a shift of the equilibrium in the forward direction (d) an increase in the Gibbs energy of the system.	On increasing the pressure on this system in equilibrium, the equilibrium tends to shift in a direction in which volume decreases, i.e., in the forward direction. Hence option (c) is the answer.
3.	Among the following acids which have the lowest pKa value? (a) CH3COOH (b) (CH3)2CH-COOH (c) HCOOH (d) CH3CH2COOH	Higher the pKa value, weaker is the acid. So the strongest acid has the lowest pKa value. Hence option (c) is the answer.
4.	Which of the following is a Lewis acid? (a) NaH (b) NF3 (c) PH3 (d) B(CH3)3	The compound which can accept a pair of electrons is known as Lewis acid. B(CH3)3 Is a Lewis acid. Hence option (d) is the answer.
5.	Which one of the following substances has the highest proton affinity? (a) H2S (b) NH3 (c) PH3 (d) H2O	The stability of the conjugate acid will give us the compound with the highest proton affinity. Here ammonia has the highest proton affinity. Hence option (b) is the answer.
6.	A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is (a) 1.8 atm (b) 3 atm (c) 0.3 atm (d) 0.18 atm	Given total pressure = 0.8 atm CO2(g) + C(s) ⇌ 2CO(g) Total pressure = 0.5-P +2P = 0.8 P = 0.8-0.5 = 0.3 KP = P2CO/PCO2 = (2P)2/(0.5-P) = (0.6)2/0.2 = 0.36/0.2 = 1.8 atm Hence option (a) is the answer.
7.	The molar solubility of Cd(OH)2 is 1.84 × 10–5 M in water. The expected solubility of Cd(OH)2 in a buffer solution of pH = 12 is (a) 1.84 × 10–9 M (b) 2.49 × 10–10 M (c) (2.49/ 1.84)× 10–9 M (d) 6.23 × 10–11 M	Given molar solubility, s = 1.84 × 10–5 Ksp = 4s3 = 4( 1.84 × 10–5)3 Cd(OH2) ⇌ Cd2+ + 2OH-. s’ represents the solubility in buffer solution pH = 12 pOH = 2 [OH–] = 10-2 s’ ×(10-2)2 = 4(1.84×10-5)3 So, s’ = 2.492 ×10-10 moles L-1 Hence option (b) is the answer.
8.	Which of the following are Lewis acids? (a) PH3 and BCl3 (b) AlCl3 and SiCl4 (c) PH3 and SiCl4 (d) BCl3 and AlCl3	The compound which can accept a pair of electrons is known as Lewis acid. BCl3 and AlCl3 have vacant orbitals and their octet is not complete. Hence these can accept electron pairs and behave as Lewis acids. Hence option (d) is the answer.
9.	For the reaction, 2SO2(g) + O2(g) ⇌ 2SO3(g), ∆H = -57.2 kJ mol–1 and Kc = 1.7 × 1016 Which of the following statements is incorrect? (a) The equilibrium will shift in the forward direction as the pressure increases. (b) The addition of inert gas at constant volume will not affect the equilibrium constant. (c) The equilibrium constant is large, suggestive of the reaction going to completion and so no catalyst is required. (d) The equilibrium constant decreases as the temperature increases.	The large value of Kc suggests that the reaction should go almost to completion. The oxidation of SO2 to SO3 is very slow. So the rate of reaction is increased by adding a catalyst. Statement c is wrong. Hence option (c) is the answer.
10.	The exothermic formation of ClF3 is represented by the equation: Cl2(g) + 3F2(g) ⇌ 2ClF3(g); ∆H = –329 kJ Which of the following will increase the quantity of ClF3 in an equilibrium mixture of Cl2, F2 and ClF3? (a) Increasing the temperature (b) Removing Cl2 (c) Increasing the volume of the container (d) Adding F2	The addition of reactants or removal of the product will favour the forward reaction. So adding F2 will increase the quantity of ClF3. Hence option (d) is the answer.
11.	When rain is accompanied by a thunderstorm, the collected rainwater will have a pH value (a) slightly lower than that of rainwater without a thunderstorm (b) slightly higher than that when the thunderstorm is not there (c) uninfluenced by the occurrence of a thunderstorm (d) which depends on the amount of dust in the air.	The temperature increases due to the thunderstorm. As temperature increases, [H+] also increases, and thus pH decreases. Hence option (a) is the answer.
12.	The pH of rain water is approximately (a) 7.5 (b) 6.5 (c) 7.0 (d) 5.6	The pH of rainwater is approximately 5.6. Hence option (d) is the answer.
13.	Species acting as both Bronsted acid and base is (a) (HSO4) – (b) Na2CO3 (c) NH3 (d) OH-	A Bronsted acid is a substance that can donate a proton to any other substance and a Bronsted base is a substance that can accept a proton from any other substance. (HSO4)– can donate and accept a proton. Hence option (a) is the answer.
14.	The correct relationship between free energy change in a reaction and the corresponding equilibrium constant Kc is (a) ∆G = RT ln Kc (b) –∆G = RT ln Kc (c) ∆G° = RT ln Kc (d) –∆G° = RT ln Kc	∆G = ∆G° + 2.303 RT logKc At equilibrium, ∆G = 0 So ∆G° = –2.303 RT logKc Hence option (d) is the answer.
15.	20 mL of 0.1 M H2SO4 solution is added to 30 mL of 0.2 M NH4OH solution. The pH of the resultant mixture is [pKb of NH4OH = 4.7] (a) 9.4 (b) 9.0 (c) 5.0 (d) 5.2	Given pKb of NH4OH = 4.7 20 mL of 0.1 M H2SO4 ⇒ nH+ = 4 30 ml 0.2 M NH4OH ⇒ nNH4OH = 6 Solution is basic buffer. pOH = pKb + log [NH4+]/[NH4OH] = 4.7 + log (4/2) = 4.7 + log 2 = 4.7+0.3 = 5 pH = 14-pOH = 14-5 = 9 Hence option (b) is the answer.

Serial No.

Question

Solution

What is the conjugate base of OH–?

(a) O2

(b) H2O

(d) O2-

When acid gives H+ then the remaining of its part is called the conjugate base.

The conjugate base of OH– is O2-.

Hence option (d) is the answer.

The increase of pressure on ice water system at constant temperature will lead to

(a) no effect on that equilibrium

(b) a decrease in the entropy of the system

(d) an increase in the Gibbs energy of the system.

On increasing the pressure on this system in equilibrium, the equilibrium tends to shift in a direction in which volume decreases, i.e., in the forward direction.

Hence option (c) is the answer.

Among the following acids which have the lowest pKa value?

(a) CH3COOH

(b) (CH3)2CH-COOH

(d) CH3CH2COOH

Higher the pKa value, weaker is the acid. So the strongest acid has the lowest pKa value.

Hence option (c) is the answer.

Which of the following is a Lewis acid?

(a) NaH

(b) NF3

(d) B(CH3)3

The compound which can accept a pair of electrons is known as Lewis acid. B(CH3)3

Is a Lewis acid.

Hence option (d) is the answer.

Which one of the following substances has the highest proton affinity?

(a) H2S

(b) NH3

(d) H2O

The stability of the conjugate acid will give us the compound with the highest proton affinity.

Here ammonia has the highest proton affinity.

Hence option (b) is the answer.

A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is

(a) 1.8 atm

(b) 3 atm

(d) 0.18 atm

Given total pressure = 0.8 atm

CO2(g) + C(s) ⇌ 2CO(g)

Total pressure = 0.5-P +2P = 0.8

P = 0.8-0.5 = 0.3

KP = P2CO/PCO2 = (2P)2/(0.5-P)

= (0.6)2/0.2

= 0.36/0.2

= 1.8 atm

Hence option (a) is the answer.

The molar solubility of Cd(OH)2 is 1.84 × 10–5 M in water. The expected solubility of Cd(OH)2 in a buffer solution of pH = 12 is

(a) 1.84 × 10–9 M

(b) 2.49 × 10–10 M

(d) 6.23 × 10–11 M

Given molar solubility, s = 1.84 × 10–5

Ksp = 4s3

= 4( 1.84 × 10–5)3

Cd(OH2) ⇌ Cd2+ + 2OH-.

s’ represents the solubility in buffer solution

pH = 12

pOH = 2

[OH–] = 10-2

s’ ×(10-2)2 = 4(1.84×10-5)3

So, s’ = 2.492 ×10-10 moles L-1

Hence option (b) is the answer.

Which of the following are Lewis acids?

(a) PH3 and BCl3

(b) AlCl3 and SiCl4

(d) BCl3 and AlCl3

The compound which can accept a pair of electrons is known as Lewis acid. BCl3 and AlCl3 have vacant orbitals and their octet is not complete. Hence these can accept electron pairs and behave as Lewis acids.

Hence option (d) is the answer.

For the reaction, 2SO2(g) + O2(g) ⇌ 2SO3(g), ∆H = -57.2 kJ mol–1 and Kc = 1.7 × 1016 Which of the following statements is incorrect?

(a) The equilibrium will shift in the forward direction as the pressure increases.

(b) The addition of inert gas at constant volume will not affect the equilibrium constant.

(d) The equilibrium constant decreases as the temperature increases.

The large value of Kc suggests that the reaction should go almost to completion. The oxidation of SO2 to SO3 is very slow. So the rate of reaction is increased by adding a catalyst. Statement c is wrong.

Hence option (c) is the answer.

10.

The exothermic formation of ClF3 is represented by the equation:

Cl2(g) + 3F2(g) ⇌ 2ClF3(g); ∆H = –329 kJ Which of the following will increase the quantity of ClF3 in an equilibrium mixture of Cl2, F2 and ClF3?

(a) Increasing the temperature

(b) Removing Cl2

(d) Adding F2

The addition of reactants or removal of the product will favour the forward reaction.

So adding F2 will increase the quantity of ClF3.

Hence option (d) is the answer.

11.

When rain is accompanied by a thunderstorm, the collected rainwater will have a pH value

(a) slightly lower than that of rainwater without a thunderstorm

(b) slightly higher than that when the thunderstorm is not there

(d) which depends on the amount of dust in the air.

The temperature increases due to the thunderstorm. As temperature increases, [H+] also increases, and thus pH decreases.

Hence option (a) is the answer.

12.

The pH of rain water is approximately

(a) 7.5

(b) 6.5

(d) 5.6

The pH of rainwater is approximately 5.6.

Hence option (d) is the answer.

13.

Species acting as both Bronsted acid and base is

(a) (HSO4) –

(b) Na2CO3

(d) OH-

A Bronsted acid is a substance that can donate a proton to any other substance and a Bronsted base is a substance that can accept a proton from any other substance. (HSO4)– can donate and accept a proton.

Hence option (a) is the answer.

14.

The correct relationship between free energy change in a reaction and the corresponding equilibrium constant Kc is

(a) ∆G = RT ln Kc

(b) –∆G = RT ln Kc

(d) –∆G° = RT ln Kc

∆G = ∆G° + 2.303 RT logKc

At equilibrium, ∆G = 0

So ∆G° = –2.303 RT logKc

Hence option (d) is the answer.

15.

20 mL of 0.1 M H2SO4 solution is added to 30 mL of 0.2 M NH4OH solution. The pH of the resultant mixture is [pKb of NH4OH = 4.7]

(a) 9.4

(b) 9.0

(d) 5.2

Given pKb of NH4OH = 4.7

20 mL of 0.1 M H2SO4 ⇒ nH+ = 4

30 ml 0.2 M NH4OH ⇒ nNH4OH = 6

JEE Equilibrium Previous Year Questions with Solutions

Solution is basic buffer.

pOH = pKb + log [NH4+]/[NH4OH] = 4.7 + log (4/2)

= 4.7 + log 2

= 4.7+0.3

= 5

pH = 14-pOH

= 14-5 = 9

Hence option (b) is the answer.

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FAQs

Is equilibrium important for the JEE Mains exam?

Understanding the equilibrium chapter is a key part of performing well in the JEE Main exam. The Equilibrium chapter helps students learn about how chemical reactions reach a balanced state, comprising the roles of reaction rates, concentrations, and external forces.

Which is the hardest topic in the JEE Mains exam?

Mathematics is often considered to be the most challenging subject in the JEE Main exam, but its difficulty level varies from person to person. Some students may find other subjects more challenging based upon their strengths and weaknesses.

Is 80 marks in chemistry easy to score in the JEE Mains exam?

In the JEE Main Chemistry section, the questions asked are generally straightforward and the difficulty level is relatively lower. With the correct preparation approach in place, candidates can easily score 80+ in the JEE Mains Chemistry section.

Is the JEE Mains chemistry section easy?

Chemistry is considered to be the rank decider for the majority of candidates in any engineering entrance examination because it mostly deals with the basics and. In comparison to the Mathematics and Physics questions asked in the JEE Main exam, Chemistry questions are more theoretical than calculative, making it relatively easier for candidates to secure good marks.

Is NCERT enough for the JEE Mains 2025 preparation?

NCERT books are very important to prepare for the JEE Main 2025 exam. The JEE Main exam follows the CBSE syllabus for 11th and 12th classes, so NCERT books are considered to be the most reliable and necessary study materials. NCERT books explain basic fundamental concepts in a simple and clear way.

Is it easy to score 99 percentile in the JEE Mains exam?

To score a 99+ percentile in the JEE Main 2025 exam, candidates require more than just knowing the JEE Main syllabus, as it required consistent practice, strategic revision, time management, and the ability to solve complex problems under pressure. However, with the appropriate resources, one can achieve these requirements.

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All sections namely Mathematics, Physics, and Chemistry are mandatory in the JEE Main exam, with questions covering the entire JEE Main syllabus. To be eligible for admission, candidates must achieve the minimum qualifying marks in each subject and the overall aggregate.

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Candidates must keep in mind that there is no requirement of 75% marks in class 12 for appearing in the JEE Mains exam. The JEE Main eligibility criteria of minimum 75% marks in class 12 is required at the time of securing admission across NITs, IIITs and GFTIs. Candidates can apply and appear in the JEE Mains 2025 exam irrespective of their class 12 marks.

Is PYQ enough for the JEE Mains preparation?

Previous year question papers are an important source for the JEE Main preparations. However, relying solely upon PYQs is not enough to fully prepare for the JEE Mains exam. Candidates must also focus on covering the entire syllabus, developing strong conceptual understanding, practising time management, and attempting mock tests.