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Consider the reaction: 2 HI(g) ⇄ H2(g) + I2(g) It is found that, when equilibrium is reached at a certain temperature, HI is 35.4 percent dissociated. Calculate the equilibrium constant Kc for the reaction at this temperature.

Respuesta :

jufsanabriasa

Answer:

Kc = 0,075

Explanation:

For a reaction:

A + B ⇄ xC + yD

Kc = [tex]\frac{[C]^x[D]^y}{[A][B]}[/tex]

For the reaction:

2 HI(g) ⇄ H₂(g) + I₂(g)

Kc = [tex]\frac{[I_{2}][H_{2}]}{[HI]^2}[/tex] (1)

In equilibrium:

[HI] = 1- 2x

[I₂] = x [H₂] = x

If 35,4% is dissociated:

[HI] = 1-2x = 0,646 ⇒ x = 0,177

[I₂] = 0,177 [H₂] = 0,177

Replacing in (1):

Kc = [tex]\frac{[0,177][0,177]}{[0,646]^2}[/tex]

kc = 0,075

I hope it helps!

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