Question

If $K_c$ for the formation of HI from $H_2$ and $I_2$ is 48, then $K_c$ for decomposition of 1 mole of HI is ................

• A. $\frac{1}{\sqrt{48}}$
• B. $48$
• C. $0.144$
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{1}{\sqrt{48}}$
C $0.144$

If $K_c$ for the formation of $HI$ from $H_2$ and $I_2$ is $48$,

$H_2+I_2\to 2HI$
$K_c=\frac{[HI{]}^2}{\left[H_2\right]\left[I_2\right]}=48$
Then, for the decomposition of $1$ mole of $HI$, we get

$HI\to \frac{1}{2}{H}_2+\frac{1}{2}{I}_2$
$K_{c2}=\frac{\left[H_2{]}^{1/2}\right[I_2{]}^{1/2}}{\left[HI\right]}$
Hence,$K_{c2}=\frac{1}{\sqrt{K_c}}$
$\therefore K_{c2}=\frac{1}{\sqrt{48}}=0.144$

Hence, both options A and C are correct.