2 2 3 1 SO g O g SO g 2 3 1 1 2 2 2 [SO ] K [SO ][O ] 3 2 2 1 SO (g) SO (g) O 2 1 2 2 2 3 1 [SO ][O ] 1 K [SO ] K 3 2 2 2SO g 2SO g O g 2 2 2 2 2 2 3 1 2 1 2 [SO ] [O ] 1 K [SO ] K 1 K K >> Thus, the rate constant can be given as: 2 1 2 1 K = K Hence, the correct answer is option (b)
Take chapter-wise tests, track your progress, and master Chemical Equilibrium.
Start Free Practice →