At a given temperature and pressure, the equilibrium constant values for the equilibria are given below: 3A2 + B2 2A3B, K1 A3B 32A2 + 12B2, K2 The relation between K1 and K2 is :
- K12 = 2K2
- K2 = K12
- K1 = 1 K2
- K2 = 1 K1
Answer
(4) K2 = 1 K1
Solution
Reversing the first equilibrium: 2A3B 3A2 + B2; K = 1K1
Dividing this equation by 2 gives the second equilibrium, so its constant is the square root:
K2 = ( 1K1 )1/2
K2 = 1 K1
Answer: (4)