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(C) Let the initial concentration of $X$ be $c$. Then,the initial concentration of $W$ is $2c$. Let the concentration of $Y$ at equilibrium be $4x$. S

Vedclass · Accepted Answer

$2.666$

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(C) Let the initial concentration of $X$ be $c$. Then,the initial concentration of $W$ is $2c$. Let the concentration of $Y$ at equilibrium be $4x$. Since the stoichiometry of the reaction $W + X \rightleftharpoons Y + Z$ is $1:1:1:1$,the concentration of $Z$ at equilibrium is also $4x$. The amount of $W$ and $X$ reacted is $4x$. Equilibrium concentrations are: $[W] = 2c - 4x$,$[X] = c - 4x$,$[Y] = 4x$,$[Z] = 4x$. Given that at equilibrium $[Y] = 4[X]$,we have $4x = 4(c - 4x)$,which simplifies to $x = c - 4x$,so $c = 5x$. Substituting $c = 5x$ into the equilibrium concentrations: $[W] = 2(5x) - 4x = 6x$,$[X] = 5x - 4x = x$,$[Y] = 4x$,$[Z] = 4x$. The equilibrium constant $K_c$ is given by $K_c = \frac{[Y][Z]}{[W][X]} = \frac{(4x)(4x)}{(6x)(x)} = \frac{16x^2}{6x^2} = \frac{16}{6} = 2.666$.

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