Question

The reaction A(B) = 2B(g) has an equilibrium constant of K = 0.045. What is the equilibrium constant for the reaction B(g) =1/2A
Express your answer using two significant figures.
I

Answer 1

Answer:

The  $K_c$ for the reaction $B(g) = \frac{1}{2}A$ will be 4.69.

Explanation:

The given equation is A(B) = 2B(g)

to evaluate equilibrium constant for $B(g) = \frac{1}{2}A$

            $K_c=[B]^2[A]$

                 = 0.045

The reverse will be $2B\leftrightharpoons A$

Then,      $K_c = \frac{[A]}{[B]^2}$

                    =  $\frac{1}{0.045}$

                    = $22m^{-1}$

The equilibrium constant for $B(g) = \frac{1}{2}A$ will be

               $K_c = \sqrt{K_c}$

                    $=\sqrt{22}$

                    = 4.69

Therefore, $K_c$ for the reaction $B(g) = \frac{1}{2}A$ will be 4.69.