Question

$8$ mole of a gas ${AB}_3$ are introduced into a $1.0{dm}^3$ vessel. It dissociates as:
$2A{B}_3\left(g\right)\rightleftharpoons A_2\left(g\right)+3B_2\left(g\right)$
at equilibrium, $2$ mole of $A_2$ are found to be present. The equilibrium constant of the reaction is:

• A. $2{mol}^2{L}^{-2}$
• B. $3{mol}^2{L}^{-2}$
• C. $27{mol}^2{L}^{-2}$
• D. $36{mol}^2{L}^{-2}$

Answer 1

The correct option is C $27{mol}^2{L}^{-2}$

$8$ mole of a gas ${AB}_3$ are introduced into a $1.0{dm}^3$ vessel. It dissociates as:
$2A{B}_3\left(g\right)\rightleftharpoons A_2\left(g\right)+3B_2\left(g\right)$
At equilibrium, $2$ mole of $A_2$ are found to be present. The equilibrium constant of the reaction is $27{mol}^2{L}^{-2}$.
Since, total volume is $1.0{dm}^3$ , the number of moles is equal to molar concentration.
As per the reaction stoichiometry, 2 moles of $A{B}_3$ dissociate to form 1 mole of $A_2$ and 3 moles of $B_2$.
At equilibrium, $2$ mole of $A_2$ are found to be present. They will be formed by dissociation of 4 moles of $A{B}_3$
4 moles of $A{B}_3$ dissociate to form 2 mole of $A_2$ and 6 moles of $B_2$.
Out of $8$ moles of $A{B}_3$, $4$ moles dissociate and $8-4=4$ moles remains.
The equilibrium constant of the reaction is $K=\frac{\left[A2\right][B2{]}^3}{[AB3{]}^2}$
$K=\frac{\left[2\right][6{]}^3}{[4{]}^2}$
$K=27{mol}^2{L}^{-2}$