Question

In a chemical equilibrium ${A}+{B} \rightleftharpoons {C}+{D}$ ,when one mole of each of the two reactants are mixed, 0.6 mole of each product are formed. The equilibrium constant is calculated as
(A) 1
(B) 0.303
(C) 2.25
(D) 4/9

Answer 1

Hint:Since we are asked to calculate the equilibrium constants, make sure that you use the correct values of concentration of the reactants. Whenever the reactant is one mole in amount, you can treat every amount of mole in the problem as mole fraction and solve.

Formula used:${K_{eq}} = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$ , where [C] and [D] represent the molar concentration of products and [A] and [B] represent the molar concentration of reactants. The exponents ‘a’, ‘b’, ‘c’, and ‘d’ represent the stoichiometric coefficients of the reactants and products in the reaction.

Complete step-by-step solution:Since the stoichiometric coefficients for every term in the reaction is 1, the values of ‘a’, ‘b’, ‘c’ and ‘d’ can be equated to one and the formula simplifies as
${K_{eq}} = \dfrac{{[C][D]}}{{[A][B]}}$ ………….(1)
Since the volume of reaction mixture is not mentioned, we can take it as 1 and solve the numerical.
At the beginning of the reaction, the amount of both reactants A and B was 1 mole. So we can easily treat the amount of reactants and products given in moles as mole fraction.
Upon reaction and reaching equilibrium, the amount of product formed was 0.6 moles. Treating 0.6 mole as mole fraction, it would simply mean that upon completion of reaction, 0.4 moles (1 – 0.6) of reactants were left unreacted.
Putting [C] = [D] = 0.6 moles and [A] = [B] = 0.4 moles in equation (1), we get:
${K_{eq}} = \dfrac{{0.6 \times 0.6}}{{0.4 \times 0.4}}$
$\Rightarrow {K_{eq}} = \dfrac{{{{0.6}^2}}}{{{{0.4}^2}}}$
$\Rightarrow {K_{eq}} = \dfrac{{0.36}}{{0.16}}$
$\Rightarrow {K_{eq}} = \dfrac{{{{0.6}^2}}}{{{{0.4}^2}}}$
$\therefore {K_{eq}} = 2.25$

Hence, option (C) is the correct answer.

Note:Whenever you come across any such question, make sure that you make use of the standard formula mentioned above. If a question is specifying that the reactants and products have some specific state, such as gaseous, the equation changes slightly, but it is derived from the equation mentioned above. And since the concentrations are molar, if the volume is not given, you can consider unity and solve the numerical.