In the reaction A + 2B rightleftharpoons 2C + | vedclass

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(A) The reaction is $A + 2B \rightleftharpoons 2C + D$.Let the initial concentration of $A$ be $x \ M$ and $B$ be $1.5x \ M$.At equilibrium,let $a$ mo

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$4$

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(A) The reaction is $A + 2B ightleftharpoons 2C + D$.Let the initial concentration of $A$ be $x \ M$ and $B$ be $1.5x \ M$.At equilibrium,let $a$ moles of $A$ react.Initial: $[A] = x, [B] = 1.5x, [C] = 0, [D] = 0$.Equilibrium: $[A] = x-a, [B] = 1.5x-2a, [C] = 2a, [D] = a$.Given that at equilibrium,$[A] = [B]$,so $x-a = 1.5x-2a$.Solving for $a$: $a = 0.5x$.Substituting $a$ into equilibrium concentrations: $[A] = x-0.5x = 0.5x$,$[B] = 1.5x-2(0.5x) = 0.5x$,$[C] = 2(0.5x) = x$,$[D] = 0.5x$.The equilibrium constant $K_c = \frac{[C]^2 [D]}{[A] [B]^2} = \frac{(x)^2 (0.5x)}{(0.5x) (0.5x)^2}$.$K_c = \frac{x^2 	imes 0.5x}{0.5x 	imes 0.25x^2} = \frac{0.5x^3}{0.125x^3} = 4$.

In the reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $B$ was $1.5$ times that of $[A]$. At equilibrium,the concentrations of $A$ and $B$ became equal. The equilibrium constant $K_c$ for the reaction is:

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