H2 (g) + I2 (g) ⇌ 2 HI(g) Kc = 54.3 at 430 °C . What will be the concentrations of all species at equilibrium at this temperature if we start with the following initial concentrations: [H2 ] =0.00623 M, [I2 ] = 0.00414 M, and [HI] = 0.0424 M?

[H₂]  = 6.74×10⁻³ M

[I₂]  = 4.65×10⁻³ M

[HI] = 0.0413 M

Explanation:

The equilibrium is:

H₂(g) + I₂(g) ⇌ 2 HI(g)

Expression for Kc = [HI]² / [H₂] .  [I₂]            54.3

We analyse the equation:

                 H₂(g)      +     I₂(g)      ⇌      2 HI(g)

Initially   0.00623      0.00414              0.0424

React            x                  x                        2x

x amount has reacted, therefore by stoichiometry 2x has been added to the HI, that we have in the beginning.

Eq         0.00623-x    0.00414-x             0.0424+2x

We make the expression for Kc:

Kc = (0.0424+2x)² / (0.00623-x) . (0.00414-x)  = 54.3

This is quadractic funcion:

54.3 = (1.79×10⁻³ + 0.1696x + 4x²) / (2.58×10⁻⁵- 0.01037x + x²)          

54.3 (2.58×10⁻⁵- 0.01037x + x²) = 1.79×10⁻³ + 0.1696x + 4x²

1.40×10⁻³- 0.563x + 54.3x² = 1.79×10⁻³ + 0.1696x + 4x²

-3.89×10⁻⁴ - 0.7326x +50.3x² = 0 → a = 50.3 ; b= - 0.7326 ; c = -3.89×10⁻⁴

Quadratic formula = (-b +- √(b² + 4ac))/ (2a)

x₁ = 0.015

x₂ = -5.13×10⁻⁴  . We choose x₂ as x₁ give us negative concentrations

[H₂] = 0.00623 - (-5.13×10⁻⁴) = 6.74×10⁻³ M

[I₂] = 0.00414 - (-5.13×10⁻⁴) = 4.65×10⁻ M

[HI] = 0.0424+2(-5.13×10⁻⁴) = 0.0413 M

the answer is d. 74.02