Question

One mole of a compound AB reacts with one mole of compound CD according to the equation
${\text{AB(g) + CD(g) }} \rightleftharpoons {\text{ AD(g) + CB(g)}}$
When equilibrium has been established it was found that $\dfrac{3}{4}$ mole each of reactants AB and CD had converted to AD and CB. There is no change in volume. The equilibrium constant for the reaction is:
A) $\dfrac{6}{{15}}$
B) $\dfrac{1}{9}$
C) $\dfrac{{19}}{5}$
D) 9

Answer 1

Using the initial concentrations of reactants and change in concentrations of reactant determine the equilibrium concentrations of reactants and product. The equilibrium constant is the ratio of the equilibrium concentration of products to the equilibrium concentration of the reactants raised to the power of its coefficient.

Complete step by step answer:
The state of a reversible reaction in which the concentration of the reactants and products do not change with time is known as a chemical equilibrium.
A chemical reaction is given to us:
${\text{AB(g) + CD(g) }} \rightleftharpoons {\text{ AD(g) + CB(g)}}$
We have given one mole of a compound AB reacts with one mole of compound CD.
So,
Initial moles of AB = 1 mole
Initial moles of CD = 1 mole
We have also given that when equilibrium has been established it was found that $\dfrac{3}{4}$ mole each of reactants AB and CD had converted to AD and CB.
Now, using the initial moles of reactants and reacted moles of reactant we can calculate the unreacted moles reactants at equilibrium.
Unreacted moles = Initial moles – Reacted moles
So, Unreacted moles of AB = (1-$\dfrac{3}{4}$) mole = $\dfrac{1}{4}$moles
Unreacted moles of CD = (1-$\dfrac{3}{4}$) mole = $\dfrac{1}{4}$moles
Thus,
Equilibrium moles of AB = $\dfrac{1}{4}$moles
Equilibrium moles of CD = $\dfrac{1}{4}$moles
Now, from the reaction, we can say that the mole ratio of AB: CD: AD: CB is 1:1:1:1
So, moles of product AD at equilibrium = $\dfrac{3}{4}$moles
Moles of product CB at equilibrium = $\dfrac{3}{4}$moles
Now, let us assume the volume of the solution is V liter.
So,
${\text{[AB] = }}\dfrac{{{\text{1/4 moles }}}}{{{\text{VL}}}}$
${\text{[CD] = }}\dfrac{{{\text{1/4 moles }}}}{{{\text{VL}}}}$

${\text{[AD] = }}\dfrac{{{\text{3/4 moles }}}}{{{\text{VL}}}}$
${\text{[CB] = }}\dfrac{{{\text{3/4 moles }}}}{{{\text{VL}}}}$
Now we have equilibrium concentrations of all reactants and products so we can calculate the equilibrium constant for the reaction as follows:
As equilibrium constant is the ratio of the equilibrium concentration of products to the equilibrium concentration of the reactants raised to the power of its coefficient. So, the equilibrium constant equation for the given reaction is:
${\text{K = }}\dfrac{{{\text{[AD][CB]}}}}{{{\text{[AB][CD]}}}}$
Now, substitute $\dfrac{{{\text{3/4moles }}}}{{{\text{VL}}}}$for equilibrium concentration of each of the products and $\dfrac{{{\text{1/4moles }}}}{{{\text{VL}}}}$ for equilibrium concentration of each reactant and calculate the equilibrium constant.
${\text{K = }}\dfrac{{\left( {\dfrac{{{\text{3/4moles }}}}{{{\text{VL}}}}} \right)\left( {\dfrac{{{\text{3/4moles }}}}{{{\text{VL}}}}} \right)}}{{\left( {\dfrac{{{\text{1/4moles }}}}{{{\text{VL}}}}} \right)\left( {\dfrac{{{\text{1/4moles }}}}{{{\text{VL}}}}} \right)}}$
${\text{K = 9}}$
Thus, the equilibrium constant of the reaction is 9.

Hence, the correct option is (D) 9.

Note: Equilibrium concentration of reactant calculated using unreacted moles of reactants. While reacted moles of reactants used to calculate the equilibrium concentration of the product. The equilibrium constant is the ratio of an equilibrium constant of product to reactant raise to the power of coefficient. So use the correct balance reaction.