Consider the following reactions in which all the reactants and products are present in gaseous state 2xy x2 + y2 K1 = 2.5 105 xy + 12z2 xyz K2 = 5 10-3 The value of K3 for the equilibrium 12x2 + 12y2 + 12z2 xyz is :
- 2.5 10-3
- 2.5 103
- 1.0 10-5
- 5 10-3
Answer
(3) 1.0 10-5
Solution
Reverse reaction (1) and multiply it by 12:
12x2 + 12y2 xy; K' = ( 1K1 )1/2 = 1 2.5 105 = 1500 = 2 10-3
Add reaction (2): xy + 12z2 xyz; K2 = 5 10-3
Sum: 12x2 + 12y2 + 12z2 xyz
∴ K3 = K' K2 = 2 10-3 5 10-3 = 1.0 10-5
Answer: (3)