Answer:
[PCl₅] → 4.47×10⁻³ M
[PCl₃] → 0.0897M
Explanation:
We state the equilibrium: PCl₅ (g) ⇄ PCl₃(g) + Cl₂(g)
Initially we have 0.4192 moles of PCl₅ so during the reaction, x moles of it have been reacted. In conclussion after the equilibrium we may have x moles of PCl₃ and Cl₂. In order to find the final concentration, we use the Kc, but we need molar concentrations. That's why we divide all the values with the 4.5L, volume for the vessel. The expression for Kc is:
Kc = [PCl₃] . [Cl₂] / [PCl₅]
1.80 = (x/ 4.45) . (x/ 4.45) / ((0.4192-x) /4.45)
1.80 = (x² / 4.45²) / ((0.4192-x) /4.45)
1.80 = x² / 4.45 / (0.4192-x)
1.80 (0.4192-x) = x² / 4.45
1.80 (0.4192-x) . 4.45 - x² = 0
3.36 - 8.01x - x² = 0 → Quadractic
a = -1 b = -8.01 c = 3.36
(-b + √(b² - 4ac)) / (2a)
(-(-8.01) + √((-8.01)² - 4(-1)3.36)) / (2(-1) = 0.3995
[PCl₅] = (0.4192 - 0.3995) / 4.45L → 4.47×10⁻³ M
[PCl₃] = 0.3995 /4.45L → 0.0897M
