The equilibrium constant of the reaction A 2 g + B 2 g ⇄ 2 A B g at 373 K is 50 . If 1 L of flask containing 1 m o l e of A 2 (g) is connected to 2 L flask containing 2 m o l e s B 2 g at 100 o C , the amount of A B produced at equilibrium at 100 o C would be

Step-by-step solution

A 2 g + B 2 g ⇄ 2 A B g Initially: 1 3 M 2 3 M At equilibrium: 1 3 − x M 2 3 − x M 2 x M K e q = 50 = 4 x 2 1 3 - x 2 3 - x or 50 = 36 x 2 1 - 3 x 2 - 3 x or 50 9 x 2 – 9 x + 2 = 36 x 2 or 450 x 2 – 450 x + 100 = 36 x 2 or 414 x 2 – 450 x + 100 = 0 or x = + 450 − 450 2 − 4 × 414 × 100 2 × 414 or x = + 450 - 202500 - 165600 2 × 414 or x = 450 - 36900 2 × 414 = 450 – 192.1 2 × 414 = 0.31 M ∴ moles of A B produced = 0.31 × 6 = 1.86