Chemical Equilibrium JEE Main previous year questions with solutions

5 solved JEE Main questions on Chemical Equilibrium, free to read — no sign-in needed. The full chapter has 132 questions; sign in to attempt the remaining 127 in the exam simulator.

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1. Q1JEE Main 2026 (08 Apr, Shift 2)Law of Mass Action, Equilibrium Constant (Kc and Kp) and its Application
   Solid carbon, CaO and CaCO$_3$ are mixed and allowed to attain equilibrium at T K. $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \quad K_{p_1} = 0.08$ atm $\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g) \quad K_{p_2} = 2$ atm The partial pressure of CO is ________ $\times 10^{-1}$ atm
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   Answer: 4

   For the first equilibrium: $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$ $K_{p_1} = P_{\text{CO}_2} = 0.08 \text{ atm}$ For the second equilibrium: $\text{C}(s) + \text{CO}_2(g) \rightleftharpoons 2\text{CO}(g)$ $K_{p_2} = \dfrac{(P_{\text{CO}})^2}{P_{\text{CO}_2}} = 2 \text{ atm}$ Substituting the value of $P_{\text{CO}_2}$: $\dfrac{(P_{\text{CO}})^2}{0.08} = 2$ $(P_{\text{CO}})^2 = 0.16$ $P_{\text{CO}} = 0.4 \text{ atm} = 4 \times 10^{-1} \text{ atm}$ Answer: $4$
2. Q2JEE Main 2025 (08 Apr, Shift 2)Ionic Product of Water
   The equilibrium constant for decomposition of $\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ $\mathrm{H}_2 \mathrm{O}(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\left(\Delta \mathrm{G}^{\circ}=92.34 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$ is $8.0 \times 10^{-3}$ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation $(\alpha)$ of water is ________ $\times 10^{-2}$ (nearest integer value). [Assume $\alpha$ is negligible with respect to 1 ]
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   Answer: 5

   $\begin{aligned} & \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \\ & \mathrm{t}=0 \quad 1 \mathrm{~mole} \\ & \mathrm{t}=\mathrm{t}_{\mathrm{eq}} \quad 1-\alpha \quad \alpha \quad \frac{\alpha}{2} \\ & \mathrm{n}_{\mathrm{T}}=1+\frac{\alpha}{2} \simeq 1(\alpha \ll 1) \\ & \mathrm{k}_{\mathrm{P}}=\frac{\mathrm{P}_{\mathrm{H}_2} \cdot \mathrm{P}_{\mathrm{O}_2}^{1 / 2}}{\mathrm{P}_{\mathrm{H}_2 \mathrm{O}}}=\frac{(\alpha \cdot \mathrm{P})\left(\frac{\alpha}{2} \mathrm{P}\right)^{\frac{1}{2}}}{(1-\alpha) \mathrm{P}} \\ & 8 \times 10^{-3}=\frac{\alpha^{3 / 2}}{\sqrt{2}} \\ & \alpha^{3 / 2}=8 \sqrt{2} \times 10^{-3} \\ & \alpha^3=128 \times 10^{-6} \\ & \alpha=\sqrt[3]{128} \times 10^{-2} \\ & =5.03 \times 10^{-2}\end{aligned}$
3. Q3JEE Main 2024 (08 Apr, Shift 2)Le chatelier's principle
   The equilibrium $\mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightleftharpoons 2 \mathrm{CrO}_4^{2-}$ is shifted to the right in :
   1. A.an acidic medium
   2. B.a basic medium
   3. C.a neutral medium
   4. D.a weakly acidic medium
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   Answer: (B)

   $\mathrm{Cr}_2 \mathrm{O}_7^{2-} \stackrel{\mathrm{OH}^{-}}{\rightleftharpoons} 2 \mathrm{HrO}_4^{2-}$
4. Q4JEE Main 2022 (24 Jun, Shift 1)Third law of thermodynamics
   $2{O}_{3}\left(g\right)\rightleftharpoons 3{O}_{2}\left(g\right)$ At $300K$, ozone is fifty percent dissociated. The standard free energy change at this temperature and $1atm$ pressure is $\left(-\right)......J{mol}^{-1}$. (Nearest integer) [Given: $\ln 1.35=0.3$ and $R=8.3J{K}^{-1}{mol}^{-1}$]
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   Answer: 747

   $2{O}_{3}\left(g\right)\rightleftharpoons 3{O}_{2}\left(g\right)$ Initially $1$ mole $0$ $\left(1-0.5\right)$ $\dfrac{3}{2}\times 0.5$ $0.5$ mole $\dfrac{1.5}{2}=0.75$ mole ${K}_{p}=\dfrac{{\left({P}_{{O}_{2}}\right)}^{3}}{{\left({P}_{{O}_{3}}\right)}^{2}}=\dfrac{{\left(\dfrac{0.75}{1.25}\right)}^{3}}{{\left(\dfrac{0.5}{1.25}\right)}^{2}}=\dfrac{{\left(\dfrac{3}{5}\right)}^{3}}{{\left(\dfrac{2}{5}\right)}^{2}}$ $=\dfrac{{\left(0.6\right)}^{3}}{{\left(0.4\right)}^{2}}=\dfrac{\left(0.216\right)}{\left(0.16\right)}=1.35$ $\Delta G^{\circ}=-{RTlnK}_{p}$ $=-8.3\times 300\ln 1.35$ $=-8.3\times 300\times 0.3$ $=-747J/mole$
5. Q5JEE Main 2012 (07 May)Rate law and rate constant
   $K_1, K_2$ and $K_3$ are the equilibrium constants of the following reactions (I), (II) and (III) respectively: (I) $\mathrm{N}_2+2 \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}_2$ (II) $2 \mathrm{NO}_2 \rightleftharpoons \mathrm{N}_2+2 \mathrm{O}_2$ (III) $\mathrm{NO}_2 \rightleftharpoons \frac{1}{2} \mathrm{~N}_2+\mathrm{O}_2$ The correct relation from the following is
   1. A.$K_1=\frac{1}{K_2}=\frac{1}{K_3}$
   2. B.$K_1=\frac{1}{K_2}=\frac{1}{\left(K_3\right)^2}$
   3. C.$K_1=\sqrt{K_2}=K_3$
   4. D.$K_1=\frac{1}{K_2}=K_3$
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   Answer: (B)

   (I) $\mathrm{N}_2+2 \mathrm{O}_2 \stackrel{K_1}{\rightleftharpoons} 2 \mathrm{NO}_2$ $$K_1=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]^2}$$ (II) $2 \mathrm{NO}_2 \underset{K_2}{\rightleftharpoons} \mathrm{N}_2+2 \mathrm{O}_2$ $$K_2=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]^2}{\left[\mathrm{NO}_2\right]^2}$$ (III) $$\begin{aligned} & \mathrm{NO}_2 \underset{K_3}{\rightleftharpoons} \frac{1}{2} \mathrm{~N}_2+\mathrm{O}_2 \\ & K_3=\frac{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{O}_2\right]}{\left[\mathrm{NO}_2\right]} \\ & \therefore\left(K_3\right)^2=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]^2}{\left[\mathrm{NO}_2\right]^2} \end{aligned}$$ $\therefore$ from equation (i), (ii) and (iii) $$K_1=\frac{1}{K_2}=\frac{1}{\left(K_3\right)^2}$$