Question

Hydrogen gas, iodine vapor, hydrogen iodine are mixed in a flask and heated to 674°C. H2(g) + I2(g) ⇋ 2 HI(g) Kc = 49 at 674°C If the initial concentrations of hydrogen gas and iodine vapor are both 0.048 mol/L and the concentration of hydrogen iodine is 0.127 mol/L what is the equilibrium concentration of hydrogen gas? Enter a number to 4 decimal places.

Answer 1

[H₂] = 0.0248M

Step-by-step explanation:

For the reaction:

H₂(g) + I₂(g) ⇋ 2 HI(g)

Kc is defined as:

Kc = [HI]² / [I₂][H₂] = 49

For the initial concentrations, Q of reaction is:

Q = [0.127M]² / [0.048M][0.048M] = 7

As Kc > Q, the reaction shift to the right, doing equilibrium concentrations as:

[I₂] = 0.048M - x

[H₂] = 0.048M - x

[HI] = 0.127M + 2x

Replacing in Kc:

49 = (0.127+2X)² / (0.048-X)²

49 = 0.0161 + 0.508X + 4X² / 0.002304 - 0.096X + X²

0.1129 - 4.704X + 49X² = 0.0161 + 0.508X + 4X²

0.0968 - 5.212X + 45X² = 0

Solving for X:

X = 0.0232M

Thus, equilibrium concentration of hydrogen gas is:

[H₂] = 0.048M - x = 0.048M - 0.0232M = 0.0248M