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 124 questions; sign in to attempt the remaining 119 in the exam simulator.

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 (03 Apr, Shift 1)Le chatelier's principle
   In the following system, $\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$ at equilibrium, upon addition of xenon gas at constant $\mathrm{T} \& \mathrm{p}$, the concentration of
   1. A.$\mathrm{PCl}_5$ will increase
   2. B.$\mathrm{Cl}_2$ will decrease
   3. C.$\mathrm{PCl}_5, \mathrm{PCl}_3 \& \mathrm{Cl}_2$ remain constant
   4. D.$\mathrm{PCl}_3$ will increase
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   Answer: (B)

   On addition of inert gas at constant P \& T, reaction moves in the direction of greater no. of moles so it will shift in forward direction, so $\left[\mathrm{PCl}_5\right]$ decrease and $\left[\mathrm{PCl}_3\right] \&\left[\mathrm{Cl}_2\right]$ will increase.
3. Q3JEE Main 2026 (08 Apr, Shift 2)Law of Mass Action, Equilibrium Constant (Kc and Kp) and its Application
   Consider the following reactions in which all the reactants and products are present in gaseous state $2xy \rightleftharpoons x_2 + y_2 \quad K_1 = 2.5\times 10^5$ $xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad K_2 = 5\times 10^{-3}$ The value of $K_3$ for the equilibrium $\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$ is:
   1. A.$2.5\times 10^{-3}$
   2. B.$2.5\times 10^{3}$
   3. C.$1.0\times 10^{-5}$
   4. D.$5\times 10^{-3}$
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   Answer: (C)

   The given reactions are: $2xy \rightleftharpoons x_2 + y_2 \quad K_1 = 2.5 \times 10^5$ $xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad K_2 = 5 \times 10^{-3}$ We need to find the equilibrium constant $K_3$ for the reaction: $\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$ Reversing the first reaction and multiplying it by $\dfrac{1}{2}$, we get: $\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 \rightleftharpoons xy$ The equilibrium constant for this modified reaction is: $K' = \left(\dfrac{1}{K_1}\right)^{\dfrac{1}{2}} = \dfrac{1}{\sqrt{K_1}}$ Adding this modified reaction to the second reaction: $\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 \rightleftharpoons xy \quad (K')$ $xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad (K_2)$ --------------------------------------------------- $\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$ The equilibrium constant $K_3$ for the overall reaction is the product of the equilibrium constants of the added reactions: $K_3 = K' \times K_2 = \dfrac{K_2}{\sqrt{K_1}}$ Substituting the given values: $K_3 = \dfrac{5 \times 10^{-3}}{\sqrt{2.5 \times 10^5}}$ $K_3 = \dfrac{5 \times 10^{-3}}{\sqrt{25 \times 10^4}}$ $K_3 = \dfrac{5 \times 10^{-3}}{5 \times 10^2}$ $K_3 = 1.0 \times 10^{-5}$ Answer: $1.0\times 10^{-5}$
4. Q4JEE Main 2026 (06 Apr, Shift 2)Law of Mass Action, Equilibrium Constant (Kc and Kp) and its Application
   In a closed flask at $600$ K, one mole of X$_2$Y$_4$(g) attains equilibrium as given below : $\text{X}_2\text{Y}_4(g) \rightleftharpoons 2\text{XY}_2(g)$ At equilibrium, $75\%$ X$_2$Y$_4$(g) was dissociated and the total pressure is $1$ atm. The magnitude of $\Delta_r G^{\ominus}$ (in kJ mol$^{-1}$) at this temperature is __________. (Nearest Integer) (Given : R $= 8.3$ J mol$^{-1}$ K$^{-1}$; $\ln 10 = 2.3$, $\log 2 = 0.3$, $\log 3 = 0.48$, $\log 5 = 0.69$, $\log 7 = 0.84$)
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   Answer: 8

   The given equilibrium reaction is: $\text{X}_2\text{Y}_4(g) \rightleftharpoons 2\text{XY}_2(g)$ Initial moles: $1$ Degree of dissociation, $\alpha = 0.75$ Moles of $\text{X}_2\text{Y}_4$ at equilibrium $= 1 - \alpha = 1 - 0.75 = 0.25$ Moles of $\text{XY}_2$ at equilibrium $= 2\alpha = 2 \times 0.75 = 1.5$ Total moles at equilibrium $= 0.25 + 1.5 = 1.75$ Given total pressure, $P = 1$ atm. Partial pressure of $\text{X}_2\text{Y}_4$, $P_{\text{X}_2\text{Y}_4} = \dfrac{0.25}{1.75} \times 1 = \dfrac{1}{7}$ atm Partial pressure of $\text{XY}_2$, $P_{\text{XY}_2} = \dfrac{1.5}{1.75} \times 1 = \dfrac{6}{7}$ atm The equilibrium constant $K_p$ is given by: $K_p = \dfrac{(P_{\text{XY}_2})^2}{P_{\text{X}_2\text{Y}_4}} = \dfrac{(\dfrac{6}{7})^2}{\dfrac{1}{7}} = \dfrac{36}{7}$ The standard Gibbs free energy change is: $\Delta_r G^{\ominus} = -RT \ln K_p = -2.3 RT \log K_p$ Calculating $\log K_p$: $\log \left(\dfrac{36}{7}\right) = \log 36 - \log 7 = \log (2^2 \times 3^2) - \log 7$ $\log \left(\dfrac{36}{7}\right) = 2 \log 2 + 2 \log 3 - \log 7$ $\log \left(\dfrac{36}{7}\right) = 2(0.3) + 2(0.48) - 0.84 = 0.6 + 0.96 - 0.84 = 0.72$ Substituting the values into the $\Delta_r G^{\ominus}$ equation: $\Delta_r G^{\ominus} = -2.3 \times 8.3 \times 600 \times 0.72$ $\Delta_r G^{\ominus} = -8246.88 \text{ J mol}^{-1} = -8.24688 \text{ kJ mol}^{-1}$ The magnitude of $\Delta_r G^{\ominus}$ is $8.24688 \text{ kJ mol}^{-1}$. Rounding to the nearest integer, we get $8$. Answer: $8$
5. Q5JEE Main 2026 (06 Apr, Shift 1)Law of Mass Action, Equilibrium Constant (Kc and Kp) and its Application
   One mole each of He and $A(g)$ are taken in a $10$ L closed flask and heated to $400$ K to establish the following equilibrium. $A(g) \rightleftharpoons B(g)$. $K_c$ for this reaction at $400$ K is $4.0$. The partial pressures (in atm) of He and $B(g)$ are respectively (at equilibrium) (Assume He, $A(g)$ and $B(g)$ behave as ideal gases) (Given: $R = 0.082$ L atm K$^{-1}$ mol$^{-1}$)
   1. A.$3.28, 2.624$
   2. B.$2.624, 3.28$
   3. C.$3.28, 0.656$
   4. D.$0.656, 6.56$
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   Answer: (A)

   The given reaction is $A(g) \rightleftharpoons B(g)$. Let the initial moles of $A$ be $1$ and at equilibrium, let $x$ moles of $A$ dissociate. Moles at equilibrium: $n_A = 1 - x$ $n_B = x$ Since $\Delta n_g = 0$, the equilibrium constant $K_c$ can be written in terms of moles: $K_c = \dfrac{n_B}{n_A} = \dfrac{x}{1 - x} = 4.0$ $x = 4 - 4x \Rightarrow 5x = 4 \Rightarrow x = 0.8$ At equilibrium, the moles of the gases are: $n_{He} = 1$ $n_B = 0.8$ Using the ideal gas equation $P = \dfrac{nRT}{V}$, the partial pressures are calculated as follows: Partial pressure of He: $P_{He} = \dfrac{1 \times 0.082 \times 400}{10} = 3.28$ atm Partial pressure of $B(g)$: $P_B = \dfrac{0.8 \times 0.082 \times 400}{10} = 2.624$ atm Answer: $3.28, 2.624$