Question

$4$ moles of $A$ are mixed with $4$ moles of $B$ when $2$ moles of $C$ are formed at equilibrium according to the reaction $A+B\rightleftharpoons C+D$.

The value of the equilibrium constant is :

• A. $4$
• B. $1$
• C. $1/2$
• D. $1/4$

Answer 1

The correct option is B $1$

$\underset{44-x}{A}+\underset{44-x}{B}\rightleftharpoons \underset{0x}{C}+\underset{0x}{D}$

$\therefore \left[A\right]=4-2=2$

$\left[B\right]=4-2=2$
$\left[C\right]=2,\left[D\right]=2$

The value of the equilibrium constant is, $K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=\frac{2\times 2}{2\times 2}=1$

The correct option is $B$