During an experiment, \( 1.15 \mathrm{~mol}^{\text {of }} \mathrm{H}_{2} \) and \( 0.574 \mathrm{~mol}^{\text {of }} \mathrm{I}_{2} \) were placed into a 1.913 liter vessel where the reaction \[ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftarrows 2 \mathrm{HI}(\mathrm{~g}) \] came to equilibrium. For this reaction, \( \mathrm{K}_{\mathrm{C}}=49.5 \) at the temperature of the experiment. What were the equilibrium concentrations of \( \mathrm{H}_{2}, \mathrm{I}_{2} \), and HI ? Correct. \[ \left[H_{2}\right]=0.322 \] eTextbook and Media Incorrect. \[ \left[\mathrm{I}_{2}\right]=\mathbf{i} 0.0203 \mathrm{M} \] Correct, \[ [H I]= \] 0.559 M

To find the equilibrium concentrations, start by calculating the initial concentrations of \( H_2 \) and \( I_2 \). The concentration of \( H_2 \) is \( \frac{1.15 \, \text{mol}}{1.913 \, \text{L}} \approx 0.601 \, \text{M} \) and for \( I_2 \) it is \( \frac{0.574 \, \text{mol}}{1.913 \, \text{L}} \approx 0.300 \, \text{M} \). Based on the stoichiometry of the balanced equation, at equilibrium, we can set up the expression using \( K_c \) to find the change in concentration, leading to the final concentrations mentioned. When working with equilibrium reactions, a common mistake is overlooking the stoichiometric coefficients in the change calculations. Ensure that you properly account for how much of each reactant is consumed or produced. It’s crucial to adjust your initial concentrations correctly based on the molar ratios to avoid making errors in your equilibrium expressions!