Question

For a reactions, the equilibrium constant $\left(K\right)$ values are given:
$A\rightleftharpoons B;K_1=2B\rightleftharpoons C;K_2=4C\rightleftharpoons D;K_3=3$
Then calculate the equilibrium constant for the reaction $A\rightleftharpoons D$:

• A. $48$
• B. $6$
• C. $24$
• D. $12$

Answer 1

The correct option is C $24$

The three reactions are:
$A\rightleftharpoons B;K_1=2B\rightleftharpoons C;K_2=4C\rightleftharpoons D;K_3=3$

The reaction $A\rightleftharpoons D$ is obtained by adding the three given reactions. So, equilibrium constant $K$ for it will be the product of all three equlibrium constants.
$\therefore K=K_1\times K_2\times K_3=2\times 4\times 3=24$

$\therefore K=2\times 4\times 3=24$