Math  /  Data & Statistics

Question2HI(g)H2(g)+I2(g)2 \mathrm{HI}(g) \rightleftarrows \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \begin{tabular}{|l|l|l|} \hline Substance & Initial Concentration (M)(M) & Equilibrium Concentration (M)(M) \\ \hline HI & 0.30 & 0.24 \\ \hline H2\mathrm{H}_{2} & 0.00 & 0.030 \\ \hline I2\mathrm{I}_{2} & 0.00 & ?? \\ \hline \end{tabular}
In an experiment involving the reaction shown above, a sample of pure HI was placed inside a rigid container at a certain temperature. The table above provides the initial and equilibrium concentrations for some of the substances in the reaction. Based on the data, which of the following is the value of the equilibrium constant ( KeqK_{e q} ) for the reaction, and why? (A) Keq=2.5×101K_{e q}=2.5 \times 10^{-1}, because [I2]eq=2×[HI]eq\left[\mathrm{I}_{2}\right]_{e q}=2 \times[\mathrm{HI}]_{e q}. (B) Keq=6.3×102K_{e q}=6.3 \times 10^{-2}, because [I2]eq=12[HI]eq\left[\mathrm{I}_{2}\right]_{e q}=\frac{1}{2}[\mathrm{HI}]_{e q}. (C) Keq=1.6×102K_{e q}=1.6 \times 10^{-2}, because [I2]eq=[H2]eq\left[\mathrm{I}_{2}\right]_{e q}=\left[\mathrm{H}_{2}\right]_{e q}. (D) Keq=3.1×102K_{e q}=3.1 \times 10^{-2}, because [I2]eq=2×[H2]eq\left[\mathrm{I}_{2}\right]_{e q}=2 \times\left[\mathrm{H}_{2}\right]_{e q}.

Studdy Solution
Compare the calculated Keq0.015625 K_{eq} \approx 0.015625 with the given options:
The closest option is (C) Keq=1.6×102 K_{eq} = 1.6 \times 10^{-2} .
The correct answer is:
(C) Keq=1.6×102 K_{eq} = 1.6 \times 10^{-2} , because [I2]eq=[H2]eq[\mathrm{I}_{2}]_{eq} = [\mathrm{H}_{2}]_{eq}.

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