Question

For the reaction $A + 2B \rightleftharpoons 2C$at equilibrium $\left[ C \right] = 1.4{\text{M}}$. ${\left[ A \right]_0} = 1{\text{M}},{\left[ B \right]_0} = 2{\text{M}},{\left[ C \right]_0} = 3{\text{M}}$. The value of ${K_c}$is:
(Given: ${\text{volume}} = 1{\text{L}},{\left[ {} \right]_0}$is initial concentration)
A. $0.084$
B. $8.4$
C. 48
D. None of these

Answer 1

To answer this question, you must recall the formula of equilibrium constant of a reaction in terms of the concentration of reactants, that is ${K_c}$. It is given by the concentration at equilibrium of products divided by the concentration at equilibrium of reactants raised to the power of their respective stoichiometric coefficients.
Formula used:
${K_c} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}}$
Where ${K_c}$is equilibrium constant of the reaction
$\left[ C \right]$is the concentration of C at equilibrium
$\left[ B \right]$is the concentration of B at equilibrium
$\left[ A \right]$ is the concentration of A at equilibrium

Complete step by step answer:
For the given reaction, $A + 2B \rightleftharpoons 2C$,
We are given the initial concentrations ${\left[ A \right]_0} = 1{\text{M}},{\left[ B \right]_0} = 2{\text{M}},{\left[ C \right]_0} = 3{\text{M}}$.
Let the concentrations at equilibrium be,
$\left[ A \right] = \left( {1 - x} \right){\text{M }};\;\left[ B \right] = \left( {2 - 2x} \right){\text{M }};\;\left[ C \right] = \left( {3 + 2x} \right){\text{M}}$
We know that, $\left[ C \right] = 1.4{\text{M}}$
Thus, we can find the value of $x$as,
$\left[ C \right] = 1.4{\text{M}} = 3 + 2x$
$\Rightarrow x = - 0.8$
Using this value, we can find the values of concentrations of A and B at equilibrium as,
$\left[ A \right] = 1 - \left( { - 0.8} \right) = 1.8{\text{M}}$ and $\left[ B \right] = 2 - 2\left( { - 0.8} \right) = 3.6{\text{M}}$.
Substituting these values in the equilibrium constant formula, we get,
${K_c} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}}$
Substituting the values:
$\Rightarrow {K_c} = \dfrac{{{{\left( {3 + 2x} \right)}^2}}}{{\left( {1 - x} \right){{\left( {2 - 2x} \right)}^2}}} = \dfrac{{{{\left( {1.4} \right)}^2}}}{{\left( {1.8} \right){{\left( {3.6} \right)}^2}}}$
Solving this, we get:
${K_c} = 0.084$

Thus, the correct answer is A.

Note:
The numerical value of an equilibrium constant is obtained by allowing a reaction to proceed to equilibrium and then measuring the concentrations of each substance involved in that reaction. Since the concentrations are measured at equilibrium, the equilibrium constant remains the same for a given reaction independent of initial concentrations of the reactants and products. This knowledge allows us to derive a model expression that can serve as a standard for any reaction.
This basic standard form of an equilibrium constant is given as
${K_{eq}} = \dfrac{{{{\left( {{a_C}} \right)}^c}}}{{{{\left( {{a_A}} \right)}^a}{{\left( {{a_B}} \right)}^b}}}$ for a reaction: $aA + bB \rightleftharpoons cC$
If $K > 1$ then equilibrium favours the formation of products, reaction proceeds in the forward direction.
If $K < 1$ then equilibrium favours the formation of reactants, reaction proceeds in the backward direction.