JEE Main 2026 — Chemistry Physical Chemistry

The given reactions are:

$$2xy \rightleftharpoons x_2 + y_2 \quad K_1 = 2.5 \times 10^5$$2xy⇌x2​+y2​K1​=2.5×105

$$xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad K_2 = 5 \times 10^{-3}$$xy+21​z2​⇌xyzK2​=5×10−3

We need to find the equilibrium constant $K_3$K3​ for the reaction:

$$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$$21​x2​+21​y2​+21​z2​⇌xyz

Reversing the first reaction and multiplying it by $\dfrac{1}{2}$21​, we get:

$$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 \rightleftharpoons xy$$21​x2​+21​y2​⇌xy

The equilibrium constant for this modified reaction is:

$$K' = \left(\dfrac{1}{K_1}\right)^{\dfrac{1}{2}} = \dfrac{1}{\sqrt{K_1}}$$K′=(K1​1​)21​=K1​​1​

Adding this modified reaction to the second reaction:

$$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 \rightleftharpoons xy \quad (K')$$21​x2​+21​y2​⇌xy(K′)

$$xy + \dfrac{1}{2}z_2 \rightleftharpoons xyz \quad (K_2)$$xy+21​z2​⇌xyz(K2​)

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$$\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons xyz$$21​x2​+21​y2​+21​z2​⇌xyz

The equilibrium constant $K_3$K3​ for the overall reaction is the product of the equilibrium constants of the added reactions:

$$K_3 = K' \times K_2 = \dfrac{K_2}{\sqrt{K_1}}$$K3​=K′×K2​=K1​​K2​​

Substituting the given values:

$$K_3 = \dfrac{5 \times 10^{-3}}{\sqrt{2.5 \times 10^5}}$$K3​=2.5×105​5×10−3​

$$K_3 = \dfrac{5 \times 10^{-3}}{\sqrt{25 \times 10^4}}$$K3​=25×104​5×10−3​

$$K_3 = \dfrac{5 \times 10^{-3}}{5 \times 10^2}$$K3​=5×1025×10−3​

$$K_3 = 1.0 \times 10^{-5}$$K3​=1.0×10−5

Answer: $1.0\times 10^{-5}$1.0×10−5