The value of equilibrium constant of the reaction HI(g) 12H2(g) + 12I2(g) is 8.0. The equilibrium constant of the reaction H2(g) + I2(g) 2HI(g) will be
- 16
- 1/8
- 1/16
- 1/64
Answer
(4) 1/64
Solution
Given: HI 12H2 + 12I2; K1 = 8
Multiplying by 2: 2HI H2 + I2; K2 = K12 = 64
Reversing: H2 + I2 2HI; K3 = 1K2 = 164
Answer: (4)