$A + B \rightleftharpoons C + D$ Initially moles of $A\ and\ B$ are equal. At equilibrium, moles of C are three times that of A. the equilibrium constant of the reaction will be

Question

Dr. Abubacker · Accepted Answer

(d)
$A\ \ \ \ \ \  + \ \ \ \ \ \ \ \ B\ \ \ \ \ \ \  ightleftharpoons \ \ \ \ \ \ C\ \ \ \  + \ \ \ \ \ D$
1 1 0 0 Initially
$(1 - 3x)$ $(1 - 3x)$ $3x$ $3x$ At equilibrium (given)
At equilibrium, the remaining moles of $A$ is $x$, because $3x$ moles of $C$ are produced.
$\Rightarrow$ $1 - 3x = x$
$	herefore$ $x = \frac{1}{4}$
Equilibrium constant,
$K_{c} = \frac{\lbrack Cbrack\lbrack Dbrack}{\lbrack Abrack\lbrack Bbrack} = \frac{3x.3x}{(1 - 3x)^{2}}$
On putting the value of $x$, we get,
$K_{c} = \frac{9 	imes \frac{1}{16}}{1 + \frac{9}{16} - \frac{6}{4}} = \frac{9}{1} = 9$

Q. $A + B \rightleftharpoons C + D$ Initially moles of $A\ and\ B$ are equal. At equilibrium, moles of C are three times that of A. the equilibrium constant of the reaction will be

Explanation

Reviewed by Dr. Abubacker

Want more practice?