Question

For given reaction, $A+B\rightleftharpoons C+D$ the initial concentrations of A and B are equal.

At equilibrium, the concentration of C is twice that of A, the equilibrium constant of the reactions is:

• A. $\frac{9}{4}$
• B. $4$
• C. $\frac{1}{9}$
• D. $\frac{4}{9}$

Answer 1

Answer: (B)

$4$

For given reaction, $A+B\rightleftharpoons C+D$ the initial concentrations of A and B are equal. At equilibrium, the concentration of C is twice that of A. the equilibrium constant of the reactions is 4.

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

$\therefore K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=\frac{2\times 2}{1\times 1}=4$