$A + B = C + D$. If initially the concentration of $A\ \ and\ B$ are both equal but at equilibrium, concentration of $D$ will be twice of that of $A$, then what will be the equilibrium constant of reaction?

Question

Dr. Abubacker · Accepted Answer

(d)
$A + B = C + D$
$x$ $x$ 0 0 At initial
$2x$ $2x$ At equilibrium
$K_{c} = \frac{\lbrack Cbrack\lbrack Dbrack}{\lbrack Abrack\lbrack Bbrack} = \frac{2x 	imes 2x}{x 	imes x} = 4$

Q. $A + B = C + D$. If initially the concentration of $A\ \ and\ B$ are both equal but at equilibrium, concentration of $D$ will be twice of that of $A$, then what will be the equilibrium constant of reaction?

Explanation

Reviewed by Dr. Abubacker

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