Question

At 700 K, equilibrium constant for the reaction:
\(H_2 (g) + I_2 (g) ⇋ 2HI (g)\)
is 54.8. If 0.5 mol L^–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?

Answer 1

Given:

Reaction:
H₂(g) + I₂(g) ⇌ 2HI(g)

Equilibrium constant, K_c = 54.8
Equilibrium concentration of HI = 0.50 mol L⁻¹

Initially, only HI was present.

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Step 1: Assume degree of dissociation

Let the concentration of H₂ and I₂ formed at equilibrium be x mol L⁻¹ each.

According to the reaction:
2HI ⇌ H₂ + I₂

Decrease in HI = 2x

Equilibrium concentrations:
[HI] = 0.50 M
[H₂] = x M
[I₂] = x M

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Step 2: Write equilibrium constant expression

K_c = [HI]² / ([H₂][I₂])

54.8 = (0.50)² / (x × x)

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Step 3: Solve for x

x² = (0.25) / 54.8

x² = 0.00456

x = 0.067 mol L⁻¹

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Final Answer:

Equilibrium concentration of H₂ = 0.067 mol L⁻¹
Equilibrium concentration of I₂ = 0.067 mol L⁻¹