Question

$4$ moles of $A$ are mixed with $4$ moles of $B$ when $2$ moles of $C$ are formed at equilibrium.

What is the equilibrium constant of the given reaction: $A+B\rightleftharpoons C+D$

• A. $4$
• B. $1$
• C. $\sqrt{2}$
• D. $\sqrt{4}$

Answer 1

The correct option is B $1$

$A+B\rightleftharpoons C+D$

Initially $4moles$ $4moles$ $0$ $0$

at eqm $4-\alpha$ $4-\alpha$ $\alpha$ $\alpha$

$2$ moles of $C$ are formed, $\therefore \alpha =2$

The equilibrium constant,

$K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$

$=\frac{2\times 2}{[4-2]\times [4-2]}=\frac{2\times 2}{2\times 2}=1$

So, the correct option is $B$