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?

[HI] = 0.784M

[I₂] = 0.108M

[H₂] = 0.108M

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

h let me think

answer: a tank of height h is filled with water and sits on the ground. water squirts from a hole at a height y above the ground and has a range r. for two y values, 0 and h, r is zero. determine the value of y for which the range will be a maximum.

explanation: