Answer:
[PCl₅] = 0.5646M
[PCl₃] = 0.1174M
[Cl₂] = 0.1174M
Explanation:
In the reaction:
PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)
K equilibrium is defined as:
<em>K = 2.44x10⁻² = [PCl₃] [Cl₂] / [PCl₅]</em>
The initial moles of each compound when volume is 15.3L are:
PCl₅ = 0.300mol/L×15.3L = 4.59mol
Cl₂ = 8.55x10⁻²mol/L×15.3L = 1.308mol
PCl₃ = 8.55x10⁻²mol/L×15.3L = 1.308mol
At 8.64L, the new concentrations are:
[PCl₅] = 4.59mol / 8.64L = 0.531M
[PCl₃] = 1.308mol / 8.64L = 0.151M
[Cl₂] = 1.308mol / 8.64L = 0.151M
At these conditions, reaction quotient, Q, is:
Q = [0.151M] [0.151M] / [0.531M]
Q = 4.29x10⁻²
As Q > K, <em>the reaction will shift to the left producing more reactant, </em>that means equilibrium concentrations are:
[PCl₅] = 0.531M + X
[PCl₃] = 0.151M - X
[Cl₂] = 0.151M - X
<em>Where X is reaction coordinate.</em>
Replacing in K expression:
2.44x10⁻² = [0.151M - X] [0.151M - X] / [0.531M + X]
1.296x10⁻² + 2.44x10⁻²X = 0.0228 - 0.302X + X²
<em>0 = 9.84x10⁻³ - 0.3264X + X²</em>
Solving for X:
X = 0.293 → False solution. Produce negative concentrations
<em>X = 0.0336M → Right solution.</em>
Replacing:
[PCl₅] = 0.531M + 0.0336
[PCl₃] = 0.151M - 0.0336
[Cl₂] = 0.151M - 0.0336
<h3>[PCl₅] = 0.5646M</h3><h3>[PCl₃] = 0.1174M</h3><h3>[Cl₂] = 0.1174M</h3>
