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Consider the following reaction: H2(g)+I2(g)⇌2HI(g) Complete the following table. Assume that all concentrations are equilibrium concentrations in M. T(∘C) [H2] [I2] [HI] [Kc] 25 0.0355 0.0388 0.922 − 340 − 4.60×10−2M 0.394M 9.6 445 4.90×10−2M 4.76×10−2M − 50.2 Find Kc at 25∘C. Find [H2] at 340∘C. Find [HI] at 445∘C.

Respuesta :

Edufirst

These are three questions.

Answer:

  • Question 1) Find Kc at 25°C: 617

  • Question 2) Find [H₂] at 340°C: 0.352 M

  • Question 3) Find [HI] at 445°C: 0.342 M

Explanation:

I will start by copying the table for better understanding of the data and the work to do.

T(∘C)       [H2]                    [I2]                     [HI]               [Kc]

25           0.0355               0.0388              0.922            -

340                                    4.60×10⁻²M       0.394M         9.6

445         4.90×10−2M      4.76×10⁻²M           −                50.2

The chemical equation that represents the equilibrium is also given:

  • H₂ (g) + I₂(g) ⇄ 2HI(g)

From that equilibrium equation, you can fnd the expression of the equlibrium constant, Kc, as:

[tex]Kc=[HI]^2/([H_2][I_2])[/tex]

The calculations and explanations to answer the three questions are in the pdf file attached.

Please, click on the image of the file to open it.

okpalawalter8

We have that for the Question "H2(g)+I2(g)⇌2HI(g) Complete the following table.  Find Kc at 25∘C. Find [H2] at 340∘C. Find [HI] at 445∘C. "

It can be said that

  • Kc at 25∘C = [tex]617.16[/tex]
  • H2 at 340∘C = [tex]0.357M[/tex]
  • HI at 445∘C = [tex]0.337M[/tex]

From the question we are given the table

T(∘C)      [H2]                  [I2]                [HI]             [ Kc]

25          0.0355           0.0388            0.922             −

340          −                  4.60×10−2M      0.394M         9.6

445    4.90×10−2M    4.76×10−2M          −                50.2

At [tex]25^0c[/tex]

[tex]Kc = \frac{HI^2}{H2*I2}\\\\Kc = \frac{0.922^2}{0.0355*0.0388}\\\\Kc = 617.16[/tex]

At [tex]340^0c[/tex]

[tex]Kc = \frac{HI^2}{H2*I2}\\\\9.6 = \frac{0.393^2}{H2*4.50*10^{-2}}\\\\H2 = 0.357M[/tex]

At [tex]445^0C[/tex]

[tex]Kc = \frac{HI^2}{H2*I2}\\\\50.2 = \frac{HI^2}{4.90*10^{-2}*4.64*10^{-2}}\\\\HI = 0.337M[/tex]

For more information on this visit

https://brainly.com/question/17756498

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