A reaction, A ( g ) + 2 B ( g ) ⇌ 2 C ( g ) + D ( g ) was studied using an initial concentration of B which was 1.5 times that of A . But the equilibrium concentrations of A and B were found to be equ

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A reaction, A ( g ) + 2 B ( g ) ⇌ 2 C ( g ) + D ( g ) was studied using an initial concentration of B which was 1.5 times that of A . But the equilibrium concentrations of A and B were found to be equal. The value of K p ​ for the equilibrium is

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A(g) + 2B(g) ⇌ 2C(g) + D(g) Initial concentrations: x 1.5 x 0 0 At equilibrium: x − y 1.5 x − 2 y 2 y y Given: x − y = 1.5 x − 2 y . Solving for y : y = 1.5 x − x y = 0.5 x Thus, [A] = x − 0.5 x = 0.5 x [B] = 1.5 x − 2 ( 0.5 x ) = 1.5 x − x = 0.5 x [C] = 2 ( 0.5 x ) = x [D] = 0.5 x K c ​ = K P ​ = ( 0.5 x ) ( 0.5 x ) 2 ( x ) 2 ( 0.5 x ) ​ = ( 0.5 x ) ( 0.25 x 2 ) x 2 ( 0.5 x ) ​ = 0.125 x 3 0.5 x 3 ​ = 4

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