Question

x mole of A are mixed with x moles of B, and at equilibrium for the reaction
$\mathop {\text{A}}\limits_{aq} + \mathop {\text{B}}\limits_{aq} \rightleftharpoons \mathop {\text{C}}\limits_{aq} + \mathop {\text{D}}\limits_{aq}$
$\dfrac{{\text{x}}}{2}$ moles of C are formed. The equilibrium constant ${{\text{K}}_{\text{c}}}$ for the reaction will be:

Answer 1

The moles of reactant consumed and number of moles of product formed can be calculated using the stoichiometric coefficient and these concentration can be used to find equilibrium constant or ${{\text{K}}_{\text{c}}}$.

Complete step by step solution:
In reversible reactions when equilibrium is reached the rate of forward reaction becomes equal to rate of backward reaction. That is the rate at which reactant converting into product become equals to the rate at which product converting back to react. The equilibrium constant is defined as the product or multiplication of concentration of the product raised to the power to their respective stoichiometric coefficient to the product of concentration of reactant raise to the power of their stoichiometric coefficient. From the given reaction it is clear that one mole of reacted A and 1 mole of the reactant B reacts to form one mole of the product C and D each.
It is given to us that x moles of reactant A and B reacts. Now since the concentration of product formed is $\dfrac{{\text{x}}}{2}$ which is given to us. For each mole of A and B reacts to form one mole of C and D. Similarly to form $\dfrac{{\text{x}}}{2}$ moles of C , $\dfrac{{\text{x}}}{2}$ moles of A and B will be consumed. Hence the reaction at equilibrium will be:
${\text{time }}\mathop {{\text{A }}}\limits_{aq} {\text{ }} + {\text{ }}\mathop {\text{B}}\limits_{aq} {\text{ }} \rightleftharpoons {\text{ }}\mathop {\text{C}}\limits_{aq} {\text{ }} + {\text{ }}\mathop {\text{D}}\limits_{aq}$
${\text{initially x }} + {\text{ x }} \rightleftharpoons {\text{ }}0{\text{ }} + {\text{ }}0$
${\text{equilibrium x}} - \dfrac{{\text{x}}}{2}{\text{ }} + {\text{ x}} - \dfrac{{\text{x}}}{2}{\text{ }} \rightleftharpoons {\text{ }}\dfrac{{\text{x}}}{2}{\text{ }} + {\text{ }}\dfrac{{\text{x}}}{2}$
Using the definition of equilibrium constant people find the value of ${{\text{K}}_{\text{c}}}$
${{\text{K}}_{\text{c}}} = \dfrac{{{{\left[ {\text{C}} \right]}^1}{{\left[ {\text{D}} \right]}^1}}}{{{{\left[ {\text{A}} \right]}^1}{{\left[ {\text{B}} \right]}^1}}}$
${{\text{K}}_{\text{c}}} = \dfrac{{\left[ {\dfrac{{\text{x}}}{2}} \right]\left[ {\dfrac{{\text{x}}}{2}} \right]}}{{\left[ {\dfrac{{\text{x}}}{2}} \right]\left[ {\dfrac{{\text{x}}}{2}} \right]}} = 1$

Hence, the value of equilibrium constant is 1.

Note:
The equilibrium constant is a fixed quantity for a particular reaction. The concept of equilibrium constant comes into existence when the reaction has reached equilibrium. We cannot use equilibrium constant in initial stages of the reaction. It is a function of temperature.