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

STEP 1

1. The reaction is at equilibrium.
2. The equilibrium constant Keq K_{eq} is calculated using the concentrations of products and reactants at equilibrium.
3. The stoichiometry of the reaction is given by the balanced equation: 2HI(g)H2(g)+I2(g) 2 \mathrm{HI}(g) \rightleftarrows \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) .

STEP 2

1. Determine the equilibrium concentration of I2\mathrm{I}_{2}.
2. Write the expression for the equilibrium constant Keq K_{eq} .
3. Calculate Keq K_{eq} using the equilibrium concentrations.
4. Compare the calculated Keq K_{eq} with the given options.

STEP 3

Determine the change in concentration of HI\mathrm{HI} from initial to equilibrium:
Δ[HI]=[HI]initial[HI]eq=0.300.24=0.06M \Delta[\mathrm{HI}] = [\mathrm{HI}]_{initial} - [\mathrm{HI}]_{eq} = 0.30 - 0.24 = 0.06 \, M

STEP 4

Since the stoichiometry of the reaction is 2:1:12:1:1, the change in concentration of H2\mathrm{H}_{2} and I2\mathrm{I}_{2} is half of Δ[HI]\Delta[\mathrm{HI}]:
Δ[H2]=Δ[I2]=12×0.06=0.03M \Delta[\mathrm{H}_{2}] = \Delta[\mathrm{I}_{2}] = \frac{1}{2} \times 0.06 = 0.03 \, M

STEP 5

Calculate the equilibrium concentration of I2\mathrm{I}_{2}:
[I2]eq=0.00+0.03=0.03M [\mathrm{I}_{2}]_{eq} = 0.00 + 0.03 = 0.03 \, M

STEP 6

Write the expression for the equilibrium constant Keq K_{eq} :
Keq=[H2]eq×[I2]eq[HI]eq2 K_{eq} = \frac{[\mathrm{H}_{2}]_{eq} \times [\mathrm{I}_{2}]_{eq}}{[\mathrm{HI}]_{eq}^2}

STEP 7

Substitute the equilibrium concentrations into the expression for Keq K_{eq} :
Keq=(0.03)×(0.03)(0.24)2=0.00090.05760.015625 K_{eq} = \frac{(0.03) \times (0.03)}{(0.24)^2} = \frac{0.0009}{0.0576} \approx 0.015625

STEP 8

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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