Question

For reaction $A + 2B \rightleftharpoons 2C$
Initially 2 mole of A and B are taken in 10 lit flask at equilibrium 1 mole of C is formed then calculate ${K_c}$ for the given reaction:
A. $6.66$
B. $3.33$
C. $2.22$
D. $1.11$

Answer 1

We all know that an equilibrium in a chemical reaction is the state where the forward and the backward reactions are generally equal and reversible. The equilibrium constant denoted asis defined as the ratio of the concentration of products to the concentration of reactants each raised to their particular stoichiometric coefficient powers.

Complete step by step answer:
We know that the equilibrium constant is the ratio of product concentrations to the reactants concentrations each raise to their stoichiometric coefficients. Let us take an example and consider this equation-$aA + bB \rightleftharpoons cC + dD$, we can write the equilibrium constant of the given equation which can be calculated as:${K_c} = \dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$
In the same way we can write the rate law for the given equation-, ${K_c} = \dfrac{{{{[C]}^2}}}{{[A]{{[B]}^2}}}$
Now, to calculate the value of, first we need to understand the conditions at initial stage and at equilibrium, for this let us write it as:
$A\;\;\; + \;\;\;\;2B\;\;\;\; \rightleftharpoons 2C$

$0.1$	$0$	$0$
$0.1 - x$	$0$	$0$

And as it is given in the question that 1 mole of C is formed at equilibrium, so we can calculate the value of $x$ using it,$2x = 1 \Rightarrow x = 0.5$ similarly the moles of rest can be calculated which will be: $A = 1.5$ and ${\text{B = 1}}$.
Now, the concentrations at equilibrium can be written as well and they will be the ratio of amount over the volume and can be written as: $A = \dfrac{{1.5}}{{10}}$, $B = \dfrac{1}{{10}}$and ${\text{C = }}\dfrac{1}{{10}}$. Finally we can put all the values of A, B and C in the formula to calculate and we will get-
${K_c} = \dfrac{{{{[C]}^2}}}{{[A]{{[B]}^2}}} \\\ {\text{ = }}\dfrac{{{{[1/10]}^2}}}{{[1.5/10]{{[1/10]}^2}}} \\\ {\text{ = }}\dfrac{{100}}{{15}} \\\ {\text{ = 6}}{\text{.667}}$

**Therefore, our correct answer is option (A).

Note:**
When the temperature is increased the value of equilibrium constant decreases and the equilibrium is shifted to higher concentration so to bring the equilibrium state to the original position value of Kc decreases. Volume has no effect on the equilibrium constant and when pressure is applied then the reactions tend to shift in the direction of less number of moles. If concentration of a reactant is increased then also the equilibrium will move in forward direction.