Question

For the reactions :-

$A\rightleftharpoons B;K_c=2,$

$B\rightleftharpoons C;K_c=4,$

$C\rightleftharpoons D;K_c=6$

What is the value of $K_c$ for the reaction?

$A\rightleftharpoons D$

Answer 1

For the reaction-

$A\rightleftharpoons B$

$K_c=\frac{\left[B\right]}{\left[A\right]}=2\left(\text{Given}\right)$

$\Rightarrow \frac{\left[B\right]}{\left[A\right]}=2.....\left(1\right)$

Similarly,

For the reaction-

$B\rightleftharpoons C$

$\frac{\left[C\right]}{\left[B\right]}=4.....\left(2\right)$

For the reaction-

$C\rightleftharpoons D$

$\frac{\left[D\right]}{\left[C\right]}=6.....\left(3\right)$

Multiplying equation $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get

$\frac{\left[D\right]}{\left[A\right]}=2\times 4\times 6=48$

$\therefore \frac{\left[D\right]}{\left[A\right]}=48.....\left(4\right)$

Now, for the reaction,

$A\rightleftharpoons D$

$K_c=\frac{\left[D\right]}{\left[A\right]}.....\left(5\right)$

From equation $\left(4\right)\&\left(5\right)$, we get

$K_c=48$

Hence for the reaction $A\rightleftharpoons D$; the value of $K_c$ is $48$.