For PCl₅(g) PCl₃(g) + Cl₂(g) , Kc = 0.04 at 500 K. If 0.5 mol of PCl₅ is placed in a 2 L vessel with 0.1 mol of Cl₂ , wh

Practice question

Question

For PCl₅(g) <=> PCl₃(g) + Cl₂(g) , Kc = 0.04 at 500 K. If 0.5 mol of PCl₅ is placed in a 2 L vessel with 0.1 mol of Cl₂ , what is [PCl₃] at equilibrium?

Options

Choose one · Correct answer highlighted

0.1 M Correct 0.2 M 0.05 M 0.15 M

Explanation

Initial: [PCl₅] = (0.5/2) = 0.25 M , [Cl₂] = (0.1/2) = 0.05 M , [PCl₃] = 0 . Let x be [PCl₃] at equilibrium, so [PCl₅] = 0.25 - x , [Cl₂] = 0.05 + x . Kc = ([PCl₃][Cl₂]/[PCl₅]) = (x (0.05 + x)/0.25 - x) = 0.04 . Solving, x(0.05 + x) = 0.04(0.25 - x) , 0.05x + x² = 0.01 - 0.04x , x² + 0.09x - 0.01 = 0 , x = (-0.09 pm sqrt0.0081 + 0.04/2) , x ≈ 0.1 M .

Share this question