Question

In the following equilibria :-
(1) $A+2B\rightleftharpoons C;K_{eq}=K_1$
(2) $C+D\rightleftharpoons 3A;K_{eq}=K_2$
(3) $6B+D\rightleftharpoons 2C;K_{eq}=K_3$
Hence:

• A. $K_1^3{K}_2^2=K_3$
• B. $K_1^3{K}_2=K_3$
• C. $3K_1+K_2=K_3$
• D. $K_1^3/K_2^2=K_3$

Answer 1

The correct option is B $K_1^3{K}_2=K_3$

$A+2B\rightleftharpoons C;K_{eq}=K_1$ ___(1)
$C+D\rightleftharpoons 3A;K_{eq}=K_2$ ___(2)
$6B+D\rightleftharpoons 2C;K_{eq}=K_3$ ___(3)
$\left[\right(1)\times (3\left)\right]+\left[\right(2\left)\right]=\left(3\right)$
$K_1^3\times K_2=K_3$