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JEE Main Previous Year Questions (2025): Chemical Equilibrium
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Page 1 JEE Main Previous Year Qs (2025): Chemical Equilibrium Q1: A vessel at 1000 K contains ???? ?? with a pressure of 0.5 atm . Some of ???? ?? is converted into ???? on addition of graphite. If total pressure at equilibrium is ?? . ?? ?????? , then ?? ?? is : JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 0.18 atm B. 0.3 atm C. 3 atm D. 1.8 atm Ans: D Solution: The reaction is: CO 2 ( ?? )+ C( ?? )? 2CO ( ?? ) Initially, the pressure of CO 2 is 0.5 atm , and the pressure of CO is 0 atm . Let ' x ' be the change in pressure of CO 2 at equilibrium. Since the stoichiometric coefficient of CO is twice that of CO 2 , the pressure of CO formed at equilibrium is 2 x . At equilibrium: Pressure of CO 2 = 0.5 - ?? Pressure of CO = 2?? The total pressure at equilibrium is given as 0.8 atm . Therefore: ( 0.5 - ?? )+ 2?? = 0.8 0.5 + ?? = 0.8 ?? = 0.8 - 0.5 = 0.3 So, at equilibrium: Pressure of CO 2 = 0.5 - 0.3 = 0.2 atm Pressure of CO = 2( 0.3)= 0.6 atm The equilibrium constant ?? ?? is given by: ?? ?? = ?? CO 2 ?? CO 2 ?? ?? = ( 0.6) 2 0.2 = 0.36 0.2 = 1.8 atm Therefore, the value of ?? ?? is 1.8 atm . Q2: Consider the reaction ?? ?? ?? ( ?? )? ?? ?? ( ?? )+ ?? ?? ?? ?? ( ?? ) The equation representing correct relationship between the degree of dissociation ( ?? ) of ?? ?? ?? ( ?? ) with its equilibrium constant Kp is ____ . Page 2 JEE Main Previous Year Qs (2025): Chemical Equilibrium Q1: A vessel at 1000 K contains ???? ?? with a pressure of 0.5 atm . Some of ???? ?? is converted into ???? on addition of graphite. If total pressure at equilibrium is ?? . ?? ?????? , then ?? ?? is : JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 0.18 atm B. 0.3 atm C. 3 atm D. 1.8 atm Ans: D Solution: The reaction is: CO 2 ( ?? )+ C( ?? )? 2CO ( ?? ) Initially, the pressure of CO 2 is 0.5 atm , and the pressure of CO is 0 atm . Let ' x ' be the change in pressure of CO 2 at equilibrium. Since the stoichiometric coefficient of CO is twice that of CO 2 , the pressure of CO formed at equilibrium is 2 x . At equilibrium: Pressure of CO 2 = 0.5 - ?? Pressure of CO = 2?? The total pressure at equilibrium is given as 0.8 atm . Therefore: ( 0.5 - ?? )+ 2?? = 0.8 0.5 + ?? = 0.8 ?? = 0.8 - 0.5 = 0.3 So, at equilibrium: Pressure of CO 2 = 0.5 - 0.3 = 0.2 atm Pressure of CO = 2( 0.3)= 0.6 atm The equilibrium constant ?? ?? is given by: ?? ?? = ?? CO 2 ?? CO 2 ?? ?? = ( 0.6) 2 0.2 = 0.36 0.2 = 1.8 atm Therefore, the value of ?? ?? is 1.8 atm . Q2: Consider the reaction ?? ?? ?? ( ?? )? ?? ?? ( ?? )+ ?? ?? ?? ?? ( ?? ) The equation representing correct relationship between the degree of dissociation ( ?? ) of ?? ?? ?? ( ?? ) with its equilibrium constant Kp is ____ . Assume ?? to be very very small. JEE Main 2025 (Online) 23rd January Evening Shift Options: A. ?? = v Kp p 3 B. ?? = v Kp 2p 3 C. ?? = v 2?? ?? 2 p 3 D. ?? = v 2?? p p 3 Ans: C Solution: To determine the relationship between the degree of dissociation ( ?? ) of X 2 Y( g) and its equilibrium constant ?? ?? , let's analyze the reaction: X 2 Y( g)? X 2 ( g)+ 1 2 Y 2 ( g) Initial and Change in Moles Initially, we start with 1 mole of X 2 Y. At equilibrium: X 2 Y is ( 1 - ?? ) moles. X 2 is ?? moles. 1 2 Y 2 is ?? 2 moles. Partial Pressures The total pressure ?? total is given by: P X 2 Y = 1 - ?? 1 + ?? 2 × P P X 2 = ?? 1 + ?? 2 × P P Y 2 = ?? /2 1 + ?? 2 × P Equilibrium Constant Expression The equilibrium constant ?? ?? can be written in terms of partial pressures: ?? ?? = ( ?? 1 + ?? 2 · P) · ( ?? 2 1 + ?? 2 · P) 1/2 1 - ?? 1 + ?? 2 · P Page 3 JEE Main Previous Year Qs (2025): Chemical Equilibrium Q1: A vessel at 1000 K contains ???? ?? with a pressure of 0.5 atm . Some of ???? ?? is converted into ???? on addition of graphite. If total pressure at equilibrium is ?? . ?? ?????? , then ?? ?? is : JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 0.18 atm B. 0.3 atm C. 3 atm D. 1.8 atm Ans: D Solution: The reaction is: CO 2 ( ?? )+ C( ?? )? 2CO ( ?? ) Initially, the pressure of CO 2 is 0.5 atm , and the pressure of CO is 0 atm . Let ' x ' be the change in pressure of CO 2 at equilibrium. Since the stoichiometric coefficient of CO is twice that of CO 2 , the pressure of CO formed at equilibrium is 2 x . At equilibrium: Pressure of CO 2 = 0.5 - ?? Pressure of CO = 2?? The total pressure at equilibrium is given as 0.8 atm . Therefore: ( 0.5 - ?? )+ 2?? = 0.8 0.5 + ?? = 0.8 ?? = 0.8 - 0.5 = 0.3 So, at equilibrium: Pressure of CO 2 = 0.5 - 0.3 = 0.2 atm Pressure of CO = 2( 0.3)= 0.6 atm The equilibrium constant ?? ?? is given by: ?? ?? = ?? CO 2 ?? CO 2 ?? ?? = ( 0.6) 2 0.2 = 0.36 0.2 = 1.8 atm Therefore, the value of ?? ?? is 1.8 atm . Q2: Consider the reaction ?? ?? ?? ( ?? )? ?? ?? ( ?? )+ ?? ?? ?? ?? ( ?? ) The equation representing correct relationship between the degree of dissociation ( ?? ) of ?? ?? ?? ( ?? ) with its equilibrium constant Kp is ____ . Assume ?? to be very very small. JEE Main 2025 (Online) 23rd January Evening Shift Options: A. ?? = v Kp p 3 B. ?? = v Kp 2p 3 C. ?? = v 2?? ?? 2 p 3 D. ?? = v 2?? p p 3 Ans: C Solution: To determine the relationship between the degree of dissociation ( ?? ) of X 2 Y( g) and its equilibrium constant ?? ?? , let's analyze the reaction: X 2 Y( g)? X 2 ( g)+ 1 2 Y 2 ( g) Initial and Change in Moles Initially, we start with 1 mole of X 2 Y. At equilibrium: X 2 Y is ( 1 - ?? ) moles. X 2 is ?? moles. 1 2 Y 2 is ?? 2 moles. Partial Pressures The total pressure ?? total is given by: P X 2 Y = 1 - ?? 1 + ?? 2 × P P X 2 = ?? 1 + ?? 2 × P P Y 2 = ?? /2 1 + ?? 2 × P Equilibrium Constant Expression The equilibrium constant ?? ?? can be written in terms of partial pressures: ?? ?? = ( ?? 1 + ?? 2 · P) · ( ?? 2 1 + ?? 2 · P) 1/2 1 - ?? 1 + ?? 2 · P Simplifying the expression, considering ?? to be very small, results in: ?? ?? = ( ?? 1 - ?? )( ?? 2 ( 1 + ?? 2 ) ) 1/2 · P 1 2 For small ?? , ( 1 - ?? )˜ 1, thus: ?? ?? = ?? 3/2 1 3 · P 1 2 ?? 3/2 = ?? ?? · 2 1/2 P 1/2 Finally, by cubing both sides: ?? 3 = 2 · ?? ?? 2 P ?? = ( 2 · ?? ?? 2 P ) 1/3 Thus, the degree of dissociation ?? relates to ?? ?? by the equation: ?? = ( 2 · ?? ?? 2 P ) 1/3 Q3: For the reaction, ?? ?? ( ?? )+ ?? ?? ( ?? )? ?????? ( ?? ) Attainment of equilibrium is predicted correctly by : JEE Main 2025 (Online) 24th January Evening Shift Options: A. B. Page 4 JEE Main Previous Year Qs (2025): Chemical Equilibrium Q1: A vessel at 1000 K contains ???? ?? with a pressure of 0.5 atm . Some of ???? ?? is converted into ???? on addition of graphite. If total pressure at equilibrium is ?? . ?? ?????? , then ?? ?? is : JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 0.18 atm B. 0.3 atm C. 3 atm D. 1.8 atm Ans: D Solution: The reaction is: CO 2 ( ?? )+ C( ?? )? 2CO ( ?? ) Initially, the pressure of CO 2 is 0.5 atm , and the pressure of CO is 0 atm . Let ' x ' be the change in pressure of CO 2 at equilibrium. Since the stoichiometric coefficient of CO is twice that of CO 2 , the pressure of CO formed at equilibrium is 2 x . At equilibrium: Pressure of CO 2 = 0.5 - ?? Pressure of CO = 2?? The total pressure at equilibrium is given as 0.8 atm . Therefore: ( 0.5 - ?? )+ 2?? = 0.8 0.5 + ?? = 0.8 ?? = 0.8 - 0.5 = 0.3 So, at equilibrium: Pressure of CO 2 = 0.5 - 0.3 = 0.2 atm Pressure of CO = 2( 0.3)= 0.6 atm The equilibrium constant ?? ?? is given by: ?? ?? = ?? CO 2 ?? CO 2 ?? ?? = ( 0.6) 2 0.2 = 0.36 0.2 = 1.8 atm Therefore, the value of ?? ?? is 1.8 atm . Q2: Consider the reaction ?? ?? ?? ( ?? )? ?? ?? ( ?? )+ ?? ?? ?? ?? ( ?? ) The equation representing correct relationship between the degree of dissociation ( ?? ) of ?? ?? ?? ( ?? ) with its equilibrium constant Kp is ____ . Assume ?? to be very very small. JEE Main 2025 (Online) 23rd January Evening Shift Options: A. ?? = v Kp p 3 B. ?? = v Kp 2p 3 C. ?? = v 2?? ?? 2 p 3 D. ?? = v 2?? p p 3 Ans: C Solution: To determine the relationship between the degree of dissociation ( ?? ) of X 2 Y( g) and its equilibrium constant ?? ?? , let's analyze the reaction: X 2 Y( g)? X 2 ( g)+ 1 2 Y 2 ( g) Initial and Change in Moles Initially, we start with 1 mole of X 2 Y. At equilibrium: X 2 Y is ( 1 - ?? ) moles. X 2 is ?? moles. 1 2 Y 2 is ?? 2 moles. Partial Pressures The total pressure ?? total is given by: P X 2 Y = 1 - ?? 1 + ?? 2 × P P X 2 = ?? 1 + ?? 2 × P P Y 2 = ?? /2 1 + ?? 2 × P Equilibrium Constant Expression The equilibrium constant ?? ?? can be written in terms of partial pressures: ?? ?? = ( ?? 1 + ?? 2 · P) · ( ?? 2 1 + ?? 2 · P) 1/2 1 - ?? 1 + ?? 2 · P Simplifying the expression, considering ?? to be very small, results in: ?? ?? = ( ?? 1 - ?? )( ?? 2 ( 1 + ?? 2 ) ) 1/2 · P 1 2 For small ?? , ( 1 - ?? )˜ 1, thus: ?? ?? = ?? 3/2 1 3 · P 1 2 ?? 3/2 = ?? ?? · 2 1/2 P 1/2 Finally, by cubing both sides: ?? 3 = 2 · ?? ?? 2 P ?? = ( 2 · ?? ?? 2 P ) 1/3 Thus, the degree of dissociation ?? relates to ?? ?? by the equation: ?? = ( 2 · ?? ?? 2 P ) 1/3 Q3: For the reaction, ?? ?? ( ?? )+ ?? ?? ( ?? )? ?????? ( ?? ) Attainment of equilibrium is predicted correctly by : JEE Main 2025 (Online) 24th January Evening Shift Options: A. B. C. D. Ans: C Solution: H 2 + I 2 ? 2HI Concentration of H 2 and I 2 decreases untill equilibrium condition and concentration of HI increases till equilibrium condition and after equilibrium concentration of all the reactant and products remain constant.Correct option (2) Q4: At temperature ?? , compound ?? ?? ?? ( ?? ) dissociates as ?? ?? ?? ( ?? ) ? ?? ?? ( ?? ) + ?? ?? ?? ?? ( ?? ) having degree of dissociation ?? (small compared to unity). The correct expression for ?? in terms of ?? ?? and ?? is: JEE Main 2025 (Online) 29th January Morning Shift Options: A. v ?? ?? B. v 2?? p 2 p 3 C. v 2?? ?? ?? 3 D. v 2?? ?? ?? 4 Ans: B Solution: ?? ?? 2 ( ?? )? ?? ?? ( ?? ) + 1 2 ?? 2 ( ?? ) Degree of dissociation ? ?? (small compared to unity) Equilibrium constant interms of pressure ? ?? ?? Partial pressure of each gas ? ?? ?? ?? 2 , ?? ???? and ?? ?? 2 Total pressure ? ?? ICE Table: Page 5 JEE Main Previous Year Qs (2025): Chemical Equilibrium Q1: A vessel at 1000 K contains ???? ?? with a pressure of 0.5 atm . Some of ???? ?? is converted into ???? on addition of graphite. If total pressure at equilibrium is ?? . ?? ?????? , then ?? ?? is : JEE Main 2025 (Online) 22nd January Morning Shift Options: A. 0.18 atm B. 0.3 atm C. 3 atm D. 1.8 atm Ans: D Solution: The reaction is: CO 2 ( ?? )+ C( ?? )? 2CO ( ?? ) Initially, the pressure of CO 2 is 0.5 atm , and the pressure of CO is 0 atm . Let ' x ' be the change in pressure of CO 2 at equilibrium. Since the stoichiometric coefficient of CO is twice that of CO 2 , the pressure of CO formed at equilibrium is 2 x . At equilibrium: Pressure of CO 2 = 0.5 - ?? Pressure of CO = 2?? The total pressure at equilibrium is given as 0.8 atm . Therefore: ( 0.5 - ?? )+ 2?? = 0.8 0.5 + ?? = 0.8 ?? = 0.8 - 0.5 = 0.3 So, at equilibrium: Pressure of CO 2 = 0.5 - 0.3 = 0.2 atm Pressure of CO = 2( 0.3)= 0.6 atm The equilibrium constant ?? ?? is given by: ?? ?? = ?? CO 2 ?? CO 2 ?? ?? = ( 0.6) 2 0.2 = 0.36 0.2 = 1.8 atm Therefore, the value of ?? ?? is 1.8 atm . Q2: Consider the reaction ?? ?? ?? ( ?? )? ?? ?? ( ?? )+ ?? ?? ?? ?? ( ?? ) The equation representing correct relationship between the degree of dissociation ( ?? ) of ?? ?? ?? ( ?? ) with its equilibrium constant Kp is ____ . Assume ?? to be very very small. JEE Main 2025 (Online) 23rd January Evening Shift Options: A. ?? = v Kp p 3 B. ?? = v Kp 2p 3 C. ?? = v 2?? ?? 2 p 3 D. ?? = v 2?? p p 3 Ans: C Solution: To determine the relationship between the degree of dissociation ( ?? ) of X 2 Y( g) and its equilibrium constant ?? ?? , let's analyze the reaction: X 2 Y( g)? X 2 ( g)+ 1 2 Y 2 ( g) Initial and Change in Moles Initially, we start with 1 mole of X 2 Y. At equilibrium: X 2 Y is ( 1 - ?? ) moles. X 2 is ?? moles. 1 2 Y 2 is ?? 2 moles. Partial Pressures The total pressure ?? total is given by: P X 2 Y = 1 - ?? 1 + ?? 2 × P P X 2 = ?? 1 + ?? 2 × P P Y 2 = ?? /2 1 + ?? 2 × P Equilibrium Constant Expression The equilibrium constant ?? ?? can be written in terms of partial pressures: ?? ?? = ( ?? 1 + ?? 2 · P) · ( ?? 2 1 + ?? 2 · P) 1/2 1 - ?? 1 + ?? 2 · P Simplifying the expression, considering ?? to be very small, results in: ?? ?? = ( ?? 1 - ?? )( ?? 2 ( 1 + ?? 2 ) ) 1/2 · P 1 2 For small ?? , ( 1 - ?? )˜ 1, thus: ?? ?? = ?? 3/2 1 3 · P 1 2 ?? 3/2 = ?? ?? · 2 1/2 P 1/2 Finally, by cubing both sides: ?? 3 = 2 · ?? ?? 2 P ?? = ( 2 · ?? ?? 2 P ) 1/3 Thus, the degree of dissociation ?? relates to ?? ?? by the equation: ?? = ( 2 · ?? ?? 2 P ) 1/3 Q3: For the reaction, ?? ?? ( ?? )+ ?? ?? ( ?? )? ?????? ( ?? ) Attainment of equilibrium is predicted correctly by : JEE Main 2025 (Online) 24th January Evening Shift Options: A. B. C. D. Ans: C Solution: H 2 + I 2 ? 2HI Concentration of H 2 and I 2 decreases untill equilibrium condition and concentration of HI increases till equilibrium condition and after equilibrium concentration of all the reactant and products remain constant.Correct option (2) Q4: At temperature ?? , compound ?? ?? ?? ( ?? ) dissociates as ?? ?? ?? ( ?? ) ? ?? ?? ( ?? ) + ?? ?? ?? ?? ( ?? ) having degree of dissociation ?? (small compared to unity). The correct expression for ?? in terms of ?? ?? and ?? is: JEE Main 2025 (Online) 29th January Morning Shift Options: A. v ?? ?? B. v 2?? p 2 p 3 C. v 2?? ?? ?? 3 D. v 2?? ?? ?? 4 Ans: B Solution: ?? ?? 2 ( ?? )? ?? ?? ( ?? ) + 1 2 ?? 2 ( ?? ) Degree of dissociation ? ?? (small compared to unity) Equilibrium constant interms of pressure ? ?? ?? Partial pressure of each gas ? ?? ?? ?? 2 , ?? ???? and ?? ?? 2 Total pressure ? ?? ICE Table: ?B 2( g) ? ?B ( g) + 1 2 B 2( g) Initial moles Change ?? +?? + 1 2 ?? moles Equilibrium 1 ?? ?? 1 2 ?? moles Total moles at = 1 ?? + ?? - 1 2 ?? = 1 - 1 2 ?? equilibrium partial pressure at equilibrium = moles at equilibrium total moles at equilibrium × total pressure Partial pressure of ?? ?? 2 ( ?? ) ?? ?? ?? 2 = ( 1-?? 1+ ?? 2 )?? Partial pressure of ???? ( ?? ) ?? ???? = ( ?? 1+ ?? 2 )?? Partial pressure of ?? 2 ( ?? ) ?? ?? 2 = ( ?? 2 1+ ?? 2 ) ?? ?? ?? for the equation ?? 2 ( ?? )? ?? ?? ( ?? ) + 1 2 ?? 2 ( ?? ) can be written asRead More
FAQs on JEE Main Previous Year Questions (2025): Chemical Equilibrium
| 1. What is the definition of chemical equilibrium? |  |
Ans.Chemical equilibrium is a state in a reversible chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time. At this point, the system is in a dynamic state, meaning that reactions continue to occur, but the concentrations remain constant.
| 2. How does Le Chatelier's principle apply to changes in concentration? | |
Ans.Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions (such as concentration, temperature, or pressure), the system will adjust to counteract that change and restore a new equilibrium. For example, if the concentration of a reactant is increased, the equilibrium will shift towards the product side to reduce the concentration of that reactant.
| 3. What factors affect the position of equilibrium in a chemical reaction? | |
Ans.The position of equilibrium in a chemical reaction can be affected by several factors, including changes in concentration of reactants or products, changes in temperature, and changes in pressure (for reactions involving gases). Increasing temperature generally favors the endothermic reaction, while increasing pressure favors the side of the reaction with fewer gas molecules.
| 4. What is the equilibrium constant (K) and its significance? | |
Ans.The equilibrium constant (K) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their respective coefficients in the balanced equation. It is significant because it provides insight into the extent of a reaction; a large value of K indicates that the products are favored at equilibrium, while a small value indicates that reactants are favored.
| 5. How do temperature changes affect the equilibrium constant? | |
Ans.Temperature changes can significantly affect the equilibrium constant (K) of a reaction. For endothermic reactions, increasing the temperature increases the value of K, favoring the formation of products. Conversely, for exothermic reactions, increasing the temperature decreases the value of K, favoring the reactants. This relationship is described by the van 't Hoff equation, which quantitatively relates changes in K to changes in temperature.
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