JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

The given reactions are: 2xy x_2 + y_2 K_1 = 2.5 10^5 xy + 1 2 z_2 xyz K_2 = 5 10^ -3 We need to find the equilibrium constant K_3 for the reaction: 1 2 x_2 + 1 2 y_2 + 1 2 z_2 xyz Reversing the first reaction and multiplying it by 1 2 , we get: 1 2 x_2 + 1 2 y_2 xy The equilibrium constant for this modified reaction is: K' = ( 1 K_1 )^ 1 2 = 1 K_1 Adding this modified reaction to the second reaction: 1 2 x_2 + 1 2 y_2 xy (K') xy + 1 2 z_2 xyz (K_2) --------------------------------------------------- 1 2 x_2 + 1 2 y_2 + 1 2 z_2 xyz The equilibrium constant K_3 for the overall reaction is the product of the equilibrium constants of the added reactions: K_3 = K' K_2 = K_2 K_1 Substituting the given values: K_3 = 5 10^ -3 2.5 10^5 K_3 = 5 10^ -3 25 10^4 K_3 = 5 10^ -3 5 10^2 K_3 = 1.0 10^ -5 Answer: 1.0 10^ -5