An equilibrium mixture of the three gases in a 7.40 L container at 535 K contains 0.313 M PCl5, 0.119 M PCl3 and 0.119 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the volume of the container is increased to 16.6 L?

PCl₅: 0.1192M

PCl₃: 0.0734M

Cl₂: 0.0734M

Explanation:

Based on the reaction:

PCl₅ ⇄ PCl₃ + Cl₂

K of reaction is defined as:

K = [PCl₃] [Cl₂] / [PCl₅]

Replacing with concentrations of gases in equilibrium:

K = [0.119] [0.119] / [0.313]

K = 0.0452

Moles of each gas is:

PCl₅: 7.40L × (0.313mol / L) = 2.3162 moles

PCl₃: 7.40L × (0.119mol / L) = 0.8806 moles

Cl₂: 7.40L × (0.119mol / L) = 0.8806 moles

When the volume of the container is increased the system will produce more moles, (That is, produce more products) in order to keep constant the pressure of the container -Le Chatelier's principle-. In equilibrium, molarity of each gas is:

PCl₅: 2.3162 moles - x / 16.6L

PCl₃: 0.8806 moles + x / 16.6L

Cl₂: 0.8806 moles + x / 16.6L

Where X is reaction coordinate.

Replacing in K formula:

0.0452 = [0.05305 + X/16.6] [0.05305 + X/16.6] / [0.13953 - X/16.6]

6.3068x10⁻³ - 2.723x10⁻³ X = 0.0028143 + 0.00639157 X + 0.00362897 X²

0 = -0.0034925 + 0.00911457X + 0.00362897 X²

Solving for X:

X = -2.8 → False answer. There is no negative concentrations

X = 0.3378 moles

Replacing:

PCl₅: 2.3162 moles - 0.3378 / 16.6L = 0.1192M

PCl₃: 0.8806 moles + 0.3378 / 16.6L = 0.0734M

Cl₂: 0.8806 moles + 0.3378 / 16.6L = 0.0734M

Beeing these concentrations the concentrations in equilibrium of the three gases

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