How many moles of HI are present at equilibrium when 2.0 moles of H2 is mixed with 1.0 moles of I2 in a 0.50 L container and allowed to react at 448oC. At this temperature K = 50.

1.9 mol.

Explanation:

For the reaction:

H₂ + I₂ → 2HI,

The initial concentrations:

[H₂] = no. of moles/volume of the container = (2.0 mol) / (0.50 L) = 4.0 mol/L. [I₂] = no. of moles/volume of the container = (1.0 mol) / (0.50 L) = 2.0 mol/L.

The equilibrium constant (K) = [HI]²/[H₂][I₂],

Write down the equation and the start concentrations. Let (x) be the finish concentrations:

H₂ + I₂ → 2HI

start: 4.0 M 2.0 M 0 M

finish: 4.0-x 2.0-x 2x

∴ The equilibrium constant (K) = [HI]²/[H₂][I₂] = (2x) ² / (4.0 - x) (2.0 - x).

∴ 50 = (2x) ² / (4.0 - x) (2.0 - x).

∴ 50 = (4x²) / (8.0 - 6.0 x + x²).

∴ 400 - 300 x + 50 x² = 4x².

∴ 46x² - 300 x + 400 = 0 (standard form)

Use the quadratic equation to solve for (x):

x = ( - b ± √ (b² - 4ac)) / (2a),

In this case a = + 46, b = - 300 and c = + 400.

x = ( + 300) ± √ (( - 300) ² - 4 (46) (400)) / (2 (46))

x = ( + 300) ± √ (9000 - 73600) / (92)

x = ( + 300) ± 127) / (92)

∴ x = 4.7 and 1.9

When x = 4.7 is not possible, because 4.0 - 4.7 (the concentration of H2 would be negative).

So, x = 1.9.

∴ [HI] = 2x

If x = 1.9,

∴ [HI] = 2 (1.9) = 3.8 mol/L.

∵ [HI] = no. of moles/volume of the container.

∴ no. of moles of (HI) = [HI]*volume of the container = (3.8 mol/L) (0.5 L) = 1.9 mol.