Question

The equilibrium constant Kc for the reaction below is 82.3 at a certain temperature.

H2(g) + I2(g) double arrows 2HI (g)

If you start with .355 M of hydrogen iodide, what will the concentrations of HI, H2, I2 be at equilbrium?

Answer 1

Reaction equation
H₂(g) + I₂(g) ⇄ 2 HI (g)

Equilibrium expression:
Kc = [HI] / ( [H₂] ∙[I₂] )
with
Kc = 82.3

ICE-Table
.............. [H₂]............. [I₂]............. [HI]
I............... 0................ 0............. 0.355
C........... + x............. + x............. - 2∙x
E.............. x................ x........ 0.355 - 2∙x

When you substitute the equilibrium concentrations from the last row to the equilibrium expression you get:
82.3 = (0.355 - 2∙x)² / (x∙x)
<=>
82.3∙x² = 0.355² -2*0.3.55*2∙x + 4∙x²
<=>
a∙x² + b∙x + c = 0
with
a = 82.3 - 4 = 78.3
b = 2 ∙ 0.355 ∙2 = 1.42
c = - 0.355² = - 0.126025
The solutions to this quadratic equation are:
x = ( - b ± √(b² - 4∙a∙c)) / (2∙a)
=>
x = - 0.0501or
x =0.0320

First solution is infeasible, because it would lead to negative concentrations for hydrogen and iodine.
So the concentrations in the equilibrium mixture are:
H₂] = 0.032 M
[I₂] = 0.032 M
[HI] = 0.355 M - 2 ∙ 0.032 M = 0.291 M