Chemical Equilibrium - JEE Mains Previous Year Questions

eSaral Academic and Editorial Team
Summary
Practice JEE Main Chemical Equilibrium previous year questions covering equilibrium constants (Kc, Kp), degree of dissociation, reaction quotients, gaseous equilibria, and numerical problem-solving with detailed solutions for effective exam preparation.

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JEE Main Previous Year Papers Questions of Chemistry With Solutions are available at eSaral. Simulator Previous Years AIEEE/JEE Mains Questions [esquestion] A vessel at 1000 K contains $\mathrm{CO}_{2}$ with a pressure of 0.5 atm. Some of the $\mathrm{CO}_{2}$ is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :- (1) 0.3 atm (2) 0.18 atm (3) 1.8 atm (4) 3 atm #tag# [AIEEE-2011] #sol# (3) [/esquestion] [esquestion] The equilibrium constant $\left(\mathrm{K}_{\mathrm{C}}\right)$ for the reaction $\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g})$ at temperature T is $4 \times 10^{-4}$ The value of $\mathrm{K}_{\mathrm{c}}$ for the reaction. $\mathrm{NO}(\mathrm{g}) \longrightarrow 1 / 2 \mathrm{N}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g})$ at the same temperature is :- (1) 50.0 (2) 0.02 (3) $2.5 \times 10^{2}$ (4) $4 \times 10^{-4}$ #tag# [AIEEE-2012] #sol# (1) $\mathrm{K}_{\mathrm{c}}=\frac{1}{\sqrt{\mathrm{K}_{\mathrm{c}}}}$ [/esquestion] [esquestion] 8 mol of $\mathrm{AB}_{3}(\mathrm{g})$ are introduced into a 1.0 $\mathrm{d} \mathrm{m}^{3}$ vessel. If it dissociates as $2 \mathrm{AB}_{3}(\mathrm{g}) \square \quad \mathrm{A}_{2}(\mathrm{g})+3 \mathrm{B}_{2}(\mathrm{g})$ At equilibrium, 2mol of $\mathrm{A}_{2}$ are found to be present. The equilibrium constant of this reaction is :- (1) 36 (2) 3 (3) 27 (4) 2 #tag# [JEE-MAINS(online)-2012] #sol# (3) [/esquestion] [esquestion] The value of Kp for the equilibrium reaction $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \square 2 \mathrm{NO}_{2}(\mathrm{g})$ is 2 The percentage dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})$ at a pressure of 0.5 atm is (1) 71 (2) 50 (3) 88 (4) 25 #tag# [JEE-MAINS(online)-2012] #sol# (1) $\mathrm{K}_{\mathrm{p}}=\frac{(2 \alpha)^{2}}{(1-\alpha)} \times \frac{0.5}{(1+\alpha)}$ [/esquestion] [esquestion] $\mathrm{K}_{1}, \mathrm{K}_{2}$ and $\mathrm{K}_{3}$ are the equilibrium constants of the following reactions (I), (II) and (III), respectively (I) $\mathrm{N}_{2}+2 \mathrm{O}_{2} \square 2 \mathrm{NO}_{2}$ (II) $2 \mathrm{NO}_{2} \square \mathrm{N}_{2}+2 \mathrm{O}_{2}$ (III) $\mathrm{NO}_{2} \square \frac{1}{2} \mathrm{N}_{2}+\mathrm{O}_{2}$ The correct relation from the following is : (1) $\mathrm{K}_{1}=\sqrt{\mathrm{K}_{2}}=\mathrm{K}_{3}$ (2) $\mathrm{K}_{1}=\frac{1}{\mathrm{K}_{2}}=\frac{1}{\mathrm{K}_{3}}$ $(3) \mathrm{K}_{1}=\frac{1}{\mathrm{K}_{2}}=\mathrm{K}_{3}$ (4)$\mathrm{K}_{1}=\frac{1}{\mathrm{K}_{2}}=\frac{1}{\left(\mathrm{K}_{3}\right)^{2}}$ #tag# [JEE-MAINS(online)-2012] #sol# (4) Fact [/esquestion] [esquestion] One mole of $\mathrm{O}_{2}(\mathrm{g})$ and two moles of SO2(g) were heated in a closed vessel of one litre capacity at 1098 K. At equilibrium 1.6 moles of $\mathrm{SO}_{3}$ (g) were found. The equilibrium constant $\mathbf{K}_{C}$ of the reaction would be :- (1) 60 (2) 80 (3) 30 (4) 40 #tag# [JEE-MAINS(online)-2012] #sol# (2) [/esquestion] [esquestion] $\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \square 2 \mathrm{NH}_{3}(\mathrm{g}), \mathrm{K}_{1}$ $\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \square 2 \mathrm{NO}(\mathrm{g}), \mathrm{K}_{2} \quad(\mathrm{B})$ $\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \square \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{K}_{3} \quad(\mathrm{C})$ The equation for the equilibrium constant of the reaction $2 \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{g}) \square 2 \mathrm{NO}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}),\left(\mathrm{K}_{4}\right)$ in terms of $\mathrm{K}_{1}, \mathrm{K}_{2}$ and $\mathrm{K}_{3}$ is : (1) $\frac{\mathrm{K}_{1} \mathrm{K}_{3}^{2}}{\mathrm{K}_{2}}$ (2) $\frac{\mathrm{K}_{2} \mathrm{K}_{3}^{3}}{\mathrm{K}_{1}}$ (3) $\frac{\mathrm{K}_{1} \mathrm{K}_{2}}{\mathrm{K}_{3}}$ (4) $\mathrm{K}_{1} \mathrm{K}_{2} \mathrm{K}_{3}$ #tag# [JEE-MAINS(online)-2013] #sol# (2) Fact [/esquestion] [esquestion] In reaction $\mathrm{A}+2 \mathrm{B} \square 2 \mathrm{C}+\mathrm{D}$, initial concentration of B was 1.5 times of |A|, but at equilibrium the concentrations of A and B became equal. The equilibrium constant for the reaction is (1)4 (2) 6 (3) 12 (4) 8 #tag# [JEE-MAINS(online)-2013] #sol# (1) $\mathrm{K}_{\mathrm{c}}=4$ [/esquestion] [esquestion] For the decomposition of the compound, represented as If the reaction is started with 1 mol of the compound, the total pressure at equilibrium would be (1) $38.8 \times 10^{-2}$ atm (2) $1.94 \times 10^{-2}$ atm (3) $5.82 \times 10^{-2}$ atm (4) $7.66 \times 10^{-2}$ atm #tag# [JEE-MAINS(online)-2014] #sol# (3) [/esquestion] [esquestion] For the reaction $\mathrm{SO}_{2(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{g})} \square \mathrm{SO}_{3(\mathrm{g})},$ if $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\mathrm{x}}$ where the symbols have usual meaning then the value of x is : (assuming ideality) ( 1)$\frac{1}{2}$ ( 2) 1 (3) –1 $(4)-\frac{1}{2}$ #tag# [JEE-MAINS 2014] #sol# (4) $\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{-\frac{1}{2}}$ [/esquestion] [esquestion] The equilibrium constants at 298 K for a reaction $\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}$ is 100 If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol $\mathrm{L}^{-1}$) will be : (1) 1.182 (2) 0.182 (3) 0.818 (4) 1.818 #tag# [JEE-Mains 2016] #sol# (4) [/esquestion]








Students studying Chemical Equilibrium can keep their revision organised with the JEE Mains PYQ chapter wise index.
This Chemical Equilibrium set covers one chapter only — for entire papers exactly as they appeared on exam day, check our JEE Main exam papers.
Frequently Asked Questions
How many questions come from Chemical Equilibrium in JEE Main?
JEE Main typically includes 1 to 2 questions from Chemical Equilibrium per session, based on NTA's official question paper archives from 2011 to 2026. The chapter appears in both the January and April sessions. It is one of the more predictable scoring chapters in Physical Chemistry because question patterns repeat across years.
What is the relationship between Kp and Kc?
Kp = Kc × (RT)^Δn, where Δn is the difference between moles of gaseous products and moles of gaseous reactants in the balanced equation, R is 0.0821 L·atm·mol⁻¹·K⁻¹, and T is temperature in Kelvin. When Δn = 0, Kp equals Kc. This formula appears directly or indirectly in nearly every JEE Main Kp question.
How do I find Kc when the equilibrium is reversed or scaled?
If the forward reaction has equilibrium constant K, the reverse reaction has constant 1/K. If all coefficients in the equation are multiplied by a factor n, the new equilibrium constant is K^n. These two rules, applied together, solve the majority of "K₁, K₂, K₃ relationship" questions in JEE Main without any calculation.
What is the degree of dissociation and how is it tested in JEE Main?
Degree of dissociation (α) is the fraction of the original substance that has dissociated at equilibrium. JEE Main questions typically give you Kp and the total pressure, then ask for the percentage dissociation. Set up the equilibrium mole table in terms of α, write the Kp expression, and solve — the equation usually simplifies to a clean quadratic or, when α << 1, a simple approximation.
Is Chemical Equilibrium important for JEE Advanced as well?
Yes. JEE Advanced tests the same concepts but at greater depth — multi-step equilibria, buffer solutions (linked to ionic equilibrium), and simultaneous equilibria. The previous year questions on this page cover the JEE Main level fully. For JEE Advanced, extend your study to ionic equilibrium and common-ion effect. The NCERT Books Class 12 are the correct starting point for both levels.

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eSaral Academic and Editorial Team
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