Chemistry - Chemical Equilibrium Question with Solution | TestHub

Reaction: $A+2B\rightleftharpoons 2C+D$

Initial:

$[A{]}_0=a$

$[B{]}_0=1.5a$

$[C{]}_0=0$

$[D{]}_0=0$

Change:

$[A]=-x$

$[B]=-2x$

$[C]=+2x$

$[D]=+x$

Equilibrium:

$[A{]}_{eq}=a-x$

$[B{]}_{eq}=1.5a-2x$

$[C{]}_{eq}=2x$

$[D{]}_{eq}=x$

Given $[A{]}_{eq}=[D{]}_{eq}$:

$a-x=x⟹a=2x⟹x=0.5a$

Substitute $x$ into equilibrium concentrations:

$[A{]}_{eq}=a-0.5a=0.5a$

$[B{]}_{eq}=1.5a-2(0.5a)=1.5a-a=0.5a$

$[C{]}_{eq}=2(0.5a)=a$

$[D{]}_{eq}=0.5a$

Equilibrium constant $K_C$:

$$K_C=\frac{[C{]}_{eq}^2[D{]}_{eq}}{[A{]}_{eq}[B{]}_{eq}^2}$$

$$K_C=\frac{(a{)}^2(0.5a)}{(0.5a)(0.5a{)}^2}=\frac{a^2(0.5a)}{(0.5a{)}^3}=\frac{a^2(0.5a)}{0.125a^3}=\frac{0.5a^3}{0.125a^3}=4$$