Question

The reaction, $2{A_{(g)}} + {B_{(g)}} \rightleftharpoons 3{C_{(g)}} + {D_{(g)}}$ is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached the concentration of D is measured and found to be 0.25 M.
What is the value of the equilibrium constant for the given reaction?
(A) $\dfrac{{{{\left( {0.75} \right)}^3}\left( {0.25} \right)}}{{{{\left( {1.0} \right)}^2}\left( {1.0} \right)}}$
(B) $\dfrac{{{{\left( {0.75} \right)}^3}\left( {0.25} \right)}}{{{{\left( {0.5} \right)}^2}\left( {0.75} \right)}}$
(C) $\dfrac{{{{\left( {0.75} \right)}^3}\left( {0.25} \right)}}{{{{\left( {0.5} \right)}^2}\left( {0.25} \right)}}$
(D) $\dfrac{{{{\left( {0.75} \right)}^3}\left( {0.25} \right)}}{{{{\left( {0.75} \right)}^2}\left( {0.25} \right)}}$

Answer 1

Hint: The equilibrium constant (${K_C}$) is the ratio of molar concentration of products to reactants at equilibrium concentrations raised to the power of their stoichiometric coefficients.

Complete step by step answer:
-We will first begin by seeing what ${K_C}$ is.
${K_C}$ is basically an equilibrium constant which represents the ratio of the equilibrium concentrations of products to the equilibrium concentrations of reactants, all raised to the power of their stoichiometric coefficients.
For a reaction: $aA \rightleftharpoons bB$
The value of ${K_C}$ for the above reaction will be:
                      ${K_C} = \dfrac{{{{\left[ B \right]}^b}}}{{{{\left[ A \right]}^a}}}$
-The question gives us the following reaction:
                    $2{A_{(g)}} + {B_{(g)}} \rightleftharpoons 3{C_{(g)}} + {D_{(g)}}$

According to the question the initial concentration of A and B is 1.00 M and the equilibrium concentration of D is 0.25 M. Since from the reaction we can see that the coefficient of D is 1, then the dissociation constant of the reaction would be 0.25. So, we will now see the equilibrium concentration of A, B, C and D.

For A: Initial concentration = 1.00 M and its coefficient is 2, so its equilibrium concentration will be = $\left[ {1 - \left( {2 \times 0.25} \right)} \right]$
            = $\left[ {1 - 0.5} \right]$
           = 0.5 M

For B: Initial concentration = 1.00 M and its coefficient is 1, so its equilibrium concentration will be = $\left[ {1 - 0.25} \right]$
            = 0.75 M

 For C: It is a product and its coefficient is 3, so its equilibrium concentration will be:
            = 3 × 0.25
            = 0.75 M

For D: Its equilibrium concentration is given in the reaction to be 0.25 M
-We will now write the expression of equilibrium constant (${K_C}$) for the given reaction:
${K_C} = \dfrac{{{{\left[ C \right]}^3}\left[ D \right]}}{{{{\left[ A \right]}^2}\left[ B \right]}}$
We will now put the values of equilibrium concentrations of A, B, C and D in the above given equation:
${K_C} = \dfrac{{{{\left[ {0.75} \right]}^3}\left[ {0.25} \right]}}{{{{\left[ {0.5} \right]}^2}\left[ {0.75} \right]}}$
Finally we have obtained the equilibrium constant (${K_C}$) for the given reaction.
So, the correct answer is “Option B”.

Note: Equilibrium constant for any reaction depends on temperature and is independent of the actual quantities of reactants and products, the presence of catalyst and the presence of any inert material. It also does not depend on the concentrations, pressures and volumes of the reactants and products.