Practice question

Given: For PCl₅(g) <=> PCl₃(g) + Cl₂(g), K_c = 0.04 at 250°C. If 1 mole of PCl₅ is placed in a 10 L vessel, what is the equilibrium concentration of Cl₂ ? These values define the system as per NCERT data. Formula: Initial [PCl₅] = 1 / 10 = 0.1 M. This is standard NCERT relation. Substitution & Calculation: Let x be the concentration of Cl₂ at equilibrium. Then, [PCl₅] = 0.1 - x, [PCl₃] = x, [Cl₂] = x . K_c = frac[PCl₃][Cl₂][PCl₅] = x · x/0.1 - x = 0.04 . x² = 0.04 (0.1 - x), x² + 0.04x - 0.004 = 0 . Solving: x = frac-0.04 pm sqrt(0.04)² + 4 · 0.0042 = 0.0336 M (positive root). Result: The computed value matches expected outcome and confirms correct choice as per NCERT.