Question

If ${\mathrm{K}}_{\mathrm{c}}$ for the formation of $\mathrm{HI}$ from ${\mathrm{H}}_2$ and ${\mathrm{I}}_2$is 48 , then ${\mathrm{K}}_{\mathrm{c}}$ for decomposition of 1 mole of $\mathrm{HI}$ is________

Answer 1

1. The formation reaction is-

$$\begin{gathered} {\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{I}}_2\left(\mathrm{g}\right)\iff 2\mathrm{HI}\left(\mathrm{g}\right){\mathrm{K}}_{{}_{\mathrm{c}}}=48 \\ \mathrm{Hydrogen}\mathrm{Iodine}\mathrm{Hydrogen}\mathrm{iodide} \end{gathered}$$
$K_c$ is given as 48 for the above reaction.

2. When the above reaction is reversed, it becomes the decomposition of hydrogen iodide ($\mathrm{HI}$). The equilibrium constant (${K_c}^{\text{'}}$) for this reaction is reciprocal of the equilibrium constant for the formation reaction ($K_c$).

$$\begin{gathered} 2\mathrm{HI}\left(\mathrm{g}\right)\iff {\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{I}}_2\left(\mathrm{g}\right)K_c\text{'}=\frac{1}{K_c} \\ \mathrm{Hydrogen}\mathrm{Hydrogen}\mathrm{Iodine} \\ \mathrm{iodide} \end{gathered}$$

3. Now, to obtain the $K_c$ for the decomposition reaction of 1 mole of $\mathrm{HI}$,
$$\begin{gathered} \mathrm{HI}\left(\mathrm{g}\right)\rightleftharpoons \frac{1}{2}{\mathrm{H}}_2\left(\mathrm{g}\right)+\frac{1}{2}{\mathrm{I}}_2\left(\mathrm{g}\right) \\ \mathrm{Hydrogen}\mathrm{Hydrogen}\mathrm{Iodine} \\ \mathrm{iodide} \end{gathered}$$
$K_c^{\text{'}\text{'}}=\sqrt{K_c\text{'}}=\sqrt{\frac{1}{K_c}}$
$K_c^{\text{'}}=\frac{1}{\sqrt{48}}$
$K_c^{\text{'}}=0.144$

4. Hence, If ${\mathrm{K}}_{\mathrm{c}}$ for the formation of HI from ${\mathrm{H}}_2$ and ${\mathrm{I}}_2$is 48 , then ${\mathrm{K}}_{\mathrm{c}}$ for decomposition of 1 mole of $\mathrm{HI}$ is 0.144.