Question

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
\(2BrCl (g) ⇋ Br_2 (g) + Cl_2 (g)\)
for which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10^–3 mol L^–1, what is its molar concentration in the mixture at equilibrium?

Answer 1

Given:

Reaction:
2BrCl(g) ⇋ Br₂(g) + Cl₂(g)

Equilibrium constant:
K_c = 32 at 500 K

Initial concentration of BrCl:
[BrCl]₀ = 3.3 × 10⁻³ mol L⁻¹

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Step 1: Assume degree of dissociation

Let the decrease in concentration of BrCl be 2x mol L⁻¹ at equilibrium.

Then, increase in concentration of Br₂ = x
Increase in concentration of Cl₂ = x

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Step 2: Write equilibrium concentrations

[BrCl]_eq = 3.3 × 10⁻³ − 2x
[Br₂]_eq = x
[Cl₂]_eq = x

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Step 3: Write the equilibrium constant expression

K_c = [Br₂][Cl₂] / [BrCl]²

32 = x² / (3.3 × 10⁻³ − 2x)²

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Step 4: Solve for x

Taking square root on both sides:

√32 = x / (3.3 × 10⁻³ − 2x)

5.66 = x / (3.3 × 10⁻³ − 2x)

x = 5.66 (3.3 × 10⁻³ − 2x)

x = 1.87 × 10⁻² − 11.32x

12.32x = 1.87 × 10⁻²

x = 1.52 × 10⁻³ mol L⁻¹

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Step 5: Calculate equilibrium concentration of BrCl

[BrCl]_eq = 3.3 × 10⁻³ − 2(1.52 × 10⁻³)

[BrCl]_eq = 3.3 × 10⁻³ − 3.04 × 10⁻³

[BrCl]_eq = 2.6 × 10⁻⁴ mol L⁻¹

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Final Answer:

The molar concentration of BrCl in the equilibrium mixture is:
2.6 × 10⁻⁴ mol L⁻¹