Question

An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.323 M HI, 4.34E-2 M H2 and 4.34E-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.228 mol of HI(g) is added to the flask?

Answer 1

[HI] = 0.5239M

[H₂] = 7.05x10⁻²

[I₂] = 7.05x10⁻²

Step-by-step explanation:

The reaction of HI to produce H₂ and I₂ is:

2HI → H₂ + I₂

Where K of reaction is defined as:

K = [H₂] [I₂] / [HI]²

Replacing with the concentrations of the gases in equilibrium:

K = [4.34x10⁻²] [4.34x10⁻²] / [0.323] ²

K = 0.0181

If you add 0.228 mol = 0.228M (Because volume of the flask is 1.0L), the concentration when the system reaches the equilibrium are:

[HI] = 0.228M + 0.323M - X = 0.551M - X

[H₂] = 4.34x10⁻² + X

[I₂] = 4.34x10⁻² + X

Where X is reaction coordinate.

Replacing in K formula:

K = 0.0181 = [4.34x10⁻²+ X] [4.34x10⁻²+ X] / [0.551 - X] ²

0.0181 = 0.00188356 + 0.0868 X + X² / 0.303601 - 1.102 X + X²

0.005495 - 0.01995 + 0.0181X² = 0.00188356 + 0.0868 X + X²

0 = -0.003611 + 0.10675X + 0.9819X²

Solving for X:

X = -0.136 → False solution, there is no negative concentrations.

X = 0.0271 → Right solution.

Replacing for concentrations of each species:

[HI] = 0.5239M

[H₂] = 7.05x10⁻²

[I₂] = 7.05x10⁻²