$A(g) + 3B(g) \rightleftharpoons 4C(g)$. Initially concentration of $A$ is equal to that of $B$. The equilibrium concentrations of $A$ and $C$ are equal. $K_{c}$ is

Question

Dr. Abubacker · Accepted Answer

(c)
$\ \ A(g)\ \ \ \ \ \ \ \  + 3B(g) ightleftharpoons 4C(g)$
1 1 0 Initial concentration
$(1 - x)$ $(1 - 3x)$ $4x$ Final concentration
(at equilibrium)
According to question, $1 - x = 4x$
$	herefore$ $x = \frac{1}{5}$
For above reaction,
$K_{c} = \frac{\lbrack Cbrack^{4}}{\lbrack Abrack\lbrack Bbrack^{3}} = \frac{(4x)^{4}}{(1 - x)(1 - 3x)^{3}}$
$K_{c} = \frac{\left( 4 	imes \frac{1}{5} ight)^{4}}{\left( 1 - \frac{1}{5} ight)\left( 1 - 3 	imes \frac{1}{5} ight)^{3}} = 8.0$

Q. $A(g) + 3B(g) \rightleftharpoons 4C(g)$. Initially concentration of $A$ is equal to that of $B$. The equilibrium concentrations of $A$ and $C$ are equal. $K_{c}$ is

Explanation

Reviewed by Dr. Abubacker

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