Question

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

• 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, $K$ for it will be the product of all three equlibrium constants.
$\because K=K_1\times K_2\times K_3=2\times 4\times 3=24$

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