Question

Determine constant of chemical equilibrium of the system А + 2В ⇄ 2С + 3D, if
concentrations of reagents were equal to А = 9.0 mol/l, В =15.8 mol/l and that for reaction
products С =27.5 mol/l and D =7.2 mol/l.

Answer 1

Constant of chemical equilibrium of the system is n 125.633

Explanation:

Kc =  [C]² [D]³ / [A] [B] ²

THE equilibrium constant is given by the above equation .

now equating the values of the concentration :

= (27.5² * 7.2³) / (9 * 15.8 ² )

= 125.633