Question

A 1.00 M sample HI is placed in a 1-L vessel at 460°C, and the reaction system is allowed to come

to equilibrium. The Hl partially decomposes, forming Hz and I2. What is the equilibrium
concentration of HI if the equilibrium constant is 52.8?
H2(g) + 12(g) = 2 HI(g) at 460°C?

Answer 1

[HI] = 0.784M

[I₂] = 0.108M

[H₂] = 0.108M

Step-by-step explanation:

Based on the equilibrium reaction:

H₂(g) + I₂(g) ⇄ 2HI(g)

The equilibrium constant, K, is:

K = 52.8 = [HI]² / [H₂] [I₂]

Where [] are equilibrium concentrations of each gas.

As initial concentration of HI is 1.00M, the equilibrium concentrations of the gases is:

[HI] = 1.00M - 2X

[I₂] = X

[H₂] = X

Replacing:

52.8 = [1.00-2X]² / [X] [X]

52.8X² = 4X² - 4X + 1

0 = -48.8X² - 4X + 1

Solving for X:

X = -0.1899M. False solution, there is no negative concentrations

X = 0.108M. Right solution.

Replacing, equilibrium concentrations are:

[HI] = 1.00M - 2*0.108M

[HI] = 0.784M

[I₂] = 0.108M

[H₂] = 0.108M