In a chemical reaction A + 2B K 2C + D, the initial concentration… - Vidyadip

MCQ

In a chemical reaction $A + 2B \overset K \rightleftharpoons 2C + D$, the initial concentration of $B$ was $1.5$ times of $A$ but the equilibrium concentrations of $A$ and $B$ were found to be equal. The equilibrium, constant $(K)$ for the aforesaid chemical reaction is

✓
$4$
B
$16$
C
$0.25$
D
$1$

✓

Answer

Correct option: A.

$4$

a
$\begin{gathered} \quad \quad \quad \quad \quad \quad \quad \quad A\,\,\, + \,\,\,\,\,2B\,\,\,\,\, \rightleftharpoons \,\,\,\,2C\,\,\, + \,\,\,D \hfill \\ Initially\,\,conc.\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,1.5a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ at\,\,equllibrium\,\,\,\,a - x\,\,1.5a - 2x\,\,\,\,\,2x\,\,\,\,\,\,\,\,\,\,x \hfill \\ \end{gathered} $

$0.5a=x$

$a=2x$

${k_c} = \frac{{{{(2x)}^2}x}}{{(a - x){{(1.5a - 2x)}^2}}} = \frac{{4{x^2}.x}}{{(x){{(x)}^2}}} = 4$

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