The compounds $A$ and $B$ are mixed in equimolar proportion to form the products, $A + B \rightleftharpoons C + D$. At equilibrium, one third of $A$ and $B$ are consumed. The equilibrium constant for the reaction is

Question

Dr. Abubacker · Accepted Answer

(c)
$A\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  + \ \ \ \ B\ \ \ \ \ \  \rightleftharpoons \ \ \ C\ \ \  + \ \ \ D$
Initial $x$ mol $x$ mol
At equilibrium $x - \frac{x}{3} = \frac{2x}{3}$ $\frac{2x}{3}$ $\frac{x}{3}$ $\frac{x}{3}$
Hence,
$K_{c} = \frac{\lbrack C\rbrack\lbrack D\rbrack}{\lbrack A\rbrack\lbrack B\rbrack} = \frac{\left( \frac{x}{3V} \right)\left( \frac{x}{3V} \right)}{\left( \frac{2x}{3V} \right)\left( \frac{2x}{3V} \right)} = \frac{1}{4} = 0.25$

Q. The compounds $A$ and $B$ are mixed in equimolar proportion to form the products, $A + B \rightleftharpoons C + D$. At equilibrium, one third of $A$ and $B$ are consumed. The equilibrium constant for the reaction is

Explanation

Reviewed by Dr. Abubacker

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