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Black_prince [1.1K]
4 years ago
6

Gaseous hydrogen iodide is placed in a closed container at 425 degrees C, where it partially decomposes to hydrogen and iodine:

2HI(g)--->/<---H2(g)+I2(g). At equilibrium it is found that [HI]= 3.59
Chemistry
1 answer:
otez555 [7]4 years ago
7 0

Answer:

0.0184

Explanation:

There is some info missing. I think this is the original question.

<em>Gaseous hydrogen iodide is placed in a closed container at 425 °C, where it partially decomposes to hydrogen and iodine: 2 HI(g) ⇄ H₂(g) + I₂(g). At equilibrium, it is found that HI = 3.59 × 10⁻³ M, H₂= 4.87 × 10⁻⁴ M, and I₂= 4.87 × 10⁻⁴ M. </em>

<em> What is the value of Kc at this temperature? </em>

<em>Express the equilibrium constant to three significant digits.</em>

<em />

Let's consider the following reaction at equilibrium.

2 HI(g) ⇄ H₂(g) + I₂(g)

The concentration equilibrium constant (Kc) is the product of the concentration of the products raised to their stoichiometric coefficients divided by the product of the concentration of the reactants raised to their stoichiometric coefficients.

Kc = [H₂] × [I₂]/[HI]²

Kc = 4.87 × 10⁻⁴ × 4.87 × 10⁻⁴/ (3.59 × 10⁻³)²

Kc = 0.0184

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\Delta G^0 _{rxn} = 2 \times \Delta G_f^0  \ N_2O_{(g)} - ( 2 \times  \Delta G_f^0  \ N_2{(g)} +   \Delta G_f^0  \ O_{2(g)})

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The equilibrium constant determined from the partial pressure denoted as K_p can be expressed as :

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\Delta G =207.6 + 8.314 \times 10 ^{-3} \times 298.15  \ ln(0.045)

\Delta G =207.6 + 2.4788191 \times \ ln(0.045)

\Delta G =207.6+ (-7.687048037)

\Delta G = 199.912952  kJ

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