At a certain temperature, the equilibrium constant, kc, kc, for this reaction is 53.3. h2(g)+i2(g)↽−−⇀2hi(g)kc=53.3 h2(g)+i2(g)↽−−⇀2hi(g)kc=53.3 at this temperature, 0.400 mol h20.400 mol h2 and 0.400 mol i20.400 mol i2 were placed in a 1.00 l container to react. what concentration of hihi is present at equilibrium?

The equilibrium constant for the reaction H₂(g) + I₂(g) ⇄ 2HI(g) is 53.3, so when 0.400 moles of H₂ and I₂ are placed in a container of 1.00 L, the concentration of HI present at equilibrium is 0.62 mol/L.

The balanced reaction is:

H₂(g) + I₂(g) ⇄ 2HI(g)   (1)

The equilibrium constant for the above reaction is:

  (2)

Before the reaction, the concentration of H₂ and I₂ is:

At equilibrium we have:

H₂(g)   +   I₂(g)   ⇄   2HI(g)   (3)

0.4-x    0.4-x            2x

Entering the values of equation (3) into equation (2), we have:

After solving the above equation for x, we get:

Since the concentrations of H₂ and I₂ could not be negative, we will take the x₁ value (with the x₂ value, we have [H₂] = [I₂] = 0.400 - 0.55 = -0.15).  

Hence, the concentration of HI is:

Therefore, the concentration of HI present at equilibrium is 0.62 mol/L.

Learn more about equilibrium constants here:

I hope it helps you!  

0.628 mol·L⁻¹

Explanation:

The balanced equation is

H₂ + I₂ ⇌ 2I ₂

Data:

 Kc = 53.3

[H₂] = 0.400 mol·L⁻¹

[I₂] = 0.400 mol·L⁻¹

1. Set up an ICE table.

2. Calculate the equilibrium concentrations

[HI] = 2x mol·L⁻¹ = 2 × 0.3140 mol·L⁻¹ = 0.628 mol·L⁻¹

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