Chemistry - Chemical Equilibrium Question with Solution | TestHub

$A(g)+2B(g)\rightleftharpoons C(g)$

$[A{]}_{eq}=3M$ $[B{]}_{eq}=4M$.

Let $[C{]}_{eq}=xM$.

$K_c=\frac{[C{]}_{eq}}{[A{]}_{eq}[B{]}_{eq}^2}=\frac{x}{3\times (4{)}^2}=\frac{x}{48}$

When the volume of the vessel is doubled, the reaction moves in the backward direction. Let the decrease in the concentration of C be $yM$.

At new equilibrium:

$A(g)+2B(g)\rightleftharpoons C(g)$

$(1.5+y)(2+2y)(\frac{x}{2}-y)$

Given that $2+2y=3$, $y=\frac{1}{2}=0.5$.

$K_c=\frac{\left(\frac{x-1}{2}\right)}{(1.5+0.5){\left(3\right)}^2}=\frac{x}{48}$

$x=4M$