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A quantity of HI was sealed in a tube, heated to 425°C and held at this temperature until
equilibrium was reached. The concentration of Hl in the tube at equilibrium was found to be
0.0706 mol/L. Calculate the equilibirum concentration of H2 (and 12). For the gas-phase
reaction,
H2 + 2 = 2HIKc = 54.6 at 425°C
O a. 1.17 x 10-3M
O b. 2.34 x 10-3M
c. 9.55 x 10-3M
d. 1.85 x 10-4 M
e. 4.78 x 10 PM​

Respuesta :

ardni313

The equilibrium concentration of H₂ : c. 9.55 x 10⁻³ M

Further explanation

Given

Kc = 54.6 at 425°C

The concentration of Hl = 0.0706 mol/L

Required

The equilibrium concentration of H₂ (and I₂)

Solution

[tex]\tt Kc=\dfrac{[HI]^2}{[H_2][I_2]}\\\\54.6=\dfrac{0.0706^2}{[H_2][I_2]}[/tex]

[H₂][I₂]=[H₂]²

[tex]\tt [H_2][I_2]=\dfrac{0.0706^2}{54.6}=[H_2]=9.55\times 10^-3~M[/tex]

pstnonsonjoku

The gas phase concentration of H2 and I2 is  9.55 x 10-3M.

The equation of the reaction is;

          H2      +       I2 ------>        2HI

We also have the information that the equilibrium constant of the process is  54.6.

Now recall that;

Kc = 2HI/[H2] [I2]

We can assume that the concentration of H2 = concentration of I2

Hence; [H2] = [I2] = x

54.6 = [0.0706]^2/x^2

54.6x^2 =  [0.0706]^2

x = √ [0.0706]^2/54.6

x = 9.55 x 10-3M

Learn more about equilibrium constant: https://brainly.com/question/17960050?

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