At a certain temperature, only 50% HI is dissociated into H 2 and I...

\begin{cases}2\text{H}\text{I}\left(\text{g}\right) & ⇌ & {\text{H}_{2}\left(\text{g}\right) & + & {\text{I}_{2}\left(\text{g}\right) \\ \text{Moles} & 1 - \alpha & \dfrac{\alpha}{2} & \dfrac{\alpha}{2} \\ \text{Concentration} & \dfrac{1 - \alpha}{\text{V}} & \dfrac{\alpha}{\text{2V}} & \dfrac{\alpha}{\text{2V}}\end{cases}$<br /><br />${\text{K}}_{\text{C}} = \dfrac{\left[{\text{H}_{2}\right]\left[{\text{I}_{2}\right]}{{\left[\text{H}\text{I}\right]}^{2}} = \dfrac{\left[\dfrac{\alpha}{\text{2V}}\right]\left[\dfrac{\alpha}{\text{2V}}\right]}{{\left[\dfrac{1 - \alpha}{\text{V}}\right]}^{2}}<br /><br /> \left(\begin{cases}∴ \alpha & = & \text{0.50} \\ \dfrac{\alpha}{2} & = & \text{0.25}\end{cases}\right)

$=\dfrac{{\alpha}^{2}}{4{\left(1 - \alpha\right)}^{2}} = \dfrac{0 .50 \times 0 .50}{4 \times 0.50 \times 0 .50} = \dfrac{1}{4}$

${\text{K}}_{\text{C}} = \text{0.25}$