Question

For the reaction $2A+B\leftrightharpoons C$
$1$ moles of $A$, $2$ moles of $B$ and $4$ moles of $C$ were taken initially in $1L$ vessel. At equilibrium moles of $C$ was $0.5$ what is the equilibrium constant?

• A. $1.4\times {10}^{-3}$
• B. $1.1\times {10}^{-2}$
• C. $3.5\times {10}^{-4}$
• D. $2.8\times {10}^{-2}$

Answer 1

The correct option is A $1.4\times {10}^{-3}$

$\underset{1}{2A}+\underset{2}{B}\rightleftharpoons \underset{4}{C}$

Moles of $C$ at equilibrium= $0.5$, $C$ is decomposing here because initially $4$ moles. But now $0.5$ moles.

$2A+B\rightleftharpoons C$

$1$ $2$ $4$

$1+2\alpha$ $2+\alpha$ $4-\alpha$

$\Rightarrow 4-\alpha =0.5$

$\Rightarrow \alpha =3.5$

$K_C=\frac{\left[A{]}^2\right[B]}{C}=\frac{[H2\alpha {]}^2-[2+\alpha ]}{[4-\alpha ]}$

$=\frac{\left(8{)}^2\right(5.5)}{0.5}$

$\Rightarrow K_C=1.4\times {10}^{-3}$