Question

At the start of the reaction there are 0.714 mole of H2, 0.984 mole of I2, and 0.886 mole of HI in a 2.70 L reaction chamber. Calculate the concentrations of the gases at equilibrium.

Answer 1

(H2) = 0.714/2.70 = ? about 0.26M
(I2) = 0.984/2.70 = ?about 0.36
(HI) = 0.886/2.70 = ?about 0.33
This is as far as you an go. You need an equilibrium constant.

Answer 2

(H2) = 0.714/2.70 = ? about 0.26M
(I2) = 0.984/2.70 = ?about 0.36
(HI) = 0.886/2.70 = ?about 0.33
You need to recalculate the above and be more accurate.

............H2 + I2 ==> 2HI
I........0.26M..0.36....0.33
C..........?.....?........?
First you must decide which way the reaction will go; i.e., left or right. To do this you calculate Qrxn.
Qrxn = (HI)^2/(H2)(I2)
Q = (0.33)^2/(0.26)(0.36) = about 1.16.
Compared with K = 53 that means the products are too small and reactants too large so the reaction will go to the right.
C...........-x.......-x......+x
E.........0.26-x...0.36-x ...0.33+2x

Substitute the E line into Keq expression and solv.