Posted by Adrianne on .
The following reaction has an equilibrium constant of 0.020 at a given temperature.
2HI(g) I2(g) + H2(g)
If you have 1.00 mol HI(g) in a 0.750L container initially, how many moles of HI(g) will be present when the system reaches equilibrium?

Chemistry 
DrBob222,
Technically you don't have an equation because you omitted the arrow. You MUST include an arrow for us to know where the reactants stop and the products start.
2HI ==> H2 + I2
K = 0.020 = (H2)(I2)/(HI)^2
Set up an ICE CHART and solve.
initial:
H2 = O
I2 = 0
HI = 1.00 mole/0.750L = 1.33
change:
H2 = +x
I2 = +x
HI = 2x
equilibrium:
H2 = +x
I2 = +x
HI = 1.332x
Substitute into the K expression and solve for x.