Asked by Kristen
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
I am very lost at where to even start with this problem
Answers
Answered by
DrBob222
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.
..........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.
Answered by
Hamza
I think you mean 0.05 - 2x for the change of (2HI)