## General Chemistry: Principles and Modern Applications (10th Edition)

$[ HI ] = 0.0174 \space M$
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 = $x_1$ $[ HI ] = 2(0.00869) = 0.0174 \space M$