Calculate Kc for the reaction at this temperature. H2(g) + I2(g) <-----> 2HI(g)?

Question

In an experiment to study the formation of HI(g), H2(g) + I2(g) <-----> 2HI(g), H2(g) and I2(g) were placed in a sealed container at a certain temperature.

At equilibrium,

[H2] = 6.50*10E-5 M
[I2] = 1.06*10E-3 M
[HI] = 1.87*10E-3 M

Calculate Kc for the reaction at this temperature.

Answer 1

K= [Products]/[Reactants]
= ((1.87x10^-3)^2) (it is to the power of 2 because there is a 2 infront of HI)/(6.5x10^-5)x(1.06x10^-3)
= 5.70x10^-5
Which means at equilibrium, more reactants are being produced (H2 & I2)