Answer
$ [ HI ] = 0.0152 \space M $
$ [ I_2 ] = 0.00328 \space M $
Work Step by Step
1. Calculate all the concentrations:
$$[HI] = ( 0.0244 )/(1.50) = 0.0163 M$$
$$[H_2] = ( 0.00623 )/(1.50) = 0.00415 M$$
$$[I_2] = ( 0.00414 )/(1.50) = 0.00276 M$$
2. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HI ]& [ H_2 ]& [ I_2 ]\\
Initial& 0.0163 & 0.00415 & 0.00276 \\
Change& -2 x& +x& +x\\
Equilibrium& 0.0163 -2 x& 0.00415 +x& 0.00276 +x\\
\end{vmatrix}$$
$ [ HI ] = 0.0163 \space M - 2x$
$ [ H_2 ] = 0.00415 \space M + x$
$ [ I_2 ] = 0.00276 \space M + x$
3. Using the concentration of $ H_2 $ at equilibrium, find x:
$ 0.00415 + x = 0.00467 $
$ x = 0.00467 - 0.00415 $
$x = 5.20 \times 10^{-4} $
$ [ HI ] = 0.0163 \space M - 2*( 5.20 \times 10^{-4} ) = 0.0152 $
$ [ I_2 ] = 0.00276 \space M + 5.20 \times 10^{-4} =3.28 \times 10^{-3} $
