Answer
$ [ HI ] = 0.0174 \space M$
Work Step by Step
1. Calculate each molarity:
$$\frac{0.150 \space mol \space H_2}{15.0 \space L} = 0.0100 \space M \space H_2$$
$$\frac{0.200 \space mol \space I_2}{15.0 \space L} = 0.0133 \space M \space I_2$$
2. At equilibrium, these are the concentrations of each compound:
$ [ H_2 ] = 0.0100 \space M - x$
$ [ I_2 ] = 0.0133 \space M - x$
$ [ HI ] = 0 \space M + 2x$
$$50.2 = \frac{(2x)^2}{(0.0100 - x)(0.0133 - x)}$$
3. Solve for x:
$x_1 = 0.00869$
$x_2 = 0.0166$
But x cannot be $x_2$:
$[H_2 = 0.0100 - 0.0166 = -0.066 \space M$, and the concentration cannot be negative.
x = $x_1$
$ [ HI ] = 2(0.00869) = 0.0174 \space M$
