Question

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

Answer 1

• The reaction for the formation of $\mathrm{HI}$ from ${\mathrm{H}}_2$and ${\mathrm{I}}_2$ is:

${\mathrm{H}}_2+{\mathrm{I}}_2\rightleftharpoons 2\mathrm{HI}$ , ${\mathrm{K}}_{{\mathrm{C}}_1}=\frac{{\left[\mathrm{HI}\right]}^2}{{\left[{\mathrm{H}}_2\right]}^1{\left[{\mathrm{I}}_2\right]}^{{\displaystyle 1}}}$

• The reaction for the decomposition of 1 mole of :

$\mathrm{HI}\rightleftharpoons \frac{1}{2}{\mathrm{H}}_2+\frac{1}{2}{\mathrm{I}}_2$ , ${\mathrm{K}}_{{\mathrm{C}}_2}=\frac{{\left[{\mathrm{H}}_2\right]}^{{\displaystyle \frac{1}{2}}}{\left[{\mathrm{I}}_2\right]}^{{\displaystyle \frac{1}{2}}}}{\left[\mathrm{HI}\right]}$

• Therefore, ${\mathrm{K}}_{{\mathrm{C}}_2}=\frac{1}{\sqrt{{\mathrm{K}}_{{\mathrm{C}}_1}}}$

${\mathrm{K}}_{{\mathrm{C}}_2}=\frac{1}{\sqrt{48}}=0.144$