Question

For the reaction $2A\left( g \right) \rightleftharpoons B\left( g \right) + 3C\left( g \right)$, at a given temperature, ${K_c} = 16$. What must be the volume of the flask, if a mixture of $2$mole each of A, B and C exist in equilibrium?
A.$\dfrac{1}{4}$
B.$\dfrac{1}{2}$
C.$1$
D.None of these

Answer 1

The chemical reaction is the representation of conversion of the products into reactants. The equilibrium constant is given in the question. The number of moles of each compound are also given. Thus, the volume of the flask can be determined by using the below formula.
Formula used:
${K_c} = \dfrac{{{n_B} \times {n_c}^3}}{{{n_A}^2}} \times \dfrac{1}{{{V^2}}}$
Where ${K_c}$ is equilibrium constant
${n_A}$ is number of moles of A
${n_B}$ is number of moles of B
${n_C}$ is number of moles of C
V is the volume of flask

Complete answer:
Given chemical reaction is the conversion of reactant a into the products B and C. The number of moles of each compound in the given reaction are $2$.
The equilibrium constant of the reaction is given in terms of concentrations as $16$.
The powers are the numbers before the concentration of the compounds.
By substituting the values in the above formula,
$16 = \dfrac{{2 \times {2^3}}}{{{2^2}}} \times \dfrac{1}{{{V^2}}}$
By simplification we will get
$V = \dfrac{1}{2}$
Thus, the volume of the flask will be $\dfrac{1}{2}$
Option B is the correct one.

Note:
The number of moles of each compound in the mixture are two, the volume of the flask must be calculated from equilibrium constant. The powers of the concentration terms of the reactants and products must be taken, but the products should be taken in numerator, and the reactants moles should be taken in denominator.