Answer
$$\begin{vmatrix}
Compound& [ H_2 ]& [ F_2 ]& [ HF ]\\
Initial& 0.20 & 0.10 & 0 \\
Change& -x& -x&+ 2 x\\
Equilibrium& 0.20 -x& 0.10 -x& 2 x\\
\end{vmatrix}$$
Work Step by Step
1. Calculate the concentrations:
$$Molarity(H_2) = \frac{0.10 \space mol \space H_2}{0.50 \space L} = 0.20 \space M \space H_2$$
$$Molarity(F_2) = \frac{0.050 \space mol \space F_2}{0.50 \space L} = 0.10 \space M$$
2. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ H_2 ]& [ F_2 ]& [ HF ]\\
Initial& 0.20 & 0.10 & 0 \\
Change& -x& -x&+ 2 x\\
Equilibrium& 0.20 -x& 0.10 -x& 2 x\\
\end{vmatrix}$$
