Question

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.

Answer 1

Kc = 0,075

Step-by-step explanation:

For a reaction:

A + B ⇄ xC + yD

Kc =
`([C]^x[D]^y)/([A][B])`

For the reaction:

2 HI(g) ⇄ H₂(g) + I₂(g)

Kc =
`([I_(2)][H_(2)])/([HI]^2)` (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 =
`([0,177][0,177])/([0,646]^2)`

kc = 0,075

I hope it helps!