when 2.75 mol HI(g) placed in 1L container and allowed to dissociate into

Question

2HI(g) <==> H2(g)+I2(g)
Final H2 concin .275M. What is Ka for rxn

The answer is .0156. But I don't know why
Thanks

Answer 1

..........2HI ==> H2 + I2
initial...2.75.....0....0
change...-2x........x....x
equil....2.75-2x....x.....x

The problem tells you H2 (which I've let = x above) = 0.275; therefore, I2 must be 0.275 and HI must be 2.75-(2*0.275) = ?
Now just substitute into Ka expression and solve. I get 0.0156 also.

Answer 2

could you do it in moles instead of molars

Answer 3

and in the K eqn is the M for HI squared

Answer 4

2.75 moles in 1L = 2.75M so the solution I posted is both moles and molar.
Yes, (HI) is squared.
Ka = (H2)(I2)/(HI)^2
Ka = (0.275)(0.275)/(2.20)^2
Ka = 0.0156