Question

H2(g) + I2(g) â 2 HI(g) Kc = 64 at 400ºC 3.00 moles of H2 and 3.00 moles of I2 are placed in a 4.00 L container and the system is allowed to reach equilibrium. Calculate the concentration of HI at equilibrium.

Answer 1

1.2 mol/L

Step-by-step explanation:

The reaction given is:

H₂(g) + I₂(g) ⇄ 2HI (g)

In the begning, the molar concentrations are (number of moles divided by volume):

MH₂ = MI₂ = 3/4 = 0.75 mol/L

So, making a table for the equilibrium:

H₂(g) + I₂(g) ⇄ 2HI (g)

0.75 0.75 0 Initial

-x -x +2x Rects (stoichiometry is 1:1:2)

0.75-x 0.75-x 2x Equilibrium

For a generic reaction aA + bB ⇄ cC +dD, the equilibrium constant is:

`Kc = \frac{{D}^dx[C]^c}{[A]^ax[B]^b}`

So, for the reaction given:

`Kc = ([HI]^2)/([H2]x[I2])`

Kc = (2x)²/[(0.75-x)(0.75-x)]

64 = 4x²/(0.5625 - 1.5x + x²)

36 - 96x + 64x² = 4x²

60x² - 96x + 36 = 0 (dividing for 12)

5x² - 8x + 3 = 0

Using Baskhara:

Δ = (-8)² - 4x5x3 = 64 - 60 = 4

x =
`(8 +/√(4) )/(2x5)`

x' = (8+2)/10 = 1

x'' = (8-2)/10 = 0.6

x must be smaller then 0.75, so x = 0.6 mol

Then the molar concentration of HI in equilibrium is 2x = 1.2 mol/L