Question

Given 8 moles of a gas $A{{B}_{3}}$ attain equilibrium in a closed container of volume $1d{{m}^{3}}$ as $2A{{B}_{3}}\rightleftharpoons {{A}_{2}}(g)+3{{B}_{2}}(g)$. If at equilibrium, 2 moles of ${{A}_{2}}$ is present, then the equilibrium constant is:
A. $72mo{{l}^{2}}{{L}^{-2}}$
B. $36mo{{l}^{2}}{{L}^{-2}}$
C. $3mo{{l}^{2}}{{L}^{-2}}$
D. $27mo{{l}^{2}}{{L}^{-2}}$

Answer 1

Hint: The concept of equilibrium and equilibrium constant is to be used in this question. The degree of dissociation needs to be found by using the formula of equilibrium constant, by substituting the values given in the question.

Complete answer:
In order to understand the question, we need to learn about equilibrium. Based upon the experimental results of a number of workers, the Norwegian scientists Guldberg and Waage in 1864 gave a general statement about the influence of masses of reactants on the reaction rates, known as the law of mass action. It states that "the rate at which a substance reacts is proportional to its active mass and the rate of a chemical reaction is proportional to the product of the active masses of the reacting substances". Active mass means molar concentration i.e., the number of gram moles of the solute per unit volume. It is expressed by enclosing the formula of the substance in square brackets. For example, [A] represents the active mass of the substance A. The active mass of solids is taken as a unit. Let us consider a simple reaction:$[A]\to products$.So,$rate\,\alpha \,[A]\,\,,\,or,\,rate=k[A]$. K is the rate constant. For a chemical reaction$A+B\rightleftharpoons C+D$the equilibrium constant is represented as:
     \[{{K}_{C}}=\dfrac{[C][D]}{[A][B]}\]
Let us assume $\alpha $as the degree of dissociation. Now, from the reaction, we get:
     \[\begin{align}
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2A{{B}_{3}}(g)\rightleftharpoons {{A}_{2}}(g)+3{{B}_{2}}(g) \\
 & t=0\,\,\,\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
 & t={{t}_{eq}}\,\,\,\,\,8-2\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,3\alpha \\
\end{align}\]
Now, the expression for equilibrium constant will be:
\[{{K}_{c}}=\dfrac{[{{A}_{2}}]{{[{{B}_{2}}]}^{3}}}{{{[A{{B}_{3}}]}^{2}}}\]
Since it is given that $[{{A}_{2}}]$ is 2 moles so $\alpha =2$ So, the concentration of $A{{B}_{3}}$ and ${{B}_{2}}$ are $8-2\alpha $ and $3\alpha $ i.e 4 and 6 respectively.. Now, by substituting in the equilibrium constant expression, we get:
\[{{K}_{C}}=\dfrac{2\times {{6}^{3}}}{{{4}^{2}}}=27mo{{l}^{2}}{{L}^{-2}}\]

Hence, we obtain our correct answer as option D.

Note:
In a reversible reaction, at equilibrium the product of the molar concentration of products, each raised to the power equal to its coefficient, and is then divided by the product of the molar concentration of the reactant each raised to the power equal to its coefficient, is constant at a constant temperature and is called equilibrium constant.