Question

What is the mass action equation of $2{{{A}}_{\left( {{{aq}}} \right)}} + {{{B}}_{\left( {{{aq}}} \right)}} \rightleftharpoons 3{{{C}}_{\left( {{{aq}}} \right)}} + {{{D}}_{\left( {{s}} \right)}}$?
(A) ${{{K}}_{{{eq}}}}{{ = }}\dfrac{{{{{{[C]}}}^{{3}}}}}{{{{{{[A]}}}^{{2}}}{{ [B]}}}}$
(B) ${{{K}}_{{{eq}}}}{{ = }}\dfrac{{{{{{[C]}}}^{{3}}}{{ + [D]}}}}{{{{{{[A]}}}^{{2}}}{{[B]}}}}$
(C) ${{{K}}_{{{eq}}}}{{ = }}\dfrac{{{{{{[A]}}}^{{2}}}{{ + [B]}}}}{{{{{{[C]}}}^{{3}}}}}$
(D) ${{{K}}_{{{eq}}}}{{ = }}\dfrac{{{{{{[A]}}}^{{2}}}{{ - [B]}}}}{{{{{{[C]}}}^{{3}}}{{ - [D]}}}}$
(E) ${{{K}}_{{{eq}}}}{{ = }}\dfrac{{{{[C]}}}}{{{{[A] + [V]}}}}$

Answer 1

Hint: As we know that law of mass action defines that the rate of the reaction is directly related to the reactant concentration. We can determine the behavior of solutions with the help of law of mass action. When a reaction is at equilibrium, the ratio of concentration of reactants to that of products is a constant.

Complete step by step answer:
The law of mass action is in force at every reaction condition because it is universal law.
We take a general equation to understand it
${{aA}} + {{bB}} \rightleftharpoons {{cC}} + {{dD}}$
For this equation in which a mole of A reacts with b moles of B forms c moles of C and d moles of D and a, b, c, d is the coefficient of the above balanced reaction. For this equation the law of mass action says that when the system is in equilibrium and at a constant temperature the ratio of product to reactant is always constant. This constant value is called equilibrium constant ${{{K}}_{{{eq}}}}$.
${{{K}}_{{{eq}}}}{{ = }}\dfrac{{{{{{[C]}}}^{{c}}}{{ [D}}{{{]}}^{{d}}}}}{{{{{{[A]}}}^{{a}}}{{ [B}}{{{]}}^{{b}}}}}$
The products are in the numerator position and reactants at the denominator side and the coefficient of the reactants and product are raised to the power of their concentration. The concentration here is the concentration present at the time of equilibrium not the initial or final concentration.
For the equation in question the mass action equation will be
${{{K}}_{{{eq}}}}{{ = }}\dfrac{{{{{{[C]}}}^{{3}}}}}{{{{{{[A]}}}^{{2}}}{{ [B]}}}}$

So, the correct answer is Option A.

Note: The equilibrium condition is the middle condition of the reaction. The reaction can go in forward direction or backward direction also depending on the conditions. These conditions are determined by Le-Chatelier's principle.
The concentration here used is the concentration present during the equilibrium conditions. The concentration entity is enclosed in square brackets. The concentration of a solid reactant or product is taken as ${{1}}$ always that is why it is not included in the expression.