$C_{HI}=\dfrac{0.5}{2}=0.25(M)$

\begin{array}{||}\hline \text{}&\text{$2HI$}&\text{$\buildrel{{t^o,xt,p}}\over\rightleftharpoons$}&\text{$H_2$}&\text{$+I_2$}\\\hline \text{I}&\text{0.25M}&\text{}&\text{-}&\text{-}\\\hline \text{C:}&\text{-xM}&\text{}&\text{($+\dfrac{x}{2}$)M}&\text{($+\dfrac{x}{2}$)M}\\\hline \text{E:}&\text{(0.25-x)M}&\text{}&\text{$\dfrac{x}{2}$M}&\text{$\dfrac{x}{2}$M}\\\hline \end{array}

$K_c=\dfrac{[H_2][I_2]}{[HI]^2}$

`->`$\dfrac{\dfrac{x}{2} \times \dfrac{x}{2}}{(0.25-x)^2}=16$ (Với $x \leq 0.25)$

`=>`$x=\dfrac{2}{9}$

`->`$[HI]=0.25-\dfrac{2}{9}=\dfrac{1}{36}(M)$