At 500 K, equilibrium constant, Kc, for the following reaction is 5.

12H2(g)+12I2(g)HI(g)

What would be the equilibrium constant Kc for the reaction?

2HI(g)H2(g)+I2(g)

(1) 0.04

(2) 0.4

(3) 25

(4) 2.5

Hint: Equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction
For the reaction, 12H2(g)+12I2(g)HI(g)
Kc=[HI][H2]1/2[I2]1/2=5
Thus, for the reaction, 2HI(g)H2(g)+I2(g)
Kc1=[H2][I2][HI]2=(1Kc)2=(15)2=125=0.04