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