Question

How do you determine the equilibrium constant $ {K_c} $ for the reaction $ X + 2Y \rightleftharpoons Z $ ? The equilibrium concentrations are $ 0.216M $ for $ Z $ , $ 0.06M $ for $ X $ , and $ 0.12M $ for $ Y $ .

Answer 1

Hint: Equilibrium constant can be defined as the ratio of concentrations of products to the concentration of reactant. It was represented by $ {K_c} $ . It can be determined from the concentration of the product to the square of the concentration of $ Y $ and the concentration of $ X $ .
 $ {K_c} = \dfrac{{\left[ Z \right]}}{{\left[ X \right]{{\left[ Y \right]}^2}}} $
 $ {K_c} $ is equilibrium constant
 $ \left[ Z \right] $ is the concentration of product Z
 $ \left[ X \right] $ is the concentration of reactant X
 $ \left[ Y \right] $ is the concentration of reactant Y.

Complete answer:
Given reaction is $ X + 2Y \rightleftharpoons Z $ in which the reactants $ X $ and $ Y $ react with each other and form a product $ Z $ . The number of moles of Y taken was two. The given reaction is at equilibrium.
Thus, while writing the equilibrium constant, the concentration of the reactant $ Y $ must be taken as $ {\left[ Y \right]^2} $ where the concentration of the reactant $ X $ will be $ \left[ X \right] $ only, as only one mole of reactant $ X $ is involved in the equilibrium reaction.
Given that $ \left[ Z \right] = 0.216M $ , $ \left[ X \right] = 0.06M $ , and $ \left[ Y \right] = 0.12M $
Substitute the values of the concentrations of the both reactants and product in the formula,
 $ {K_c} = \dfrac{{\left[ {0.216} \right]}}{{\left[ {0.06} \right]{{\left[ {0.12} \right]}^2}}} $
By simplifying the above values,
 $ {K_c} = 250 $
Thus, the equilibrium constant $ {K_c} $ for the reaction $ X + 2Y \rightleftharpoons Z $ where the equilibrium concentrations are $ 0.216M $ for $ Z $ , $ 0.06M $ for $ X $ , and $ 0.12M $ for $ Y $ is $ 250 $ .

Note:
The equilibrium concentration of the reactants and products must be written by considering the moles of reactants and products in the balanced equation only. For the reactions that were at equilibrium (which was represented by double heads half arrow) only, the equilibrium constant can be determined.