A reaction is, $A + B \rightarrow C + D$. Initially we start with equal concentrations of $A$and $B$. At equilibrium, we find the moles of $C$ are two times of $A$. What is the equilibrium constant of the reaction?

Question

Dr. Abubacker · Accepted Answer

(b)
$A + B ightleftharpoons C + D$
$a\ \ \ \ \ a\ \ \ \ \ \ 2a\ \ \ 2a$ at equilibrium
$	herefore\ \ K_{c} = \frac{\lbrack Cbrack\lbrack Dbrack}{\lbrack Abrack\lbrack Bbrack}$
$= \frac{2a 	imes 2a}{a 	imes a} = 4$

Q. A reaction is, $A + B \rightarrow C + D$. Initially we start with equal concentrations of $A$and $B$. At equilibrium, we find the moles of $C$ are two times of $A$. What is the equilibrium constant of the reaction?

Explanation

Reviewed by Dr. Abubacker

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