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.
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. |
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