Question

$A+2B\rightleftharpoons 2C+D$ initial concentration of $B$ was $1.5$ times that of $A$, but the equilibrium concentration of $A$ and $B$ are found to be equal. The equilibrium constant for the reaction is:

• A. $4$
• B. $8$
• C. $12$
• D. $16$

Answer 1

The correct option is A $4$

$A+2B\rightleftharpoons 2C+D$
Initial $1$ $1.5$ $0$ $0$
Equb.$1-x$ $1.5-2x$ $2x$ $x$

At equilibrium $\left[A\right]=\left[B\right]$
$1-x=1.5-2x$
$x=0.5$

$\therefore$ $K_c=\frac{{\left[C\right]}^2\left[D\right]}{\left[A\right]{\left[B\right]}^2}$

$\frac{{\left(2x\right)}^2\left(x\right)}{(1-x){(1.5-2x)}^2}$

$=\frac{1\times 1\times 0.5}{0.5\times 0.5\times 0.5}=4$

Hence, option A is correct