Hydrogen gas and iodine gas react via the equation, H2 + I2 2HI, and K=76 at 600K. If 0.05 mole of HI is placed in a flask at 600K, what are the equilibrium concentrations of HI, I2, H2?

Question

Hydrogen gas and iodine gas react via the equation, H2 + I2 <--> 2HI, and K=76 at 600K. If 0.05 mole of HI is placed in a flask at 600K, what are the equilibrium concentrations of HI, I2, H2?

I am very lost at where to even start with this problem

DrBob222 · Answer

Are you sure there is no volume given. The following will give you mols at equilibrium.
..........2HI ==> H2 + I2
initial...0.05.....0....0
change....=2x......x.....x
equil...0.05-x.....x.....x

Kc = 76 = (H2)(I2)/(HI)^2
Solve for x
Then x/volume = concn.

Hamza · Answer

I think you mean 0.05 - 2x for the change of (2HI)