Question

Assume that in the reaction of the formation of gaseous hydrogen fluoride from hydrogen and fluorine, 3.00 mole H₂ and 6.00 mole of F₂ are mixed in a 3.00 L flask. Assume that the equilibrium constant for the synthesis of hydrogen fluoride is 1.15 X 10². Calculate the equilibrium concentrations of all species.

Answer 1

H₂ + F₂      2HF

K = [HF] / { [H₂] [F₂] }

[H₂] = 3 mole / 3 L = 1 M

[F₂] = 6 mole / 3 L = 2 M

K = 1.15 x 10²

H₂ + F₂    2HF
IC: 1 2 0
C: -x -x +2x
EC: 1-x 2-x 2x

K = [HF] / { [H₂] [F₂] }

1.15 x 10² = (2x)² / { (1 - x) (2 - x) }

1.15 x 10² = 4x² / (2 - x - 2x + x²)

1.15 x 10² = 4x² / (2 - 3x + x²)

115 (2 - 3x + x²)  = 4x²

230 - 345x + 115x² = 4x²

111x² - 345x + 230 = 0

Solving the quadratic equation, we get

x = 2.14 and x = 0.97

Since x value cannot be greater than 2, so x = 2.14 is discarded. Hence we have

x = 0.97

So, equilibirium concentrations are

[HF] = 2x = 2 ( 0.97) = 1.94 M

[H₂] = 1-x = 1 - 0.97 = 0.03 M

[F₂] = 2-x = 2 - 0.97 = 1.03 M