Question

A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be \(0.5×10^{–1}\ mol L^{–1}\). If value of K_c is \(8.3×10^{–3}\), what are the concentrations of PCl₃ and Cl₂ at equilibrium?
\(PCl_5 (g) ⇋ PCl_3 (g) + Cl_2(g)\)

Answer 1

Given:

Reaction:
PCl₅(g) ⇋ PCl₃(g) + Cl₂(g)

Temperature = 473 K

Equilibrium constant:
K_c = 8.3 × 10⁻³

Concentration of PCl₅ at equilibrium:
[PCl₅]_eq = 0.5 × 10⁻¹ mol L⁻¹ = 0.05 mol L⁻¹

Initially, the vessel contained only pure PCl₅.

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Step 1: Assume degree of dissociation

Let the initial concentration of PCl₅ be C mol L⁻¹.

Let the amount dissociated be x mol L⁻¹.

At equilibrium:

[PCl₅] = C − x = 0.05
[PCl₃] = x
[Cl₂] = x

Hence, C = 0.05 + x

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Step 2: Write the expression for K_c

K_c = [ PCl₃ ][ Cl₂ ] / [ PCl₅ ]

8.3 × 10⁻³ = x² / 0.05

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Step 3: Solve for x

x² = 8.3 × 10⁻³ × 0.05

x² = 4.15 × 10⁻⁴

x = √(4.15 × 10⁻⁴)

x = 2.04 × 10⁻² mol L⁻¹

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Step 4: Calculate equilibrium concentrations

[PCl₃]_eq = x = 2.04 × 10⁻² mol L⁻¹

[Cl₂]_eq = x = 2.04 × 10⁻² mol L⁻¹

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Final Answer:

At equilibrium:

[PCl₃] = 2.04 × 10⁻² mol L⁻¹
[Cl₂] = 2.04 × 10⁻² mol L⁻¹