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:

H₂ + I₂ → 2HI,

The initial concentrations:

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.

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .